Answer:
(a) The time the book is in the air stays the same.
(b) The horizontal distance the book travels doubles.
(c) The speed of the book just before it reaches the floor increases but not doubles.
Explanation:
The following kinematics equations will be used to solve this question:
[tex]v_y = v_{y_0} + a_yt\\y - y_0 = v_{y_0}t + \frac{1}{2}a_yt^2\\v_y^2 = v_{y_0}^2 + 2a_y(y - y_0)[/tex]
(a) Initially, the y-component of the velocity is zero. So, the x-component of the velocity is doubled.
We will use the second equation for both cases:
[tex]0 - y_0 = v_{y_0}t - \frac{1}{2}(g)t^2 = 0 - \frac{1}{2}gt^2\\0 - y_0 = v_{y_0}t_2 -\frac{1}{2}gt_2^2 = 0 - \frac{1}{2}gt_2^2\\t = t_2[/tex]
Since, the initial velocity in the y-direction is zero, and the height of the table is constant. The time it takes from the edge of the table to the floor is the same.
(b) For the horizontal distance the book travels, we should use the second equation again, and keep in mind that the acceleration in the x-direction is zero.
[tex]x - 0 = v_{x_0}t + \frac{1}{2}a_xt^2 = v_0t\\x_2 = 2v_0t_2 = 2v_0t[/tex]
Hence, the horizontal distance doubles.
(c) The vertical velocity does not change, since the initial velocity in the y-direction is zero. The horizontal velocity does not change along the motion, since the acceleration in the x-direction is zero. However, since the initial velocity in the x-direction doubles, the final velocity in the x-direction doubles as well.
[tex]v_x = v_{x_0} + a_xt = v_{x_0}[/tex]
However, this does not mean that final speed of the book will double. Because the speed of the object is calculated as follows:
[tex]v_{\rm final_1} = \sqrt{v_x^2 + v_y^2} = \sqrt{v_0^2 + v_y^2}\\v_{\rm final_2} = \sqrt{v_{x_2}^2 + v_y^2} = \sqrt{(2v_0)^2 + v_y^2}[/tex]
As can be seen from above, the final speed increases but not doubles.
Doubling the initial horizontal velocity of a book in projectile motion would result in the same time in the air, double the distance covered, and a higher (but not doubled) speed when it hits the floor.
Explanation:This question explores the concepts of projectile motion in physics. When we double the initial horizontal velocity of the book from v0 to 2v0, the following changes occur:
Time the book is in the air: This will stay the same. The time a projectile is in the air is determined by the vertical component of its motion alone. Thus, the horizontal velocity does not affect the time.Horizontal distance the book travels while it is in the air: This will double. The horizontal distance (d) a projectile travels is determined by the horizontal velocity (v) and the time it spends in the air (t). So, d=v*t. If we double the initial horizontal velocity, the distance will double as well.Speed of the book just before it reaches the floor: This will increase. The speed a projectile hits the floor with is determined by both its horizontal and vertical speeds. Doubling the initial horizontal velocity will result in a greater overall speed when it hits the floor, but not necessarily double the original speed.Learn more about Projectile Motion here:https://brainly.com/question/20627626
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In reaching her destination, a backpacker walks with an average velocity of 1.20 m/s, due west. This average velocity results, because she hikes for 5.63 km with an average velocity of 2.33 m/s due west, turns around, and hikes with an average velocity of 0.374 m/s due east. How far east did she walk (in kilometers)?
Answer:
[tex] x_2 = 648.46\ m[/tex]
Explanation:
given,
Average velocity due west = 1.20 m/s
case 1
Distance moved in west, x₁ = 5.63 km
speed due west, v₁ = 2.33 m/s
Case 2
Distance moved in east = x₂
speed due east, v₂ = 0.374 m/s
total distance = x₁ + x₂ = 5.62
total time = t₁ + t₂ = 5630/2.33 + x₂/0.374
now,
[tex]average\ velocity = \dfrac{total\ distance}{total\ time}[/tex]
[tex]-1.20= \dfrac{-5630 + x_2}{\dfrac{5630}{2.33}+\dfrac{x_2}{0.374}}[/tex]
negative sign is used because we want the distance in east but velocity is in west.
[tex] - 2900 - 3.21 x_2 = -5630 + x_2[/tex]
[tex]-4.21 x_2 = -2730[/tex]
[tex] x_2 = 648.46\ m[/tex]
The distance she walked in east is equal to 648.46 m
The solar system is of largely uniform composition. (T/F)
Answer:
False
Explanation:
Sun mass is dominating in Solar system as compared to other planets, asteroids and comets. Sun itself accounting for the 99.9% of the mass of the solar system. Hence the gravitational force exerted by the Sun dominates the other objects in the solar system. So we can conclude that solar system has non-uniform composition. The given statement is false
When a sound wave moves through a medium such as air, the motion of the molecules of the medium is in what direction (with respect to the motion of the sound wave)? Group of answer choicesa. Perpendicularb. Parallalc. Anit-paralleld. Both choices B and C ara valid
Answer:
Both choices B and C are valid
Explanation:
Sound wave are Mechanical wave. Air (or viscus fluid) is the medium of propagation. Sound is produced by the back and forth vibration of the object. Consider the vibration of object is from left to right then this back and forth vibrations of object displaces the molecules of the medium both rightward and leftward (to and fro) to the direction of the energy transport forming compression and rarefaction. This shows that the motion of the molecules of the medium is both parallel (and anti-parallel) to the direction of the sound wave propagation.
