Answer
given,
distance of fall = h
initial speed = 0 m/s
it travels 0.540 h in the last 1.00 s
Speed of the fall = [tex]\dfrac{Distance}{time}[/tex]
Speed = [tex]\dfrac{0.540\ h-0}{1}[/tex]
S = 0.540h m/s
time of the fall
using equation of motion
v = u + g t
0.540 h = 0 + 9.8 t
t = 0.055h s
The time of fall is equal to 0.055h s
height of fall = ?
again using equation of motion
v² = u² + 2 g s
(0.540h)² = 0 + 2 x 9.8 x s
s = 0.0148 h²
Hence, the time of fall of the object = 0.055h s.
the distance of fall of object = 0.0148 h²
The object has been falling for a total time of 3.13 seconds and the total height from which it fell is 47.9 meters, given that it covered 0.540 of its total height in the last 1 second of free fall.
Explanation:Solution for Time and Height of a Free-Falling ObjectLet's tackle this physics problem step by step to find both the time the object has been falling and the total height from which it fell. We'll use the known acceleration due to gravity (g = 9.81 m/s²) and apply kinematic equations.
Given that the object travels 0.540h in the last 1.00 s, we can set up two equations using the kinematic formula h = ½gt², where h is the height, g is the acceleration due to gravity, and t is the time in seconds:
Total height fallen (h) after time t: h = ½g(t²)Height fallen in the last 1 second (0.540h): 0.540h = ½g(t²) - ½g((t-1)²)We solve the second equation for t, which will then be used to find the total height using the first equation.
Solving the second equation: 0.540h = ½g(t²) - ½g((t-1)²)
Expand and simplify the equation: 0.540h = ½g(t²) - ½g(t² - 2t + 1)
0.540 = t² - ½g(t² - 2t + 1)
Now, assuming g is approximately 9.8 m/s², we can plug in values and solve for t:
0.540 = t² - 0.5(9.81)(t² - 2t + 1)
After solving for t, we find:
(a) Total time of fall, t = 3.13 s(b) Total height fallen, h = 0.5(9.81)(3.13²) = 47.9 mThis calculation assumes that air resistance is negligible and the acceleration due to gravity is constant.
Which of the following statement(s) about energy and phase is/are correct? Select all that apply. Choose one or more: A. While only one phase is present, adding or removing energy changes PE but not KE. B. While only one phase is present, adding or removing energy changes KE but not PE. C. During a phase change, adding or removing energy changes KE but not PE. D. During a phase change, adding or removing energy changes PE but not KE.
In a single phase, the addition or removal of energy changes Kinetic Energy not Potential Energy. However, during a phase change, this energy addition or subtraction results in a change in Potential Energy, not Kinetic Energy.
Explanation:The subject of this question is energy and phase, particularly in the context of Potential Energy (PE) and Kinetic Energy (KE). When only one phase is present, adding or removing energy will mainly change the KE, not the PE. This is because the energy is utilized to speed up or slow down the particles, thus changing their kinetic energy. However, during a phase change, adding or removing energy changes PE but not KE as it alters the state rather than the speed of the particles. Statement B is the one that is accurate while only one phase is present, whereas the correct option for the phase change scenario is option D.
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The front of an aircraft hanger is being built in the shape of a parabola, which is 48 ft. wide, and has a maximum height of 18 ft., AND must have a rectangular doorway that is 8 ft. tall. What is the maximum width of the doorway? (Round your answer to one decimal place.)
Answer:
maximum width of the doorway = 35.77ft
Explanation:
The detailed calculation and derivation from first principle is as shown in the attachment
Answer:
the maximum width is x= 4√2 ft = 5.656 ft
Explanation:
for the parabola
y= a*x² + b*x + c
where y= height and x= width
an aircraft hangar should be symmetric with respect to the y axis , then
y(-x)=y(x) → a*x² + b*x + c = a*x² - b*x + c →-2*b*x =0 → b=0
it also should be pointing downwards → a is negative
, then the parabola would be
y= c- a*x²
since c= maximum height = 18 ft
then for y=0 , x= 48 ft/2 = 24 ft → 0 = 18 ft - a*(24 ft)² → a= 1/32 ft⁻¹
then
y= 18 ft- 1/32 ft⁻¹ *x²
since the doorway cannot go beyond the parabola , the maximum possible doorway is obtained when the doorway touches the parabola.
then for a height y= 8 ft
8 ft = 18 ft- 1/32 ft⁻¹ *x²
x= 4√2 ft = 5.656 ft
Is a nucleus that absorbs at 4.13 δ more shielded or less shielded than a nucleus that absorbs at 11.45 δ?
Explanation:
A nucleus that absorbs 11.45δ is less shielded than a nucleus that absorbs at 4.13δ.
the nucleus that absorbs at 11.45δ requires weaker applied field strength to come into resonance than the nucleus that absorbs at 4.13δ.
A nucleus absorbing at 4.13 δ is more shielded than one absorbing at 11.45 δ because it is in a weaker local magnetic field and resonates at a lower frequency.
Explanation:In the context of nuclear magnetic resonance (NMR) spectroscopy, the chemical shift is given in units of delta (δ) which represents the resonance frequency of a nucleus relative to a standard reference compound. When a nucleus is surrounded by a dense cloud of electrons, it is considered to be ‘shielded’. A shielded nucleus is influenced by a smaller local magnetic field because the electrons repel some of the external magnetic field. As a result, shielded nuclei resonate at a lower frequency (higher δ values) when compared to de-shielded nuclei. Therefore, a nucleus that absorbs at 4.13 δ is more shielded than a nucleus that absorbs at 11.45 δ because the latter is in a stronger local magnetic field and is, hence, de-shielded.
You have a remote-controlled car that has been programmed to have velocity v⃗ =(−3ti^+2t2j^)m/s, where t is in s. At t = 0 s, the car is at r⃗ 0=(3.0i^+2.0j^)mWhat is the x component of the car's position vector at 10 s?What is the y component of the car's position vector at 10 s?What is the x component of the car's acceleration vector at 10 s?What is the y component of the car's acceleration vector at 10 s?
Answer:
The y-component of the car's position vector is 670m/s.
The x-component of the acceleration vector is -3, and the y-component is 40.
Explanation:
The displacement vector of the car with velocity
[tex]\boldsymbol{v}= (-3t\boldsymbol{i}+2t^2\boldsymbol{j})m/s[/tex]
is the integral of the velocity.
