Answer:
a) v₀ = 32.64 m / s , b) x = 59.68 m
Explanation:
a) This is a projectile launching exercise, we the distance and height of the cliff
x = v₀ₓ t
y = [tex]v_{oy}[/tex] t - ½ g t²
We look for the components of speed with trigonometry
sin 43 = v_{oy} / v₀
cos 43 = v₀ₓ / v₀
v_{oy} = v₀ sin 43
v₀ₓ = v₀ cos 43
Let's look for time in the first equation and substitute in the second
t = x / v₀ cos 43
y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²
y = x tan 43 - ½ g x² / v₀² cos² 43
1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²
v₀² = g x² / [(x tan 43 –y) 2 cos² 43]
Let's calculate
v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]
v₀ = √ (35280 / 33.11)
v₀ = 32.64 m / s
.b) we use the vertical distance equation with the speed found
y = [tex]v_{oy}[/tex] t - ½ g t²
.y = v₀ sin43 t - ½ g t²
25 = 32.64 sin 43 t - ½ 9.8 t²
4.9 t² - 22.26 t + 25 = 0
t² - 4.54 t + 5.10 = 0
We solve the second degree equation
t = (4.54 ±√(4.54 2 - 4 5.1)) / 2
t = (4.54 ± 0.46) / 2
t₁ = 2.50 s
t₂ = 2.04 s
The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled
x = v₀ₓ t
x = v₀ cos 43 t
x = 32.64 cos 43 2.50
x = 59.68 m
The minimum muzzle velocity of a cannon to clear a 25.0 m tall cliff from 60.0 m away involves using projectile motion equations to find the initial velocity components, while ensuring the shell reaches the required height and distance.
Explanation:To solve for the minimum muzzle velocity for a cannon to shoot a shell past a 25.0 m cliff from 60.0 m away, we can use projectile motion equations. First, we separate the initial velocity into its horizontal (vx) and vertical (vy) components:
vx = v0 × cos(θ)vy = v0 × sin(θ)The shell must reach a height of at least 25.0 m to clear the cliff. We use the equation of motion in the vertical direction, considering the initial vertical velocity (vy) and the displacement (s = 25 m), to find the time (t) it takes to reach the top of the cliff. Then, using the horizontal velocity (vx), we can calculate how far the shell would travel horizontally in that time, ensuring that it covers at least 60.0 m. With the horizontal distance (d) and time (t) determined, we can calculate the shell's trajectory past the cliff edge.
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You push a shopping cart filled with groceries (total mass = 20 kg) by applying a force to the cart 30° from the horizontal. If the force you apply has a magnitude of 86 N, what is the cart’s acceleration? Assume negligible friction.You push a shopping cart filled with groceries (total mass = 20 kg) by applying a force to the cart 30° from the horizontal. If the force you apply has a magnitude of 86 N, what is the cart’s acceleration? Assume negligible friction.2.2 m/s25.5 m/s23.7 m/s24.3 m/s2
Answer:
3.72 m/s²
Explanation:
Horizontal component of the force applied = Fcosθ = 86 cos 30 = 74.478 N
Force = mass × acceleration = 74.478 N
ma = 74.478 N
a = 74.478 / 20 since friction can be neglected = 3.72 m/s²
Answer:
[tex]3.7m/s^{2}[/tex]
Explanation:
The forces acting on the cart is displayed in the attached file below.
Since the motion is only along the horizontal, we can conclude that the force bringing about that motion is the horizontal force.And from second newton law of motion, we deduce that
F=ma,
F=force, m=mass and a=acceleration
from the given question,
F=86N, mass,m=20kg,
Hence we can write the horizontal component of the force as
[tex]F_{X}=Fcos\alpha \\F_{x}=86cos30\\F_{x}=74.48\\Hence \\a=\frac{F_{x}}{m} \\a=\frac{74.48}{20}\\ a=3.7m/s^{2}[/tex]
You throw a baseball directly upward at time t = 0 at an initial speed of 13.5 m/s.
What is the maximum height the ball reaches above where it leaves your hand? Ignore air resistance and take g = 9.80 m/s².