Final answer:
The motion of the particles of a medium in a sound wave is parallel to the direction of the wave motion.
Explanation:
A sound wave is a longitudinal wave, which means that the motion of the particles of the medium is parallel to the direction of the wave motion. In other words, the particles of the medium vibrate or oscillate back and forth in the same direction that the sound wave is traveling. This can be compared to compressing and stretching a coiled spring. As the wave propagates through the medium, it creates zones of compression and rarefaction, causing the air molecules to move in the same direction as the sound wave.
Problem 12.6 A hockey player hits a puck so that it comes to rest 10 s after sliding 100 ft on the ice. Determine (a) the initial velocity of the puck, (b) the coefficient of friction between the puck and the ice.
Answer:
a)The initial velocity of the puck is 20 ft/s.
b)The coefficient of friction is 0.062.
Explanation:
Hi there!
a)For this problem let's use the equations of position and velocity of an object moving in a straight line with constant acceleration:
x = x0 + v0 · t + 1/2 · a · t²
v = v0 + a · t
Where:
x = position of the puck after a time t.
x0 = initial position.
v0 = initial velocity.
a = acceleration.
t = time.
v = velocity of the puck at a time t.
Let's place the origin of the frame of reference at the point where the puck is hit so that x0 = 0.
We know that at t = 10 s the velocity of the puck is zero (v = 0) and its position is 100 ft (x = 100 ft):
100 ft = v0 · 10 s + 1/2 · a · (10 s)²
0 = v0 + a · 10 s
We have a system of two equations with two unknowns, so, we can solve the system.
Solving for v0 in the second equation:
0 = v0 + a · 10 s
v0 = -a · 10 s
Replacing v0 in the first equation:
100 ft = (-a · 10 s) · 10 s + 1/2 · a · (10 s)²
100 ft = -50 s² · a
100 ft / -50 s² = a
a = -2.0 ft/s²
Then the initial velocity of the puck will be:
v0 = -a · 10 s
v0 = -(-2.0 ft/s²) · 10 s
v0 = 20 ft/s
The initial velocity of the puck is 20 ft/s.
b) The friction force is calculated as follows:
Fr = N · μ
Where:
Fr = friction force.
N = normal force.
μ = coefficient of friction.
Since the only vertical forces acting on the puck are the weight of the puck and the normal force and since the puck is not being accelerated in the vertical direction, then, the normal force is equal to the weight of the puck. The weight (W) is calculated as follows:
W = m · g
Where "m" is the mass of the puck and "g" is the acceleration due to gravity (32.2 ft/s²).
Then the friction force can be calculated as follows:
Fr = m · g · μ
Since the acceleration of the puck is provided only by the friction force, then, due to Newton's second law:
Fr = m · a
Where "m" is the mass of the puck and "a" its acceleration. Then:
Fr = m · g · μ
Fr = m · a
m · g · μ = m · a
μ = a/g
μ = 2.0 ft/s² / 32.2 ft/s²
μ = 0.062
The coefficient of friction is 0.062.
If an electric current of 8.50 A flows for 3.75 hours through an electrolytic cell containing copper-sulfate (CuSO4) solution, then how much copper is deposited on the cathode (the negative electrode) of the cell? (Copper ions carry two units of positive elementary charge, and the atomic mass of copper is 63.5 g/mol.)
Answer:
75.5g
Explanation:
From the ionic equation, we can write
[tex]CU^{2+}+SO^{2-}_{4}\\[/tex]
next we find the number of charge
Note Q=it
for i=8.5A, t=3.75 to secs 3.75*60*60=13500secs
hence
[tex]Q=8.5*13500\\Q=114750C[/tex]
Since one faraday represent one mole of electron which equal 96500C
Hence the number of mole produced by 114750C is
114750/96500=1.2mol
The mass of copper produced is
[tex]mol=\frac{mass}{molar mass} \\mass=mole*molar mass\\mass=1.2*63.5\\mass=75.5g[/tex]
Hence the amount of copper produced is 75.5g
Answer:
75.5 g.
Explanation:
Dissociation equation:
Cu2+ + SO4^2- --> CuSO4
Q = I * t
Where,
I = current
= 8.5 A
t = time
= 3.75 hours
= 13500 s
Q = 8.5 * 13500
= 114750 C
1 faraday represent one mole of electron which equal 96500C
Number of mole of Cu2+
= 114750/96500
= 1.19 mol.
Mass = number of moles * molar mass
= 1.19 * 63.5
= 75.5 g.
A square is 1.0 m on a side. Point charges of +4.0 µC are placed in two diagonally opposite corners. In the other two corners are placed charges of +3.0 µC and -3.0 µC. What is the potential (relative to infinity) at the midpoint of the square?
Answer:
[tex]V = 1.44\times 10^{5}~V[/tex]
Explanation:
The electric potential can be found by using the following formula
[tex]V = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}[/tex]
Applying this formula to each charge gives the total potential.