Integrating [tex]\boldsymbol{v}[/tex] we get the displacement vector [tex]\boldsymbol{d}[/tex]:
[tex]\boldsymbol{d}=(-\dfrac{3}{2}t^2\boldsymbol{i}+\dfrac{2}{3}t^3\boldsymbol{j} )[/tex]
Now if the initial position if the car is
[tex]\boldsymbol{r}= (3.0\boldsymbol{i}+2.0\boldsymbol{j})[/tex]
then the displacement of the car at time [tex]t[/tex] is
[tex]\boldsymbol{d(t)}= \boldsymbol{r+d}[/tex]
[tex]\boxed{\boldsymbol{d(t)}=(-\dfrac{3}{2}t^2+3.0\boldsymbol{i}+\dfrac{2}{3}t^3+2.0\boldsymbol{j} )}[/tex]
Now at [tex]t=10s[/tex], we have
[tex]\boxed{\boldsymbol{d(t)}=(-147\boldsymbol{i}+670\boldsymbol{j} )}m[/tex]
The y-component of the car's position vector is 670m/s.
The acceleration vector is the derivative of the velocity vector:
[tex]\boldsymbol{a(t)}=\dfrac{d\boldsymbol{v(t)}}{dt} =(-3\boldsymbol{i}+4t\boldsymbol{j})[/tex]
and at [tex]t=10s[/tex] it is
[tex]\boldsymbol{a(t)}=(-3\boldsymbol{i}+40\boldsymbol{j})m/s^2[/tex]
The x-component of the acceleration vector is -3, and the y-component is 40.
The x and y components of the car's position at 10 s are -147.0 m and 668.67 m, respectively. The x and y components of the car's acceleration at 10 s are -3 m/s² and 40 m/s², respectively.
The problem involves determining the position and acceleration components of a remote-controlled car from given velocity functions over time.
1.) X Component of Car's Position at 10 s:
Given the velocity component, vx = -3t m/s, we need to integrate it with respect to time to find the position (x). The initial position x0 is 3.0 m.
x(t) = x0 + ∫vx dt = 3.0 + ∫(-3t) dt = 3.0 + (-3/2) t²When t = 10 s:
x(10) = 3.0 + (-3/2)(10)² = 3.0 - 150 = -147.0 m2.) Y Component of Car's Position at 10 s:
Given the velocity component, vy = 2t² m/s, integrating it with respect to time gives the position (y). The initial position y0 is 2.0 m.
y(t) = y0 + ∫vy dt = 2.0 + ∫(2t²) dt = 2.0 + (2/3) t³When t = 10 s:
y(10) = 2.0 + (2/3)(10)³ = 2.0 + 666.67 = 668.67 m3.) X Component of Car's Acceleration at 10 s:
Given the velocity component, vx = -3t m/s, the acceleration is the time derivative of velocity.
ax = dvx/dt = d(-3t)/dt = -3 m/s²Hence, at t = 10 s:
ax (10) = -3 m/s²4.) Y Component of Car's Acceleration at 10 s:
Given the velocity component, vy = 2t² m/s, the acceleration is the time derivative of velocity.
ay = dvy/dt = d(2t²)/dt = 4t m/s²Hence, at t = 10 s:
ay (10) = 4(10) = 40 m/s²A driver starts from rest on a straight test track that has markers every 0.14 km. The driver presses on the accelerator and for the entire period of the test holds the car at constant acceleration. The car passes the 0.14 km post at 8.0 s after starting the test.
(a) What was the car's acceleration?
(b) What was the car's speed as it passed the 0.14 km post?
Final answer:
To find the car's acceleration, use the kinematic equation v = u + at. The car's acceleration is 2.5 m/s^2. To find the car's speed as it passes the 0.14 km post, plug the values into the kinematic equation v = u + at. The car's speed is 20 m/s.
Explanation:
To find the car's acceleration, we can use the kinematic equation:
v = u + at
where v is the final velocity, u is the initial velocity (which is 0 m/s since the car starts from rest), a is the acceleration, and t is the time. We are given that the car passes the 0.14 km post at 8.0 s after starting, which means the car travels a distance of 0.14 km in 8.0 s. Converting 0.14 km to meters gives us 140 m. Plugging the values into the equation, we have:
20 = 0 + a * 8.0
Simplifying, we find that the car's acceleration is 2.5 m/s^2.
To find the car's speed as it passes the 0.14 km post, we can use the kinematic equation:
v = u + at
Since the car starts from rest (u = 0 m/s) and the car's acceleration is 2.5 m/s^2 (which we just found), we can plug these values into the equation along with the time (8.0 s) to find the car's speed:
v = 0 + 2.5 * 8.0
Simplifying, we find that the car's speed as it passes the 0.14 km post is 20 m/s.
A river flows due south with a speed of 2.0 m/s. You steer a motorboat across the river; your velocity relative to the water is 4.2 m/s due east. The river is 500 m wide. (a) What is your velocity (magnitude and direction) relative to the earth? (b) How much time is required to cross the river? (c) How far south of your starting point will you reach the opposite bank?
The velocity of the boat relative to the earth is 7.2 m/s, 32.0° south of east. It will take the boat 0.118 seconds to cross the river. The boat will reach a point 0.236 meters south of the starting point.
Explanation:To find the velocity of the boat relative to the earth, we need to find the resultant of the boat's velocity relative to the water and the river's velocity relative to the earth. Using vector addition, we can find that the magnitude of the total velocity is 7.2 m/s and the direction is 32.0° south of east.
To find the time required to cross the river, we can use the formula t = d/v, where d is the width of the river and v is the horizontal component of the boat's velocity relative to the earth. Plugging in the values, we find that the time required is 0.118 seconds.
To find how far south of the starting point the boat will reach the opposite bank, we can use the formula d = v*t, where v is the vertical component of the boat's velocity relative to the earth and t is the time. Plugging in the values, we find that the boat will reach a point 0.236 meters south of the starting point.
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We use vector addition to find the magnitude of 4.65 m/s at an angle of 25.4° south of east. The time required to cross the river is approximately 119.05 seconds, and the boat will reach a point around 238.1 meters south of its starting point.
The question is about calculating the motion of a boat in a river flowing in a specific direction. Let's solve each part step-by-step.