Answer:
[tex]h=9.30m[/tex]
Explanation:
We have an uniformly accelerated motion, due to the gravitational acceleration. So, we use the kinematic equations, since the ball is throw directly upward, g is negative:
[tex]h=v_0t-\frac{gt^2}{2}[/tex]
First, we need to calculate the time taken by the ball to reach the maximum height, in this point its final speed is zero:
[tex]v_f=v_0-gt\\\\\frac{0-v_0}{-g}=t\\t=\frac{v_0}{g}\\t=\frac{13.5\frac{m}{s}}{9.8\frac{m}{s^2}}\\t=1.38s[/tex]
Now, we can calculate h:
[tex]h=v_0t-\frac{gt^2}{2}\\h=13.5\frac{m}{s}(1.38s)-\frac{9.8\frac{m}{s^2}(1.38s)^2}{2}\\h=9.30m[/tex]
how much work is required to move a 1 microcoulomb charge by a distance of 5 meters along an equipotential line of 6V?
Answer:
The work done is zero.
Solution:
As per the question:
Charge, [tex]q = 1\mu C = 1\times 10^{- 6}\ C[/tex]
Distance moved, d = 5 m
Voltage, V = 6V
Now, we know that an equipotential surface is one where the potential is same everywhere on the surface.
Suppose the the voltage at a distance d = 5 m is V'
Thus
V' = 6 V, (since the surface is equipotential)
Work done in moving a charge is given by:
[tex]W = q\Delta V[/tex]
[tex]W = q(V - V')[/tex]
[tex]W = (1\times 10^{- 6})(V - V')[/tex]
[tex]W = (1\times 10^{- 6})(6 - 6) = 0[/tex]
Thus the work done in moving a charge on an equipotential surface comes out to be zero as the potential difference is zero.
Final answer:
The work required to move a 1 microcoulomb charge by a distance of 5 meters along an equipotential line of 6V is zero because there's no change in potential energy.
Explanation:
The question relates to determining the amount of work needed to move a charge along an equipotential line. When a charge moves along an equipotential, the potential energy of the charge does not change because the voltage (potential difference) across its path remains zero. In other words, the work done on the charge is zero since work is defined as the change in potential energy, which is given by the formula W = qV, where W is work, q is charge in coulombs, and V is potential difference in volts. For movement along an equipotential line, V = 0, hence, Work = 0 Joules.
A boy throws a ball upward with a speed v0 = 12 m/s. The wind imparts a horizontal acceleration of 0.4 m/s2 to the left. At what angle θ must the ball be thrown so that it returns to the point of release? Assume that the wind does not affect the vertical motion.
Answer:
The angle is 2.33°.
Explanation:
Given that,
Speed of ball = 12 m/s
Acceleration = 0.4 m/s²
We need to calculate the time
Using formula of time of flight
[tex]t=\dfrac{2u}{g}[/tex]
[tex]t=\dfrac{2v\cos\theta}{g}[/tex]
Put the value into the formula
[tex]t=\dfrac{2\times12\cos\theta}{9.8}[/tex]
[tex]t=2.44\cos\theta[/tex]
We need to calculate the angle
Using equation of motion along vertical direction
[tex]s=ut-\dfrac{1}{2}at^2[/tex]
[tex]s=v\sin\theta\times t-\dfrac{1}{2}at^2[/tex]
Put the value in the equation
[tex]0=12\sin\theta\times2.44\cos\theta-\dfrac{1}{2}\times0.4\times(2.44\cos\theta)^2[/tex]
[tex]2\times12\sin\theta\times2.44=0.4\times(2.44)^2\cos\theta[/tex]
[tex]\tan\theta=\dfrac{0.4\times2.44}{2\times12}[/tex]
[tex]\theta=\tan^{-1}(0.04066)[/tex]
[tex]\theta=2.33^{\circ}[/tex]
Hence, The angle is 2.33°.
Light does not move infinitely fast but has a finite speed. We normally use ""c"" to indicate the speed of light in science. Write down the value for the speed of light in metric units: c = _________.
Answer:
6.71 × 10^8 mi/hr
Explanation:
Light is usually defined as an electromagnetic wave that is comprised of a definite wavelength. It is of both types, visible and invisible. The light emitted from a source usually travels at a speed of about 3 × 10^8 meter/sec. This speed of light is commonly represented by the letter 'C'.
To write it in the metric system, it has to be converted into miles/hour.