[tex]V = V_1 + V_2 + V_3 + V_4\\V = \frac{1}{4\pi\epsilon_0}\frac{4\times 10^{-6}}{(\sqrt{2}/2)^2} + \frac{1}{4\pi\epsilon_0}\frac{4\times 10^{-6}}{(\sqrt{2}/2)^2} + \frac{1}{4\pi\epsilon_0}\frac{3\times 10^{-6}}{(\sqrt{2}/2)^2} - \frac{1}{4\pi\epsilon_0}\frac{3\times 10^{-6}}{(\sqrt{2}/2)^2}\\V = \frac{16\times 10^{-6}}{4\pi\epsilon_0}\\V = 1.44\times 10^{5}~V[/tex]
Since the potential is a scalar quantity, it is safe to sum all the potentials straightforward. And since they all placed on the corners of a square, +3 and -3 μC charges cancel out each other.
What is the strength of an electric field that will balance the weight of a 9.6 g plastic sphere that has been charged to -9.2 nC ? Express your answer to two significant figures and include the appropriate units.
The strength of an electric field that will balance the weight is 1.023 × 10⁷ N/C.
What is electric field?An electric field is a physical field that surrounds electrically charged particles and acts as an attractor or repellent to all other charged particles in the vicinity. Additionally, it refers to a system of charged particles' physical field.
Electric charges and time-varying electric currents are the building blocks of electric fields.
The strength of an electric field that will balance the plastic sphere is = weight of the object/charge on the object
= ( 9.6 ×10⁻³×9.8)/(9.2×10⁻⁹) N/C
= 1.023 × 10⁷ N/C
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A pressure gage connected to a tank reads 55 kPa at a location where the atmospheric pressure is 72.1 cmHg. The density of mercury is 13,600 kg/m3 . Calculate the absolute pressure in the tank.
Answer:
Explanation:
Given
Gauge Pressure of a tank is
[tex]P_{gauge}=55\ kPa[/tex]
at that place atmospheric Pressure is [tex]h=72.1\ cm\ of\ Hg[/tex]
Density of mercury [tex]\rho _{Hg}=13600\ kg/m^3[/tex]
Atmospheric Pressure in kPa is given by
[tex]P_{atm}=\rho _{Hg}\times g\times h[/tex]
[tex]P_{atm}=13600\times 9.8\times 0.721[/tex]
[tex]P_{atm}=96.09\ kPa[/tex]
and Absolute Pressure is summation of gauge pressure and atmospheric Pressure
[tex]P_{abs}=P_{gauge}+P_{atm}[/tex]
[tex]P_{abs}=55+96.09=151.09\ kPa[/tex]
A space-based telescope can achieve a diffraction-limited angular resolution of 0.05″ for red light (wavelength 700 nm). What would the resolution of the instrument be (a) in the infrared, at 3.5 µm, and (b) in the ultraviolet, at 140 nm?
Answer:
a) [tex] \theta_2 = 0.05 * \frac{3.5}{0.7} = 0.25[/tex]
b) [tex] \theta_2 = 0.05 * \frac{140}{700} = 0.01[/tex]
Explanation:
We are comparing two wavelengths with the radius and diameter constant, and if we want to compare it, we need to use the following formula:
[tex]\frac{\theta_1}{\theta_2}= \frac{\lambda_1}{\lambda_2}[/tex]
Where [tex] \theta[/tex] represent the angular resolution and [tex]\lambda[/tex] the wavelength.
So if we have a fixed resolution and wavelength 1 and we want to find the resolution for a new condition we can solve for [tex] \theta_2[/tex] and we got
[tex] \theta_2 = \theta_1 \frac{\lambda_2}{\lambda_1}[/tex]
Part a
For this case the subindex 1 is for the color red and we know that:
[tex] \lambda_1 = 700 nm *\frac{1 \mu m}{1000 nm} = 0.7 \mu m[/tex]
And the angular resolution for the color red is specified as [tex] \theta_1 = 0.05[/tex]
And for the infrared case we know that [tex] \lambda_2 = 3.5 \mu m[/tex], so if we replace we got:
[tex] \theta_2 = 0.05 * \frac{3.5}{0.7} = 0.25[/tex]
Part b
For this case the subindex 1 is for the color red and we know that:
[tex] \lambda_1 = 700 nm[/tex]
And the angular resolution for the color red is specified as [tex] \theta_1 = 0.05[/tex]
And for the ultraviolet case we know that [tex] \lambda_2 = 140 nm[/tex], so if we replace we got:
[tex] \theta_2 = 0.05 * \frac{140}{700} = 0.01[/tex]
a large parallel plate capacitor has plate seperation of 1.00 cm and plate area of 314 cm^2. The capacitor is connected across a voltage of 20.0 V and has air betweeen the plates. How much work is done on the capacitor as the plate seperation is inceased to 2.00 cm?
Answer:
[tex]W = -2.76\times 10^{-9}~J[/tex]
Explanation:
The work done on the capacitor is equal to the difference in potential energy stored in the capacitor in two different cases.
The potential energy is given by the following formula:
[tex]U = \frac{1}{2}CV^2[/tex]
where C can be calculated using the plate separation and area.
[tex]C = \epsilon\frac{A}{d} = \epsilon\frac{0.0314}{0.01} = 3.14\epsilon[/tex]
Therefore, the potential energy in the first case is
[tex]U = \frac{1}{2}3.14\epsilon (20)^2 = 628\epsilon[/tex]
In the second case:
[tex]C_2 = \epsilon\frac{A}{d} = \epsilon\frac{0.0314}{0.02} = 1.57\epsilon\\U = \frac{1}{2}C_2 V^2 = \frac{1}{2}1.57\epsilon (20)^2 = 314\epsilon[/tex]
The permittivity of the air is very close to that of vacuum, which is 8.8 x 10^-12.