(a) The boat's velocity relative to the water is 4.2 m/s due east, and the river flows due south at 2.0 m/s. We can use the Pythagorean theorem to find the magnitude:
[tex]Magnitude = \sqrt{(velocity_x^2 + velocity_y^2)}\\Magnitude = \sqrt{(4.22 + 2.02)}\\Magnitude = \sqrt{(17.64 + 4.00)}\\Magnitude = \sqrt{21.64}\\Magnitude = 4.65 m/s[/tex]
To find the direction, we need to calculate the angle θ relative to the east:
[tex]tan(\theta) = velocity_y / velocity_x = 2.0 / 4.2\\\theta = tan-1(2.0 / 4.2)\\\theta \approx 25.4\textdegree south\right \left of\right \left east[/tex]
(b) Given the river's width is 500 m and the boat's velocity relative to the water is 4.2 m/s, we use the formula:
Time = Distance / Velocity
Time = 500 m / 4.2 m/s
Time ≈ 119.05 s
(c) To find the southward distance, we use the river's flow velocity and the time calculated in part (b):
[tex]Distance_{south} = Velocity_{river} \times Time\\Distance_{south} = 2.0 m/s \times 119.05 s\\Distance_{south} \approx 238.1 m[/tex]
A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help." When she has fallen for 3.0 s, she hears the echo of her shout from the valley floor below. The speed of sound is 340 m/s. (a) How tall is the cliff? (b) If we ignore air resistance, how fast will she be moving just before she hits the ground? (Her actual speed will be less than this, due to air resistance.)
Answer:
532.0725 m
102.17270893 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s² = g
H = Height of cliff
Distance traveled in 3 seconds
[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 3^2\\\Rightarrow s=44.145\ m[/tex]
Distance traveled by sound = 2H-44.145 m
[tex]2H-44.145=ut+\dfrac{1}{2}at^2\\\Rightarrow 2H-44.145=340\times 3\\\Rightarrow H=\dfrac{340\times 3+44.145}{2}\\\Rightarrow H=532.0725\ m[/tex]
The height of the cliff is 532.0725 m
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 532.0725+0^2}\\\Rightarrow v=102.17270893\ m/s[/tex]
Her speed just before she hits the ground is 102.17270893 m/s
A generator has a terminal voltage of 108 V when it delivers 10.2 A, and 93 V when it delivers 47.4 A. Calculate the emf. Answer in units of V
112.11 V
Explanation:The electromotive force, e.m.f, (E) of the generator is related to its terminal voltage(V) and the current (I) it delivers as follows;
E = V + (I x r) -------------------------(i)
Where;
r = the internal resistance of the generator
Now, as stated in the question;
at V = 108V, I = 10.2A
Substitute the values into equation (i) as follows;
E = 108 + (10.2 x r)
E = 108 + 10.2r ------------------(ii)
Also, as stated in the question;
at V = 93V, I = 47.4A
Substitute these values also into equation (i) as follows;
E = 93 + (47.4 x r)
E = 93 + 47.4r ------------------------(iii)
Now solve equations (ii) and (iii) simultaneously;
Subtract equation (iii) from (ii)
E = 108 + 10.2r
_
E = 93 + 47.4r
______________
0 = 15 - 37.2r ----------------(iv)
_______________
Solve for r in equation (iv)
37.2r = 15
r = 15 / 37.2
r = 0.403Ω
The internal resistance is therefore 0.403Ω
Substitute r = 0.403Ω into equation (ii);
E = 108 + 10.2(0.403)
E = 108 + 4.11
E = 112.11
Therefore, the emf is 112.11 V
The electromotive force (emf) produced in the generator is 112.1126 Volts.
Given the following data:
Terminal voltage A = 108 VCurrent A = 10.2 AmpsTerminal voltage B = 93 VCurrent A = 47.4 AmpsTo calculate the electromotive force (emf) produced in the generator:
Mathematically, the electromotive force (emf) is given by the formula:
[tex]E = V + Ir[/tex]
For the first instance A:
[tex]E = 108 + 10.2r[/tex] .....equation 1.
For the first instance B:
[tex]E = 93 + 47.4r[/tex] .....equation 2.
Next, we would equate eqn 1 and eqn 2:
[tex]108 + 10.2r = 93 + 47.4r\\\\108 - 93 = 47.4r - 10.2r\\\\15 = 37.2r\\\\r = \frac{15}{37.2}[/tex]
Internal resistance, r = 0.4032 Ohms
Now, we can calculate the electromotive force (emf) produced in the generator:
[tex]E = 108 + 10.2r\\\\E = 108 + 10.2(0.4032)\\\\E = 108 + 4.1126[/tex]
Emf, E = 112.1126 Volts.
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Henrietta is jogging on the side-walk at 3.05 m/s on the way to her physics class. Bruce realizes that she forgot her bag of bagels, so he runs to the window, which is 38.0 m above the street level and directly above the sidewalk, to throw the bag to her. He throws it horizontally 9.00 s after she has passed below the window, and she catches it on the run. Ignore air resistance. (a) With what initial speed must Bruce throw the bagels so that Henrietta can catch the bag just before it hits the ground? (b) Where is Henrietta when she catches the bagels?
Answer:
12.9121148614 m/s
35.9393048982 m
Explanation:
t = Time taken
u = Initial velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s² = a
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 38=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{38\times 2}{9.81}}\\\Rightarrow t=2.78337865516\ s[/tex]
Time taken for the bag to fall is 2.78337865516 seconds
Time she has been jogging for
9+2.78337865516 = 11.78337865516 seconds
Total distance traveled by her
[tex]s=vt\\\Rightarrow s=3.05\times 11.78337865516=35.9393048982\ m[/tex]
Henrietta is 35.9393048982 m away
Velocity of throwing
[tex]\dfrac{35.9393048982}{2.78337865516}=12.9121148614\ m/s[/tex]
The velocity of throwing is 12.9121148614 m/s
Final answer:
Bruce must throw the bagels at an initial speed of 12.92 m/s for Henrietta to catch them, and Henrietta will be 35.93 m from the point directly below Bruce's window when she catches the bagels.
Explanation:
Projectile Motion and Kinematics Problem
To find the initial speed Bruce must throw the bagels, we need to consider two aspects of projectile motion: the horizontal motion, which is constant because air resistance is neglected, and the vertical motion, which is influenced by gravity.
Firstly, we need to calculate the time it takes for the bagels to fall from the window to the ground. Using the equation for free fall h =
1/2 g t², where h is the height (38.0 m), and g is the acceleration due to gravity (9.81 m/s²), we can solve for t, the time to fall:
38.0 m = 1/2 * 9.81 m/s² * t²
t = sqrt(2 * 38.0 m / 9.81 m/s²) = sqrt(7.74) ≈ 2.78 s
Bruce throws the bagels 9.00 s after Henrietta has passed below the window. In this time, Henrietta has jogged a distance of d = speed * time = 3.05 m/s * 9.00 s = 27.45 m horizontally.