We know that,
1 minute = 60 seconds
60 minutes = 1 hour
1 kilometer = 1000 meter
1 miles = 1.6 kilometer
Now,
= [tex]\frac{3 \times\ 10^8 meter \times\ 60 sec \times\ 60 min}{1 sec \times\ 1 min \times\ 1 hr}[/tex]
= 1.08 × 10^12 m/ hr (meter/hour)
= [tex]\frac{1.08 \times\ 10^{12} meter \times\ 1 km \times\ 1 miles}{1 hr \times\ 1000 meter \times\ 1.6 km}[/tex]
= 6.71 × 10^8 mi/hr (miles/hour)
Thus, the value for speed of light (C) in metric unit is 6.71 × 10^8 mi/hr.
Final answer:
The speed of light in a vacuum is precisely known and has a fixed value of c = 2.99792458 × 10^8 m/s, roughly equal to 3.00 × 10^8 m/s. It's a fundamental constant in physics but is slower in materials, as defined by their index of refraction.
Explanation:
The speed of light in a vacuum is represented by the symbol c and has a fixed value of c = 2.99792458 × 108 m/s, which is essentially 3.00 × 108 m/s when rounded to three significant digits. This constant speed is a fundamental physical quantity and is crucial in many areas of physics including relativity and electromagnetism. It's important to note that the speed of light reduces when it passes through matter, which is characterized by the index of refraction n of the material.
By what order of magnitude is something that runs in nanoseconds faster than something that runs in milliseconds?
To solve this problem we will define the order of magnitude of both points, then we will obtain the radius and obtain the conclusion of the order of magnitude.
A nanosecond is one billionth of a second while and a millisecond is one millionth of a second
[tex]\frac{\text{millisec}}{\text{nanosec}} = \frac{10^{-3}}{10^{-9}} = 10^6[/tex]
Therefore something that runs in nanoseconds is six times faster than something that runs in milliseconds
Word magnitude means the extent of something. It is the property that determines the object is larger or smaller than the other. The object's magnitude can be arranged into class.
Nano second is the SI unit of time that is equal to bn of a second. That is 10⁻⁹ seconds. The nanosecond is a one billion th of the second, whereas an millisecond is 1000th of a second. The nanosecond process is thus 1,000,000 times faster.The correct answer is 100,00,00.
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Two light bulbs, A and B, are connected to a 120-V outlet (a constant voltage source). Light bulb A is rated at 60 W and light bulb B is rated at 100 W. Which light bulb has a greater filament resistance?
Answer:
Bulb A has a greater resistance.
Explanation:
Electric power (P) = V²/R
P = V²/R................ Equation 1
Where P = power, V = Voltage, R = Resistance.
Make R the subject of the equation
R = V²/P ................ Equation 2
For Bulb A,
Given: V = 120 V, P = 60 W.
Substitute into equation 2
R = 120²/60
R = 240 Ω
For bulb B
Given: V =120 V, P = 100 W.
Substitute into equation 2
R = 120²/100
R = 14400/100
R = 144 Ω
Hence Bulb A has a greater resistance.
bulb A has the greater filament resistance.
What is resistance?This can be defined as the ability of a conducting material to oppose the flow of electric current.
To know the bulb with the greatest filament resistance, we use the formula below.
Formula:
P = V²/R................. Equation 1making R the subject of the equation
R = V²/P............... Equation 2Where:
P = power of the bulbV = Outlet voltageR = Resistance of the filament.For Bulb A,
Given:
P = 60 WV = 120 VSubstitute these values into equation 1
R = (120²)/60R = 240 ohms.For Bulb B,
Given:
P = 100 W.V = 60 VSubstitute these values into equation 1
R = (120²)/100R = 144 ohms.Hence, From the above, it can be seen that bulb A has the greater filament resistance.
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An object’s velocity is measured to be vx(t) = α - βt2, where α = 4.00 m/s and β = 2.00 m/s3. At t = 0 the object is at x = 0. (a) Calculate the object’s position and acceleration as functions of time. (b) What is the object’s maximum positive displacement from the origin?