So, the difference in the potential energy is
[tex]W = U_2 - U_1 = \epsilon(314 - 628) = -314 \times 8.8 \times 10^{-12} = -2.76\times 10^{-9}~J[/tex]
A block of mass M M is placed on a semicircular track and released from rest at point P P , which is at vertical height H 1 H1 above the track’s lowest point. The surfaces of the track and block are considered to be rough such that a coefficient of friction exists between the track and the block. The block slides to a vertical height H 2 H2 on the other side of the track. How does H 2 H2 compare to H 1 H1 ?
Answer:
Explanation:
A block of mass M is placed on a semicircular track and released from rest at point P , which is at vertical height H₁ above the track’s lowest point.
Its initial potential energy = mgH₁
Kinetic energy = 0
Total energy = mgH₁
When block slides to a vertical height H₂ on the other side of the track
Its final potential energy = mgH₂
Kinetic energy = 0
Total final energy = mgH₂
As negative work is done by frictional force while block moves ,
final energy < initial energy
mgH₂ < mgH₁
H₂ < H₁
H₂ will be less than H₁ .
Under constant acceleration the average velocity of a particle is half the sum of its initial and final velocities. Is this still true if the acceleration is not constant? Explain.
Answer:
Explanation:
Under constant acceleration the average velocity of a particle is half the sum of its initial and final velocities. Is this still true if the acceleration is not constant? Explain.
True or False? The superposition or overlapping of two waves always results in destructive interference between the different waves.
Answer:
Explanation:
False
When two waves overlap or superimpose over each other then they either undergo Constructive or destructive interference.
waves are the disturbance created by a force and add up to gives constructive interference when they are in the same line i.e. in the same phase.
When these disturbances are in the opposite phase then they superimpose to give destructive interference where the amplitude of the resulting wave will be much smaller as compared to original waves.
You run due east at a constant speed of 3.00 m/s for a distance of 120.0 m and then continue running east at a constant speed of 5.00 m/s for another 120.0 m. For the total 240.0-m run, is your average velocity 4.00 m/s, greater than 4.00 m/s, or less than 4.00 m/s? Explain.
Answer:
Explanation:
Given
Speed while running towards east is [tex]v_1=3\ m/s[/tex]
Distance traveled in east direction [tex]x_1=120\ m[/tex]
For Another interval you run with velocity
[tex]v_2=5\ m/s[/tex]
[tex]x_2=240\ m[/tex]
Total displacement[tex]=x_1+x_2[/tex]
[tex]=120+120=240\ m[/tex]
Time for first interval
[tex]t_1=\frac{x_1}{v_1}=\frac{120}{3}[/tex]
[tex]t_1=\frac{120}{3}=40\ s[/tex]
Time for second interval
[tex]t_2=\frac{x_2}{v_2}=\frac{120}{5}=24\ s[/tex]
total time [tex]t=t_1+t_2[/tex]
[tex]t=40+24=64\ s[/tex]
average velocity [tex]v_{avg}=\frac{x_1+x_2}{t}[/tex]
[tex]v_{avg}=\frac{240}{64}=3.75\ m/s[/tex]
Therefore average velocity is less than [tex]4 m/s[/tex]
A series RL circuit with L = 3.00 H and a series RC circuit with C = 3.00 F have equal time constants. If the two circuits contain the same resistances R, (a) what is the value of R and (b) what sit the time constant?
Answer:
(a) R = 1Ω
(b) τ = 3
Explanation:
The time constants of the given circuits are as follows
[tex]\tau_{RL} = \frac{L}{R}\\\tau_{RC} = RC[/tex]
If the two circuits have equal time constants, then
[tex]\frac{L}{R} = RC\\R^2 = \frac{L}{C} = \frac{3}{3} = 1\\R = 1\Omega[/tex]
Therefore, the time constant in any of the circuits is
[tex]\tau = RC = 3[/tex]
(a). Value of resistance R is 1 ohm.
(b). The Value of time constant will be 3 second.
The time constant is defined as, time taken by the system to reach at 63.2% of its final value.
Time constant in RL circuit is, [tex]=\frac{L}{R}[/tex]
Time constant in RC circuit is, [tex]=RC[/tex]
Since, in question given that both circuit have same time constant.
So, [tex]RC=\frac{L}{R}\\\\R^{2}=\frac{L}{C}\\\\R=\sqrt{\frac{L}{C} }[/tex]
substituting L = 3H and C = 3F in above expression.
[tex]R=\sqrt{\frac{3}{3} }=1ohm[/tex]
Time constant = [tex]RC=1*3=3s[/tex]
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An electron is brought from rest infinitely far away to rest at point P located at a distance of 0.033 m from a fixed charge q. That process required 111 eV of energy from an eternal agent to perform the necessary work.
Answer:
The potential at point P is -111 Volt.
Explanation:
Given that,
Distance = 0.033 m
Work = 111 ev
Suppose what is the potential at point P?
We need to calculate the potential at point P
Using formula of potential
[tex]W=qV[/tex]
[tex]V=\dfrac{W}{q}[/tex]
Where, W = work
q = charge
Put the value into the formula
[tex]V= \dfrac{111\times1.6\times10^{-19}}{-1.6\times10^{-19}}[/tex]
[tex]V=-111\ V[/tex]
Hence, The potential at point P is -111 Volt.