Since Henrietta is already past the point directly below the window, we need to add the distance she will cover in the time it takes for the bagels to fall. This distance is additional distance = jogging speed * fall time = 3.05 m/s * 2.78 s ≈ 8.48 m.
Overall, Henrietta will be approximately 27.45 m + 8.48 m = 35.93 m from the point directly below the window when she catches the bagels.
To find the initial speed with which Bruce throws the bagels, we use the horizontal motion formula initial speed = distance / time, which gives us an initial speed of approximately 35.93 m / 2.78 s ≈ 12.92 m/s.
Bruce must throw the bagels horizontally at an initial speed of approximately 12.92 m/s for Henrietta to catch them just before they hit the ground, at a distance of approximately 35.93 m from the point directly below Bruce's window.
If instead the distance between the moon and the planet were 7 times as large (no change in mass), what would the magnitude of the force be?
Answer:
Reduced by 49 times
Explanation:
We have Newton formula for attraction force between 2 objects with mass and a distance between them:
[tex]F_G = G\frac{M_1M_2}{R^2}[/tex]
where G is the gravitational constant. [tex]M = M_1 = M_2[/tex] are the masses of the 2 objects. and R is the distance between them.
Since R squared is in the denominator of the formula, if we make it 7 times as large with no change in mass, gravitational force would be dropped by 7*7 = 49 times
To solve the problem we should know about Newton's Law of gravity.
What is Newton's Law of gravity?
According to Newton's law of gravity, there is an attractive force between any two-particle carrying mass, such that the force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
[tex]F \propto m_1m_2\\\\F \propto \dfrac{1}{R^2}[/tex]
[tex]F = G\dfrac{m_1m_2}{R^2}[/tex]
Where G is the proportionality constant and the value of G is 6.67 x 10-11 N m² / kg².
The force between the two will be [tex]\dfrac{1}{49}[/tex] time of the force before.
Given to us,
Mass of the planet = [tex]m_1[/tex]Mass of the earth = [tex]m_2[/tex]distance between the moon and the planet is 7 timesAssumption
Let's assume that the distance between the moon and the planet is d.
Values
As it is given that there is no change in the mass of the moon or the planet, therefore,
Mass of the planet = [tex]m_1[/tex]Mass of the earth = [tex]m_2[/tex]Also, it is given that the distance between them changes to 7 times, therefore,
distance between the moon and the planet =7dNewton's Law of gravitySubstitute the value Newton's Law of gravity,
[tex]F = G\dfrac{m_1m_2}{(7d)^2}\\\\\\F = G\dfrac{m_1m_2}{49d^2}[/tex]
Thus, the force between the two will be [tex]\dfrac{1}{49}[/tex] time of the force before.
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Asteroids, meteoroids, and comets are remnants of the early solar system. (T/F)
Answer: Asteroids, meteoroids, and comets are remnants of the early solar system. This Statement is TRUE.
Explanation:
METEOROID: these are small rocky or metallic objects found in outer space.
ASTEROIDS: these are also known as minor planets of the inner solar system. They are irregularly shaped object in space that orbits the Sun.
COMETS: these are dusty chunk of ice, that moves in a highly elliptical orbit about the sun.
Asteroids, meteoroids, and comets as remnants of the early solar system was further proved in nebular hypothesis
initially proposed in the eighteenth century by German philosopher Immanuel Kant and French mathematician Pierre-Simon Laplace. (The word nebula means a gaseous cloud.) According to the modern version of the theory, about 4.5 to 5 billion years ago the solar system developed out of a huge cloud of gases and dust floating through space. These materials were at first very thin and highly dispersed.
Choose true or false for each statement regarding the resistance of a wire.
1. The resistivity depends on the temperature of most metal wires.
2. While maintaining a constant voltage V, the current I increases when the length L of a wire decreases.
3. While maintaining a constant voltage V, the current I increases when the resistivity rho of a wire increases.
Answer:
1. True
2. True
3. False
Explanation:
According to Ohm's law, V = IR where V is the voltage, I is the current and R is the resistance. So if V is constant, R is inversely proportional to I
1. The resistivity of metal wires would depend on the temperature, this is true.
2. When length L of a wire decreases, the resistance R decreases as well, this would make I increase.
3. When the resistivity increases, resistance R increases as well, this would make I decrease.
The resistivity of a wire is affected by temperature, with most metals increasing in resistivity with higher temperatures. Current increases when the length of a wire decreases at constant voltage. However, current decreases when resistivity increases at constant voltage.
Explanation:When examining the resistance of a wire, there are several factors to consider that impact how the resistance changes under various conditions. The statements about resistance of a wire need to be evaluated for their truthfulness.
True: The resistivity depends on the temperature of most metal wires. The resistivity of conductors increases with increasing temperature because the atoms vibrate more of causing electrons to make more collisions.True: While maintaining a constant voltage V, the current I increases when the length L of a wire decreases. This is because resistance is directly proportional to the length of the wire.False: While maintaining a constant voltage V, the current I increases when the resistivity of a wire increases. In fact, the current decreases because resistance increases with resistivity.
A test car starts from rest on a horizontal circular track of 85-m radius and increases its speed at a uniform rate to reach 115 km/h in 13 seconds. Determine the magnitude a of the total acceleration of the car 11 seconds after the start.
To calculate the magnitude of the total acceleration of the car 11 seconds after it starts, consider both tangential and centripetal accelerations. The tangential acceleration is 2.457 m/s², and the centripetal acceleration is 8.63 m/s², resulting in a total acceleration of 8.97 m/s².
To determine the magnitude of the total acceleration of the car 11 seconds after the start, we need to consider both the tangential acceleration (uniform acceleration as it speeds up) and the centripetal acceleration (due to the car's circular motion). To calculate the tangential acceleration ([tex]a_t[/tex]), we use the final speed (v) the car reaches in 13 seconds, which is 115 km/h, and convert it to meters per second:
v = 115 km/h = (115 * 1000 m) / (3600 s) = 31.94 m/s
Since the car accelerates uniformly from rest, the tangential acceleration is given by:
[tex]a_t[/tex] = v / t = 31.94 m/s / 13 s = 2.457 m/s²
After 11 seconds, the speed of the car (v₁₁) is:
v₁₁ = at * 11 s = 2.457 m/s² * 11 s = 27.03 m/s
The centripetal acceleration ([tex]a_c[/tex]) is calculated using the formula:
[tex]a_c[/tex] = v₁₁² / r = (27.03 m/s)² / 85 m = 8.63 m/s²
To find the total acceleration, we use the Pythagorean theorem because the tangential and centripetal accelerations are perpendicular to each other:
a = √([tex]a_t[/tex]² + [tex]a_c[/tex]²) = √((2.457 m/s²)² + (8.63 m/s²)²) = √(6.036 + 74.516) = √(80.552) = 8.97 m/s².