The question is about the process of bringing an electron from rest to rest near a fixed charge q at a specific distance, and the energy required for this process. The answer explains the equation for the energy required per unit charge and the concept of electron volts (eV) as an energy unit.
Explanation:The question is asking about the process of bringing an electron from rest at an infinite distance to rest at a point located a distance of 0.033 m from a fixed charge q using an external agent. The process requires 111 eV of energy to perform the necessary work. The equation for the energy required per unit charge is given, which relates the potential energy to the charge and distance. The concept of electron volts (eV) as an energy unit is also mentioned.
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How much work does it take to slide a box 37 meters along the ground by pulling it with a 217 N force at an angle of 19° from the horizontal?
Answer:
W = 7591.56 J
Explanation:
given,
distance of the box, d = 37 m
Force for pulling the box, F = 217 N
angle of inclination with horizontal,θ = 19°
We know,
Work done is equal to product of force and the displacement.
W = F.d cos θ
W = 217 x 37 x cos 19°
W = 7591.56 J
Hence, the work done to pull the box is equal to W = 7591.56 J
Final answer:
The work done to slide the box is 7586.09 Joules.
Explanation:
To calculate the work done to slide a box along the ground, we can use the formula:
Work = Force x Distance x cos(theta)
Where:
Force = 217 N (the force applied to pull the box)
Distance = 37 meters (the distance the box is being slid)
theta = 19° (the angle between the applied force and the horizontal)
Plugging in these values into the formula, we get:
Work = 217 N x 37 m x cos(19°)
Calculating this using a calculator, we find that the work done to slide the box is approximately 7586.09 Joules.
If the wind velocity is 43 km/h due south, in what direction should the pilot set her course to travel due west? Use the same airspeed of 212 km/h
Answer:
11.48 degree N of W
Explanation:
We are given that
Wind velocity=[tex]v_w=-43[/tex]km/h
Because wind is blowing towards south
Air speed=[tex]v_a=-212km/h[/tex]
Because the captain want to move with air speed in west direction.
x component of relative velocity=-212 km/h
y-Component of relative velocity=-43km/h
Direction=[tex]\theta=tan^{-1}(\frac{y}{x})[/tex]
[tex]\theta=tan^{-1}(\frac{-43}{-212}=11.48^{\circ}[/tex]N of W
Hence, the direction in which the pilot should set her course to travel due west=11.48 degree N of W
A bird is flying due east. Its distance from a tall building is given by x(t) = 28.0 m + (12.4 m/s)t – (0.0450 m/s3)t3. What is the instantaneous velocity of the bird when t = 8.00s?
Answer:
3.76 m/s
Explanation:
Instantaneous velocity: This can be defined as the velocity of an object in a non uniform motion. The S.I unit is m/s.
v' = dx(t)/dt..................... Equation 1
Where v' = instantaneous velocity, x = distance, t = time.
Given the expression,
x(t) = 28.0 m + (12.4 m/s)t - (0.0450 m/s³)t³
x(t) = 28 + 12.4t - 0.0450t³
Differentiating x(t) with respect to t.
dx(t)/dt = 12.4 - 0.135t²
dx(t)/dt = 12.4 - 0.135t²
When t = 8.00 s.
dx(t)/dt = 12.4 - 0.135(8)²
dx(t)/dt = 12.4 - 8.64
dx(t)/dt = 3.76 m/s.
Therefore,
v' = 3.76 m/s.
Hence, the instantaneous velocity = 3.76 m/s
Final answer:
To find the bird's instantaneous velocity at t = 8.00s, we differentiate its position function to get v(t) = 12.4 m/s - 0.135 m/s^2 × t^2, then substitute t = 8.00s to find v(8.00) = 3.76 m/s east.
Explanation:
The question asks for the instantaneous velocity of a bird flying due east when t = 8.00s, given the position function x(t) = 28.0 m + (12.4 m/s)t – (0.0450 m/s3)t3. To find the instantaneous velocity, we need to differentiate the position function with respect to time (t) to get the velocity function, v(t).
First, let's differentiate x(t):
Derivative of 28.0 m is 0 since it's a constant.
Derivative of (12.4 m/s)t is 12.4 m/s, as the derivative of t is 1.
Derivative of (-0.0450 m/s3)t3 is -0.135 m/s2 × t2, using the power rule for derivatives.
So, the velocity function is v(t) = 12.4 m/s - 0.135 m/s2 × t2. To find the instantaneous velocity at t = 8.00s, we plug in t = 8.00 into the velocity function:
v(8.00) = 12.4 m/s - 0.135 m/s2 × (8.002)
Calculating this gives us:
v(8.00) = 12.4 m/s - 0.135 m/s2 × 64.00 = 12.4 m/s - 8.64 m/s = 3.76 m/s
Therefore, the instantaneous velocity of the bird when t = 8.00s is 3.76 m/s east.
Is the electric-field magnitude between the plates larger or smaller than that for the original capacitor?
Answer:
The magnitude of the electric field will decrease
Explanation:
The capacitance of a parallel plate capacitor having plate area A and plate separation d is C=ϵ0A/d.
Where ϵ0 is the permittivity of free space.
A capacitor filled with dielectric slab of dielectric constant K, will have a new capacitance C1=ϵ0kA/d
C1=K(ϵ0A/d)
C1=KC
Where C is the capacitance with no dielectric.
The new capacitance is k times the capacitance of the capacitor without dielectric slab.
This implies that the charge storing capacity of a capacitor increases k times that of the capacitor without dielectric slab.
The charge stored in the original capacitor Q=CV
The charge stored in the original capacitor after inserting dielectric Q1=C1V1
The law of conservation of energy states that the energy stored is constant:
i.e Charge stored in the original capacitor is same as charge stored after the dielectric is inserted.
Q = Q1
CV = C1V1
CV = C1V1 -------2
We derived C1=KC. Inserting this into equation 2
CV = KCV1
V1 = (CV)/KC
V/K
This implies the voltage decreases when a dielectric is used within the plate.
The relationship between electric field and potential voltage is a linear one
V= Ed
Therefore the electric field will decrease
Final answer:
The capacitance of a parallel-plate capacitor with non-uniform electric field due to increased plate separation will be less than the ideal scenario calculated by C = € A/d.
Explanation:
When the separation of the plates in a parallel-plate capacitor is not small enough to maintain a uniform electric field, the field lines will bulge outwards at the edges. This bulging implies that some of the field lines that emanate from one plate do not reach the opposite plate, reducing the effective area over which the field is acting. Consequently, this non-uniform field means that the capacitance of the capacitor will be less than the ideal calculation using the formula C = € A/d, where C is the capacitance, € is the permittivity of the medium between the plates, A is the plate area, and d is the separation between the plates.
An initially stationary 2.7 kg object accelerates horizontally and uniformly to a speed of 13 m/s in 4.0 s. (a) In that 4.0 s interval, how much work is done on the object by the force accelerating it? What is the instantaneous power due to that force (b) at the end of the interval and (c) at the end of the first half of the interval?
Explanation:
A.
Given:
V = 13 m/s
t = 4 s
Constant acceleration, a= (V-Vi)/t
= 13/4
= 3.25 m/s^2
F = mass * acceleration
= 2.7 * 3.25
= 8.775 N.
Using equations of motion,
distance,S = (13 * 4) - (1/2)(3.25)(4^2)
= 26 m
Workdone, W = force * distance
= 8.775 * 26
= 228.15 J
B.
Instantaneous power, P = Force *Velocity
= 8.775 * 13
= 114. 075 W
C.
t = 2 s,
Constant acceleration, a= (V-Vi)/t
= 13/2
= 6.5 m/s^2
Force = mass * acceleration
= 2.7 * 6.5
= 17.55 N
Instantaneous power, P = Force *Velocity
= 17.55 * 13
= 228.15 W.
= 114. 075 W.
An excited hydrogen atom emits light with a wavelength of 486.4 nm to reach the energy level for which n = 2. In which principal quantum level did the electron begin?
Answer:
The electron began in the quantum level of 4
Explanation:
Using the formula of wave number:
Wave Number = 1/λ = Rh(1/n1² - 1/n2²)
where,
Rh = Rhydberg's Constant = 1.09677 x 10^7 /m
λ = wavelength of light emitted = 486.4 nm = 486.4 x 10^-9 m
n1 = final shell = 2
n2 = initial shell = ?
Therefore,
1/486.4 x 10^-9 m = (1.09677 x 10^7 /m)(1/2² - 1/n2²)
1/4 - 1/(486.4 x 10^-9 m)(1.09677 x 10^7 /m) = 1/n2²
1/n2² = 0.06082
n2² = 16.44
n2 = 4
The principal quantum level in which the electron began is 5.
Given the following data:
Final transition = 2Wavelength = 486.4 nm = [tex]486.4 \times 10^{-9}\;m[/tex]Rydberg constant = [tex]1.09 \times 10^7\; m^{-1}[/tex]
To determine the principal quantum level in which the electron began, we would use the Rydberg equation:
Mathematically, the Rydberg equation is given by the formula:
[tex]\frac{1}{\lambda} = R(\frac{1}{n_f^2} -\frac{1}{n_i^2})[/tex]
Where:
[tex]\lambda[/tex] is the wavelength.R is the Rydberg constant.[tex]n_f[/tex] is the final transition.[tex]n_i[/tex] is the initial transition.Substituting the parameters into the formula, we have;
[tex]\frac{1}{486.4 \times 10^{-9}} = 1.09 \times 10^7(\frac{1}{2^2} -\frac{1}{n_i^2})\\\\\frac{1}{486.4 \times 10^{-9} \times 1.09 \times 10^7}=\frac{1}{4} -\frac{1}{n_i^2}\\\\\frac{1}{5.3018} =\frac{1}{4} -\frac{1}{n_i^2}\\\\\frac{1}{n_i^2}=\frac{1}{4}-\frac{1}{5.3018}\\\\\frac{1}{n_i^2}=0.25-0.1886\\\\\frac{1}{n_i^2}=0.0614\\\\n_i^2=\frac{1}{0.0614} \\\\n_i=\sqrt{16.29} \\\\n_i=4.0[/tex]
Initial transition = 4.0
Read more: https://brainly.com/question/14649374
If the frequency of an electromagnetic wave increases, does the number of waves passing by you increase, decrease, or stay the same?
Answer:
If the frequency of an electromagnetic wave increases, the number of waves passing by you increase.
Explanation:
The total number of vibration or oscillation per unit time is called frequency of a waver. It is given by :
[tex]f=\dfrac{n}{t}[/tex]
n is the number of waves passing
t is the time taken
It is clear that the frequency is directly proportional to the number of waves. Hence, if the frequency of an electromagnetic wave increases, the number of waves passing by you increase.