A sly 1.5-kg monkey and a jungle veterinarian with a blow-gun loaded with a tranquilizer dart are 25 m above the ground in trees 70 m apart. Just as the veterinarian shoots horizontally at the monkey, the monkey drops from the tree in a vain attempt to escape being hit. What must the minimum muzzle velocity of the dart be for the dart to hit the monkey before the monkey reaches the ground?
Answer:
31 m/s
Explanation:
As both the monkey and the darts are subjected to constant gravitational acceleration g = 9.8 m/s2 and both start from rest (vertically speaking). Their vertical position will always be the same. For the dart to hit the monkey, its horizontal position must be the same as the monkey's, which is unchanged before reaching the ground. Therefore, the time it takes for the dart to travel across 70 m must be less than the time it takes for the monkey to drop 25m to the ground. We can find it out using the following equation of motion
[tex]s_m = gt_m^2/2[/tex]
[tex]25 = 9.8t_m^2/2[/tex]
[tex]t_m^2 = 50/9.8 = 5.1[/tex]
[tex]t_m = \sqrt{5.1} = 2.26 s[/tex]
For the dart to takes less that 2.26 s to travel 70m, its horizontal speed must at least be 70 / 2.26 = 31 m/s
If Earth were completely blanketed with clouds and we couldn’t see the sky, could we learn about the realm beyond the clouds? What forms of radiation might penetrate the clouds and reach the ground?
The definition of waves that propagate through electric fields is called electromagnetic waves. The earth, despite being covered with clouds, can be 'affected' because waves such as sunlight or the moon have the ability to penetrate and be visible to the inhabitants of the earth. Microwaves and radio waves would be less affected by the clouds that cover the Earth.
Through these waves, you can know that there is beyond the clouds.
Ultraviolet light, microwaves and radio waves are the radiations that penetrate through the clouds and reach the Earth's surface.
Therefore, the answer is Yes, ultraviolet light, microwaves and radio waves are the forms of radiation that penetrate and reach the ground.
It is indeed possible to learn about the universe beyond the clouds due to other non-visual forms of radiation, mainly radio waves and gamma rays, which can penetrate through the clouds and reach the earth's surface.
Explanation:Yes, even if Earth were completely blanketed with clouds and we could not see the sky, we could still learn about the universe beyond the clouds. This is because, in addition to visible light which would be blocked by the clouds, the universe also emits various other forms of radiation that can penetrate the clouds and reach the ground.
Two major types of radiation that could penetrate the dense clouds are radio waves and gamma rays. Radio waves are a form of electromagnetic radiation used in many areas of science and technology, while gamma rays are highly energetic forms of radiation and are used in fields such as astronomy to get valuable information about distant celestial bodies.
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Metals are good conductors of electricity because their protons can roam freely throughout the material. True False
Answer:
Explanation:
False
Electric conductivity in metals is the result of the motion of charged particles called electrons. Atoms of metallic elements are distinguished by the availability of valence electrons. Valence electrons are the electrons in the outer shell of an atom and free to move around.
These free electrons are free to move in the lattice of metal and result in electric current when a voltage is applied.
Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, and (d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the point of release to be
y0=0.
Answer:
Explanation:
V = Deltax/Deltat
V = 15.0 m/s
Displacement:
(a) Vf = Vi + adeltat
Vf = 15.0m/s - 9.8m/s^2 x 0.500s = 10.1m/s
Displacement = (15.0m/s x 0.500s) - (0.5)(9.8m/s^2)(0.500s)^2 = 6.275m
(b) Vf = 15.0m/s - 9.8m/s^2 x 1.00s = 5.2m/s
Displacement = (15.0m/s x 1.00s) - (0.5)(9.8m/s^2)(1s)^2 = 10.1m
(c) Vf = 15.0m/s - 9.8m/s^2 x 1.50s = 14.7m/s
Displacement = (15.0m/s x 1.50s) - (0.5)(9.8m/s^2)(1.5s)^2 = 11.475m
(d) Vf = 15.0m/s - 9.8m/s^2 x 2.00s = 19.6m/s
Displacement = (15.0m/s x 2.00s) - (0.5)(9.8m/s^2)(2s)^2 = 10.4m
What is the lowest frequency that will resonate in an organ pipe 2.00 m in length, closed at one end? The speed of sound in air is 340 m/s.
Answer:
42.5 Hz.
Explanation:
The fundamental frequency of a closed pipe is given as
f₀ = v/4l....................... Equation 1
Where f₀ = lowest frequency, v = speed of sound in air, l = length of the organ pipe
Given: v = 340 m/s, l = 2.00 m.
Substitute into equation 1
f₀ = 340/(4×2)
f₀ = 340/8
f₀ = 42.5 Hz.
Hence the smallest frequency that will resonant in the organ pipe = 42.5 Hz.
Suppose electrons in a TV tube are accelerated through a potential difference of 2.00 104 V from the heated cathode (negative electrode), where they are produced, toward the screen, which also serves as the anode (positive electrode), 25.0 cm away.At what speed would the electrons impact the phosphors on the screen? Assume they accelerate from rest, and ignore relativistic effects?
Answer:
83816746.4254 m/s
Explanation:
m = Mass of electron = [tex]9.11\times 10^{-31}\ kg[/tex]
q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]
V = Voltage = [tex]2\times 10^4\ V[/tex]
The kinetic energy of the electron is
[tex]K=\dfrac{1}{2}mv^2[/tex]
Energy is given by
[tex]E=qV[/tex]
Balancing the energy
[tex]qV=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{\dfrac{2qV}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 2\times 10^4}{9.11\times 10^{-31}}}\\\Rightarrow v=83816746.4254\ m/s[/tex]
The velocity of the electrons is 83816746.4254 m/s
You are observing a spacecraft moving in a circular orbit of radius 100,000 km around a distant planet. You happen to be located in the plane of the spacecraft’s orbit. You find that the spacecraft’s radio signal varies periodically in wavelength between 2.99964 m and 3.00036 m. Assuming that the radio is broadcasting at a constant wavelength, what is the mass of the planet?