The acceleration of a bus is given by ax(t) = αt, where α = 1.2 m/s3. (a) If the bus’s velocity at time t = 1.0 s is 5.0 m/s, what is its velocity at time t = 2.0 s? (b) If the bus’s position at time t = 1.0 s is 6.0 m, what is its position at time t = 2.0 s? (c) Sketch ay-t, vy-t , and x-t graphs for the motion.
Answer:
(a). The velocity of bus at 2.0 sec is 6.8 m/s.
(b). The position of bus at 2.0 s is 11.8 m.
(c). [tex]a_{y}-t[/tex], [tex]v_{y}-t[/tex] and x-t graphs
Explanation:
Given that,
[tex]\alha=1.2\ m/s^3[/tex]
Time t = 1.0 s
Velocity = 5.0
The Acceleration equation is
[tex]a_{x(t)}=\alpha t[/tex]
We need to calculate the velocity
Using formula of acceleration
[tex]a=\dfrac{dv}{dt}[/tex]
On integrating
[tex]\int_{v_{0}}^{v}{dv}=\int_{0}^{t}{a dt}[/tex]
Put the value into the formula
[tex]v-v_{0}=1.2\int_{0}^{t}{t dt}[/tex]
[tex]v-v_{0}=0.6t^2[/tex]
[tex]v=v_{0}+0.6t^2[/tex]
Put the value into the formula
[tex]v_{0}=5.0-0.6\times(1.0)^2[/tex]
[tex]v_{0}=4.4\ m/s[/tex]
We need to calculate the velocity at 2.0 sec
Put the value of initial velocity in the equation
[tex]v=4.4+0.6\times(2.0)^2[/tex]
[tex]v=6.8\ m/s[/tex]
(b). If the bus’s position at time t = 1.0 s is 6.0 m,
We need to calculate the position
Using formula of velocity
[tex]v=\dfrac{dx}{dt}[/tex]
On integrating
[tex]\int_{x_{0}}^{x}{dx}=\int_{0}^{t}{v dt}[/tex]
[tex]x_{0}-x=\int_{0}^{t}{v_{0}dt}+\int_{0}^{t}{0.6 t^2}[/tex]
[tex]x_{0}-x=v_{0}t+\dfrac{0.6}{3}t^3[/tex]
[tex]x=x_{0}+v_{0}t+\dfrac{0.6}{3}t^3[/tex]
[tex]x_{0}=6-4.4\times1-\dfrac{0.6}{3}\times1^3[/tex]
[tex]x=1.4\ m[/tex]
The position at t = 2.0 s
[tex]x=1.4+4.4\times2.0+\dfrac{0.6}{3}\times2^3[/tex]
[tex]x=11.8\ m[/tex]
Hence, (a). The velocity of bus at 2.0 sec is 6.8 m/s.
(b). The position of bus at 2.0 s is 11.8 m.
(c). [tex]a_{y}-t[/tex], [tex]v_{y}-t[/tex] and x-t graphs
A truck is passing over a bridge with a weight limit of 50,000 pounds. When empty, the front of the truck weighs 19,800 pounds and the back of the truck weighs 12,500 pounds. How much cargo (C), in pounds, can the truck carry and still be allowed to pass over the bridge?
Answer:
W = 17,700 lb
Explanation:
given,
Weight limit of the Bridge = 50,000 lb
Weight of empty truck = 19800 lb
Weight on the back of the truck = 12500 lb
Now,
Total weight of truck + cargo
= Weight of empty truck + Weight on the back of the truck
= 19800 + 12500
= 32300 lb
Weight of cargo which is still allowed.
W = weight limit - weight of the system at present
W = 50000 - 32300
W = 17,700 lb
Weight Truck can still carry is equal to W = 17,700 lb
What is the magnitude of a point charge that would create an electric field of 1.18 N/C at points 0.822 m away?
Answer:
q = 8.85 x 10⁻¹¹ C
Explanation:
given,
Electric field, E = 1.18 N/C
distance, r = 0.822 m
Charge magnitude = ?
using formula of electric field.
[tex]E = \dfrac{kq}{r^2}[/tex]
k is the coulomb constant
[tex]q= \dfrac{Er^2}{k}[/tex]
[tex]q= \dfrac{1.18\times 0.822^2}{9\times 10^9}[/tex]
q = 8.85 x 10⁻¹¹ C
The magnitude of charge is equal to q = 8.85 x 10⁻¹¹ C
At 47 °C, what is the fraction of collisions with energy equal or greater than an activation energy of 88.20 kJ/mol?
Answer:
The fraction of collision is [tex]4.00\times10^{-15}[/tex]
Explanation:
Given that,
Temperature = 47°C
Activation energy = 88.20 KJ/mol
From Arrhenius equation,
[tex]k=Ae^{-\dfrac{E_{a}}{RT}}[/tex]
Here, [tex]e^{-\dfrac{E_{a}}{RT}}[/tex]=fraction of collision
We need to calculate the fraction of collisions
Using formula of fraction of collisions
[tex]f=e^{-\dfrac{E_{a}}{RT}}[/tex]
Where f = fraction of collision
E = activation energy
R = gas constant
T = temperature
Put the value into the formula
[tex]f=e^{-\dfrac{88.20}{8.314\times10^{-3}\times(47+273)}}[/tex]
[tex]f=4.00\times10^{-15}[/tex]
Hence, The fraction of collision is [tex]4.00\times10^{-15}[/tex]
An old light bulb draws only 54.3 W, rather than its original 60.0 W, due to evaporative thinning of its filament. By what factor is the diameter of the filament reduced, assuming uniform thinning along its length? Neglect any effects caused by temperature differences.