To solve this problem we will apply the concepts related to centripetal acceleration, which will be the same - by balance - to the force of gravity on the body. To find this acceleration we must first find the orbital velocity through the Doppler formulas for the given periodic signals. In this way:
[tex]v_{o} = c (\frac{\lambda_{max}-\bar{\lambda}}{\bar{\lambda}}})[/tex]
Here,
[tex]v_{o} =[/tex] Orbital Velocity
[tex]\lambda_{max} =[/tex] Maximal Wavelength
[tex]\bar{\lambda}} =[/tex] Average Wavelength
c = Speed of light
Replacing with our values we have that,
[tex]v_{o} = (3*10^5) (\frac{3.00036-3}{3})[/tex]
Note that the average signal is 3.000000m
[tex]v_o = 36 km/s[/tex]
Now using the definition about centripetal acceleration we have,
[tex]a_c = \frac{v^2}{r}[/tex]
Here,
v = Orbit Velocity
r = Radius of Orbit
Replacing with our values,
[tex]a = \frac{(36km/s)^2}{100000km}[/tex]
[tex]a= 0.01296km/s^2[/tex]
[tex]a = 12.96m/s^2[/tex]
Applying Newton's equation for acceleration due to gravity,
[tex]a =\frac{GM}{r^2}[/tex]
Here,
G = Universal gravitational constant
M = Mass of the planet
r = Orbit
The acceleration due to gravity is the same as the previous centripetal acceleration by equilibrium, then rearranging to find the mass we have,
[tex]M = \frac{ar^2}{G}[/tex]
[tex]M = \frac{(12.96)(100000000)^2}{ 6.67*10^{-11}}[/tex]
[tex]M = 1.943028*10^{27}kg[/tex]
Therefore the mass of the planet is [tex]1.943028*10^{27}kg[/tex]
A prism-shaped closed surface is in a constant, uniform electric field E, filling all space, pointing right.The 3 rectangular faces of the prism are labeled A, B, and C. Face A is perpendicular to the E-field. The bottom face C is parallel to E. Face B is the leaning face. (The two triangular side faces are not labeled.)Which face has the largest magnitude electric flux through it?
a) A b) B c) C d) A and B have the same magnitude flux
Answer:
The correct answer is:
d) A and B have the same magnitude flux
Explanation:
Electric flux is the property of electric field that measures the electric field lines, passing through a surface and electric flux is also directly proportional to the number of electric field lines passing through a surface.
The formula of electric flux is:
Φ = E A Cos θ
(where E is the electric field, A is the area of face and θ is the angle between the face and the electric field).
Since, faces A and B are perpendicular to the electric field and the electric field lines passing through face A also passes through face B therefore, both of these faces have larger and same magnitude of electric flux.
Since, face C is parallel to the electric field so, the electric flux is smaller at face C, because the magnitude of Cos 180 (when face is parallel) is smaller than the magnitude of Cos 90 (when face is perpendicular).
Face A, which is perpendicular to the uniform electric field, has the largest magnitude electric flux through it because the angle between the field lines and the normal to the surface is zero, maximizing the electric flux.
Explanation:The question revolves around calculating the electric flux through different faces of a prism in a uniform electric field. Electric flux (Φ) is given by the equation Φ = E ⋅ A ⋅ cos(θ), where E is the magnitude of the electric field, A is the area through which the field lines pass, and θ is the angle between the field lines and the normal (perpendicular) to the surface.
Face A is perpendicular to the electric field, which means the angle θ is 0 degrees and cos(θ) is 1. Thus the flux through Face A is maximum. For Face B, the leaning face, θ is greater than 0 degrees but less than 90 degrees, thus cos(θ) will be less than 1. Hence, flux through Face B will be less than through Face A. Face C, being parallel to the electric field, has θ as 90 degrees, and cos(90) is 0, so the flux through Face C is zero. Therefore, in comparison, Face A has the largest magnitude electric flux through it.
A used car is pushed off an 87-ft-high sheer seaside cliff with a speed of 8 ft/s. Find the speed at which the car hits the water.
Final Answer:
The speed at which the car hits the water is approximately 75.2 feet per second.
Explanation:
To find the speed at which the car hits the water, we can use one of the kinematic equations that relates the initial velocity, acceleration due to gravity, the height it fell from, and the final velocity. The kinematic equation that we need is:
[tex]\[ v^2 = u^2 + 2gh \][/tex]
Where:
- v is the final velocity,
- u is the initial velocity,
- g is the acceleration due to gravity (which we will use [tex]\( 32.174 \, \text{ft/s}^2 \)[/tex] for since we are dealing with feet),
- h is the height.
Here, we are given:
- [tex]\( u = 8 \, \text{ft/s} \)[/tex] (initial velocity)
- [tex]\( h = 87 \, \text{ft} \)[/tex] (height)
- [tex]\( g = 32.174 \, \text{ft/s}^2 \)[/tex] (acceleration due to gravity)
Let's find the final velocity \( v \) using these values.
[tex]\[ v^2 = u^2 + 2gh \][/tex]
[tex]\[ v^2 = (8 \, \text{ft/s})^2 + 2 \cdot 32.174 \, \text{ft/s}^2 \cdot 87 \, \text{ft} \][/tex]
[tex]\[ v^2 = 64 \, \text{ft}^2/\text{s}^2 + 2 \cdot 32.174 \, \text{ft/s}^2 \cdot 87 \, \text{ft} \][/tex]
[tex]\[ v^2 = 64 \, \text{ft}^2/\text{s}^2 + 5591.148 \, \text{ft}^2/\text{s}^2 \][/tex]
[tex]\[ v^2 = 5655.148 \, \text{ft}^2/\text{s}^2 \][/tex]
Now we take the square root of both sides to solve for the final velocity \( v \):
[tex]\[ v = \sqrt{5655.148} \, \text{ft/s} \][/tex]
Performing the square root calculation, we get:
[tex]\[ v \approx 75.2 \, \text{ft/s} \][/tex]
So, the speed at which the car hits the water is approximately 75.2 feet per second.
A ball is tossed with a velocity of 10 m/s directed vertically upward from a window located 20 m above the ground. Determine the following: (a) The velocity v and elevation y of the ball above the ground at any time t. (b) The highest elevation reached by the ball and its corresponding time t. (c) The time when the ball will hit the ground and the impact velocity.