Answer:
The factor of the diameter is 0.95.
Explanation:
Given that,
Power of old light bulb = 54.3 W
Power = 60 W
We know that,
The resistance is inversely proportional to the diameter.
[tex]R\propto\dfrac{1}{D}[/tex]
The power is inversely proportional to the resistance.
[tex]P\propto\dfrac{1}{R}[/tex]
[tex]P\propto D^2[/tex]
We need to calculate the factor of the diameter of the filament reduced
Using relation of power and diameter
[tex]\dfrac{P_{i}}{P_{f}}=\dfrac{D_{i}^2}{D_{f}^2}[/tex]
Put the value into the formula
[tex]\dfrac{D_{i}^2}{D_{f}^2}=\dfrac{54.3}{60}[/tex]
[tex]\dfrac{D_{i}}{D_{f}}=0.95[/tex]
[tex]D_{i}=0.95 D_{f}[/tex]
Hence, The factor of the diameter is 0.95.
Answer:
Explanation:
Po = 60 W
P = 54.3 W
Let the initial diameter of the filament is do and the final diameter of the filament is d.
Let the voltage is V and the initial resistance is Ro and the final resistance is R.
The formula for power is given by
P = V²/R
The resistance of the filament is inversely proportional to the square of the diameter of the filament. As voltage is constant so the power is
Power α diameter²
So, initial power is
Po α do² ..... (1)
Final power is
P α d² ..... (2)
Divide equation (2) by equation (1), we get
P / Po = d² / do²
54.3 / 60 = d² / do²
d² / do² = 0.905
d = 0.95 d
Thus, the diameter of the filament is reduced to a factor of 0.95 .
9) A balloon is charged with 3.4 μC (microcoulombs) of charge. A second balloon 23 cm away is charged with -5.1 μC of charge. The force of attraction / repulsion between the two charges will be: ______________________ 10) If one of the balloons has a mass of 0.084 kg, with what acceleration does it move toward or away from the other balloon? (calculate both magnitude AND direction) ________________________________________
Answer:
9. The force is a force of attraction and it is 2.95N
10. The magnitude of acceleration 35.12m/s^2 and the direction of this acceleration is away from the other balloon.
Explanation:
Parameters given:
Q1 = 3.4 * 10^-6C
Q2 = - 5.1 * 10^-6C
Distance between the two balloons = 23cm = 0.23m
9. Force acting between the two balloons is a force of attraction because they are unlike charges. Hence, the force between them is:
F = kQ1Q2/r^2
F = (9 *10^9 * 3.4 * 10^-6 * -5.1 * 10^-6)/(2.3 * 10^-1)^2
F = (1.56 * 10^-1)/(5.29 * 10^-2)
F = - 2.95N
10. Assuming that Balloon A has a mass, m, of 0.084kg, then:
F = ma
Where a = acceleration
a = F/m
a = -2.95/0.084
a = - 35.12m/s^2
The acceleration has a magnitude of 35.12m/s^2 and its direction is away from balloon B.
The negative sign shows that the balloon A is slowing down as it moves towards balloon B. Hence, it's velocity is reducing slowly.
A 1.78-m3 rigid tank contains steam at 220°C. One-third of the volume is in the liquid phase and the rest is in the vapor form. The properties of steam at 220°C are given as follows: vf = 0.001190 m3/kg and vg = 0.08609 m3/kg.
Answer:
a) P = 2319.6[kPa]; b) 2.6%
Explanation:
Since the problem data is not complete, the following information is entered:
A 1.78-m3 rigid tank contains steam at 220°C. One-third of the volume is in the liquid phase and the rest is in the vapor form. Determine (a) the pressure of the steam, and (b) the quality of the saturated mixture.
From the information provided in the problem we can say that you have a mixture of liquid and steam.
a) Using the steam tables we can see (attached image) that the saturation pressure at 220 °C is equal to:
[tex]P_{sat} =2319.6[kPa][/tex]
[tex]v_{f}=0.001190[m^{3}/hr]\\v_{g}=0.08609[m^{3}/hr]\\[/tex]
b) Since the specific volume of the gas and liquid is known, we can find the mass of each phase using the following equation:
[tex]m_{f}=\frac{V_{f} }{v_{f} } \\m_{g}=\frac{V_{g} }{v_{g} } \\where:\\V_{f}=volume of the fluid[m^3]\\v_{f}=specific volume of the fluid [m^3/kg]\\[/tex]
We know that the volume of the fluid is equal to:
[tex]V_{f}=1/3*V_{total} \\V_{total}=1.78[m^3]\\[/tex]
Now we can find the mass of the gas and the liquid.
[tex]m_{f}=\frac{1/3*1.78}{0.001190} \\m_{f}=498.6[kg]\\m_{g}=\frac{2/3*1.78}{0.08609}\\m_{g}=\ 13.78[kg][/tex]
The total mass is the sum of both
[tex]m_{total} =m_{g} + m_{fluid} \\m_{total} = 498.6 + 13.78\\m_{total} = 512.38[kg][/tex]
The quality will be equal to:
[tex]x = \frac{m_{g} }{m_{T} }\\ x= \frac{13.78}{512.38} \\x = 0.026 = 2.6%[/tex]