Answer:
Explanation:
Given
Initial velocity of ball [tex]u=10\ m/s[/tex]
height of window [tex]h=20\ m[/tex]
Using Equation of motion
[tex]y=ut+\frac{1}{2}at^2[/tex]
where u=initial velocity
t=time
a=acceleration
As ball is already is at a height of 20 m so
[tex]Y=ut+\frac{1}{2}at^2+20[/tex]
[tex]Y=10\times t+0.5\times (-9.8)t^2+20[/tex]
[tex]Y=-4.9t^2+10t+20[/tex]
(b)highest point is obtained at v=0
[tex]v^2-u^2=2as[/tex]
where
v=final velocity
u=initial velocity
a=acceleration
s=displacement
[tex](0)-10^2=2\times (-9.8)\times s[/tex]
[tex]s=\frac{100}{19.6}[/tex]
[tex]s=5.102\ m[/tex]
Highest Point will be [tex]s+20=25.102\ m[/tex]
(c)Time taken when the ball hit the ground i.e. at Y=0
[tex]-4.9t^2+10t+20=0[/tex]
[tex]t=3.28\ s[/tex]
impact velocity [tex]v=\sqrt{2\times 9.8\times 25.102}[/tex]
[tex]v=22.181\ m/s[/tex]
(a) The equation be "Y = -4.9t² + 10t + 20".
(b) The highest point be "25.102 m".
(c) The impact velocity be "22.181 m/s"
Equation of motionAccording to the question,
Ball's initial velocity, u = 10 m/s
Window's height, h = 20 m
(a) By using equation of motion,
Y = ut + [tex]\frac{1}{2}[/tex]at²
By substituting the values,
= ut + [tex]\frac{1}{2}[/tex]at² + 20
= 10 × t + 0.5 × (9.8)t² + 20
= -4.9t² + 10t + 20
(b) We know that,
→ v² - u² = 2as
here, Final velocity, v = 0
0 - (10)² = 2 × (-9.8) × s
s = [tex]\frac{100}{19.6}[/tex]
= 5.102 m
(c) Time taken will be:
→ -4.9t² + 10t + 20 = 0
t = 3.28 s
hence,
The impact velocity,
v = [tex]\sqrt{2\times 9.8\times 25.102}[/tex]
= 22.181 m/s
Thus the above response is correct.
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sewing machine needle moves up and down in simple harmonic motion with an amplitude of 0.0127 m and a frequency of 2.55 Hz. How far doesthe needlemove in one period?
If a sewing machine needle moves up and down in simple harmonic motion with an amplitude of 0.0127 m and a frequency of 2.55 Hz then the needle move in one period = 0.0508 m
If the displacement of a particle undergoing simple harmonic motion of amplitude A at a time is [tex]→x=Asin[/tex]ω[tex]t^i[/tex]
then the total displacement of the particle over one period of the oscillation from time t=0 is:= [tex](+A^i)+ (-2A^i)+(+A^i)[/tex]
= [tex]0^i.[/tex]
The total distance traveled by the particle during that one period is A+2A+A = 4AGiven:
Amplitude = 0.0127 m
frequency = 2.55 Hz
Solution:
To find the displacement of the needle in one period we need to put value in the formula:
The total distance traveled = 4A
= 4*0.0127 m
= 0.0508 m
Thus, If a sewing machine needle moves up and down in simple harmonic motion with an amplitude of 0.0127 m and a frequency of 2.55 Hz then the needle move in one period = 0.0508 m
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Final answer:
The distance moved by the sewing machine needle in one period of its simple harmonic motion is four times the amplitude, which equals 0.0508 meters.
Explanation:
The simple harmonic motion of a sewing machine needle is defined by its amplitude and frequency. The amplitude (0.0127 m) is the maximum displacement from its rest position, and the needle's frequency (2.55 Hz) indicates how often it repeats this motion in one second. To determine the distance the needle moves in one period, we need to consider that it moves from the rest position to the maximum amplitude, back through the rest position to the negative maximum amplitude, and back to rest again, completing one full cycle.
In simple harmonic motion, the distance moved in one period is four times the amplitude. So, the needle moves a distance of 4 x 0.0127 m in one period. Therefore, the needle moves 0.0508 meters in one cycle.
When you drop an object from a certain height, it takes time T to reach the ground with no air resistance. If you dropped it from three times that height, how long (in terms of T) would it take to reach the ground?
When an object is dropped from a certain height with no air resistance, it takes a specific time to reach the ground. The time it takes to reach the ground is proportional to the square root of the height. If the object is dropped from three times the original height, it will take T * sqrt(3) time to reach the ground.
Explanation:When an object is dropped from a certain height with no air resistance, it takes a specific time (T) to reach the ground. If the object is dropped from three times that height, it will take a longer time to reach the ground. The relationship between the height and the time taken is linear, assuming no air resistance.
To determine the time it would take to reach the ground from three times the original height, we need to consider that the time is proportional to the square root of the height. Let's represent the original time as T and the original height as H. Therefore, the time it would take to reach the ground from three times the original height (3H) would be:
t = T * sqrt(3)
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Arctic sea ice has declined over the past few decades causing water levels to increase. This is an interaction of which two spheres?
Biosphere and geosphere
Cryosphere and hydrosphere
Geosphere and atmosphere H
ydrosphere and geosphere
Answer:
Option (2)
Explanation:
The Cryosphere refers to the frozen water bodies on earth. This includes the glaciers, icebergs, ice sheets and the frozen water surrounding the Arctic as well as Antarctica.
The Hydrosphere refers to all the water bodies on earth including the rivers, streams, lakes, and ponds.
The given condition is based on the interaction between the cryosphere and the hydrosphere.
The frozen ice in the Antarctic and Arctic is melting rapidly due to the increase in the global warming effect. This declining ice in the polar region results in the rise in the global sea level. This can be catastrophic as many of the big cities will be flooded because of this increasing height of sea level.
Thus, the correct answer is option (2).
The decline of Arctic sea ice and its impact on water levels is an interaction between two Earth system spheres: the cryosphere and hydrosphere.
The interaction of Arctic sea ice decline and increasing water levels involves the cryosphere and hydrosphere spheres. The cryosphere refers to the frozen components of the Earth system, including ice caps, glaciers, and sea ice. The hydrosphere encompasses all the water on Earth, including oceans, lakes, and rivers.
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Calculate the molecular weight of a polyethylene molecule with n=750. Express your answer to three significant figures.
To calculate the molecular weight of a polyethylene molecule with n=750, multiply the molecular weight of the ethylene unit by n.
Explanation:To calculate the molecular weight of a polyethylene molecule with n=750, we need to know the molecular formula and the atomic weights of the elements present in the molecule. Polyethylene is made up of repeating ethylene (C2H4) units, so we can calculate the molecular weight of the polyethylene molecule by multiplying the molecular weight of the ethylene unit (28.05 g/mol) by the value of n (750).
Calculation:
Molecular weight of polyethylene = Molecular weight of ethylene unit × n = 28.05 g/mol × 750 = 21,037.5 g/mol
Therefore, the molecular weight of the polyethylene molecule with n=750 is 21,037.5 g/mol, rounded to three significant figures.
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A certain carbon monoxide molecule consists of a carbon atom mc = 12 u and an oxygen atom mo = 17 u that are separated by a distance of d = 128 pm, where "u" is an atomic unit of mass.
Part (a) write a symbolic equation for the location of the center of mass of the carbon monoxide molecule relative to the position of the oxygen atom. This expression should be in terms of the masses of the atoms and the distance between them. 50%
Part (b) Calculate the numeric value for the center of mass of carbon monoxide in units of pm. Grade Summary Deductions Potential 0% 100%
Answer:
a) x_{cm} = m₂/ (m₁ + m₂) d , b) x_{cm} = 52.97 pm
Explanation:
The expression for the center of mass is
[tex]x_{cm}[/tex] = 1 / M ∑ [tex]x_{i}[/tex] [tex]m_{i}[/tex]
Where M is the total masses, mI and xi are the mass and position of each element of the system.
Let's fix our reference system on the oxygen atom and the molecule aligned on the x-axis, let's use index 1 for oxygen and index 2 for carbon
x_{cm} = 1 / (m₁ + m₂) (0+ m₂ x₂)
Let's reduce the magnitudes to the SI system
m₁ = 17 u = 17 1,661 10⁻²⁷ kg = 28,237 10⁻²⁷ kg
m₂ = 12 u = 12 1,661 10⁻²⁷ kg = 19,932 10⁻²⁷ kg
d = 128 pm = 128 10⁻¹² m
The equation for the center of mass is
x_{cm} = m₂/ (m₁ + m₂) d
b) let's calculate the value
x_{cm} = 19.932 10⁻²⁷ /(19.932+ 28.237) 10⁻²⁷ 128 10-12
x_{cm} = 52.97 10⁻¹² m
x_{cm} = 52.97 pm
(a) The expression for the center mass of these two atoms relative to oxygen atom is [tex]X_{cm} = \frac{m_1 d_0 \ +\ m_2d}{m_1 + m_2}[/tex]
(b) The numeric value for the center of mass of carbon monoxide is 53 pm.
The given parameters;
mass of the carbon atom = 12umass of the oxygen atom, = 17 udistance between the atoms, = 128 pmThe center mass of these two atoms relative to oxygen atom is calculated as follows;
[tex]X_{cm} = \frac{m_1 d_0 \ +\ m_2d}{m_1 + m_2}[/tex]
where;
[tex]d_0[/tex] is distance of the atom in the fixed reference point (oxygen atom)(b)
The numeric value for the center of mass of carbon monoxide in units of pm is calculated as follows;
[tex]X_{cm} = \frac{17u(0) \ +\ 12u(128 \ pm)}{(12u + 17u)}\\\\X_{cm} = \frac{(12 \times 128u) \ pm}{29u} \\\\X_{cm} = 53 \ pm[/tex]
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A straight wire 0.280 m in length carries a current of 3.40 A. What are the two angles between the direction of the current and the direction of a uniform 0.0400 T magnetic field for which the magnetic force on the wire has magnitude 0.0250 N?
Answer:
The angle θ between the direction of the current and the direction of the uniform magnetic field is 41° or 139°.
Explanation:
The force on a current carrying wire is given by the following equation:
[tex]\vec{F} = I\vec{L}\times \vec{B}[/tex]
The cross-product can be written with a sine term:
[tex]F = ILB\sin(\theta)\\0.025 = (3.4)(0.28)(0.04)\sin(\theta)\\\sin(\theta) = 0.6565[/tex]
Therefore, the angle θ is 41.03° and 138.97°
The angles can be calculated using the formula sin(θ) = F / (I L B), giving two symmetrical values about 90° in the first and second quadrants because the sine function is periodic.
The question is asking for the angles at which the force on a current-carrying wire in a magnetic field is a specific magnitude. The magnitude of the force exerted on a current-carrying wire placed in a magnetic field is given by the formula F = I L B sin(θ), where F is the force, I is the current, L is the length of the wire, B is the magnetic field strength, and θ is the angle between the direction of the current and the direction of the magnetic field.
By rearranging for θ, we get the equation sin(θ) = F / (I L B). Plugging in the values from the question, we find sin(θ) = 0.0250 N / (3.40 A imes 0.280 m imes 0.0400 T). This gives us θ values that correspond to the sine of this ratio.
There are two angles that will produce the same sine value because sine is a periodic function, which are θ and 180°-θ. Therefore, the two angles between the direction of the current and the direction of the uniform magnetic field for which the force on the wire has a magnitude of 0.0250 N will be symmetrical about 90° in the first and second quadrants.
What length of tube would be required to produce a second tone under the same experimental conditions? Explain your answer.
To produce a second tone or the first overtone in a tube closed at one end, the length of the tube required is three times the length used for the fundamental frequency, resulting in a length of 1.008 m.
To understand the length required to produce a second tone or the first overtone in a tube closed at one end, it's essential to grasp the concept of harmonics in sound resonance. In such a tube, the resonant frequencies occur in odd multiples of the fundamental frequency. The first resonance the students observed, with the fundamental frequency of 256 Hz at a length of 0.336 m, corresponds to a quarter wavelength of the sound wave in the tube.
For the first overtone (second resonance), the air column in the tube must accommodate three-quarters of a wavelength, meaning the effective length will be three times larger than that of the fundamental. Thus, if the fundamental resonance occurs at a length of 0.336 m, the length for the second resonance will be:
0.336 m x 3 = 1.008 m.This calculation is based on the understanding that the second tone or first overtone in a closed tube happens at three times the length necessary for the fundamental frequency, leading to the described increase in the length of the air column.
To find the length of tube for the second resonance, halve the initial length where the first resonance occurred at a fundamental frequency of 256 Hz.
Explanation:The length required to produce a second tone under the same experimental conditions can be calculated based on the concept of resonance in a closed tube.
To find the length for the second resonance (first overtone), we know that the first resonance occurs at 0.336m for a fundamental frequency of 256 Hz. The second resonance, in this case, would occur at half the wavelength of the fundamental frequency, so the length would be half of the initial length: 0.168m.