Answer:
Explanation:
Given
[tex]v_x(t)=\alpha -\beta t^2[/tex]
[tex]\alpha =4\ m/s[/tex]
[tex]\beta =2\ m/s^3[/tex]
[tex]v_x(t)=4-2t^2[/tex]
[tex]v=\frac{\mathrm{d} x}{\mathrm{d} t}[/tex]
[tex]\int dx=\int \left ( 4-2t^2\right )dt[/tex]
[tex]x=4t-\frac{2}{3}t^3[/tex]
acceleration of object
[tex]a=\frac{\mathrm{d} v}{\mathrm{d} t}[/tex]
[tex]a=-4t[/tex]
(b)For maximum positive displacement velocity must be zero at that instant
i.e.[tex]v=0[/tex]
[tex]4-2t^2=0[/tex]
[tex]t=\pm \sqrt{2}[/tex]
substitute the value of t
[tex]x=4\times \sqrt{2}-\frac{2}{3}\times 2\sqrt{2}[/tex]
[tex]x=3.77\ m[/tex]
The definitions of acceleration and velocity allow to find the results for the questions about the motion of the particle are:
A) the function of the acceleration is: a = -4t
and the position function is: x = 4 t - ⅔ t³
B) The maximum displacement is: x = 3.77 m
Given parameters
The velocity of the body v = α-β t² with α = 4 m/s and β = 2 m/s²To find
a) position and acceleration as a function of time,
b) maximum displacement,
The acceleration of defined as the change in velocity with time.
a = [tex]\frac{dv}{dt}[/tex]
Let's calculate.
a = - 2βt
a = - 2 2 t
a = -4 t
The speed is defined by the variation of the position with respect to time.
v = [tex]\frac{dx}{dt}[/tex]
dx = v dt
We integrate.
∫ dx = ∫ v dt
x - x₀ = ∫ (α - β t²)
x-x₀ = αt - βt³/ 3
we substitute.
x = 4 t - ⅔ t³
B) To find the maximum displacement we use the first derivative to be zero.
[tex]\frac{dx}{dt}[/tex] = 0
4 - 2t² = 0
t² = 2
t = √2 = 1.414 s
Let's find the position for this time.
x = 4 √2 - ⅔ (√2)³
x = 3.77 m
In conclusion using the definitions of acceleration and velocity we can find the result for the questions about the motion of the particle are:
A) the function of the acceleration is: a = -4t
and the position function is: x = 4 t - ⅔ t³
B) The maximum displacement is: x = 3.77 m
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A 200-g block is attached to a horizontal spring and executes simple harmonic motion with a period of 0.250 s. The total energy of the system is 2.00 J. Find (a) the force constant of the spring and (b) the amplitude of the motion.
Answer:
(A) Spring constant will be 126.58 N/m
(B) Amplitude will be equal to 0.177 m
Explanation:
We have given mass of the block m = 200 gram = 0.2 kg
Time period T = 0.250 sec
Total energy is given TE = 2 J
(A) For mass spring system time period is equal to [tex]T=2\pi \sqrt{\frac{m}{K}}[/tex]
So [tex]0.250=2\times 3.14 \sqrt{\frac{0.2}{K}}[/tex]
[tex]0.0398=\sqrt{\frac{0.2}{K}}[/tex]
Now squaring both side
[tex]0.00158=\frac{0.2}{K}[/tex]
K = 126.58 N/m
So the spring constant of the spring will be 126.58 N/m
(B) Total energy is equal to [tex]TE=\frac{1}{2}KA^2[/tex], here K is spring constant and A is amplitude
So [tex]2=\frac{1}{2}\times 126.58\times A^2[/tex]
[tex]A^2=0.0316[/tex]
A = 0.177 m
So the amplitude of the wave will be equal to 0.177 m
What is the electric field strength just outside the surface of a conducting sphere carrying surface charge density 1.4 μC/m2μC/m2?
Answer:
[tex]E=158.19\frac{kN}{m}[/tex]
Explanation:
Gauss's theorem states that the flux of the electric field through a closed surface is equal to the the charge enclosed by the surface divided by the vacuum permittivity:
[tex]\int\limits{\vec{E}\cdot \vec{dS}} \,=\frac{q}{\epsilon_0}[/tex]
The direction of the electric field ([tex]\vec{E}[/tex]) just outside of a conductor is parallel to its surface ([tex]\vec{S}[/tex]):
[tex]\int\limits{\vec{E}\cdot \vec{dS}} \,=\int\limits{EdScos(0^\circ)} \,=E\int\limits{dS} \,=ES[/tex]
Recall that the surface charge density is defined as:
[tex]\sigma=\frac{q}{S}[/tex]
Now, we get the electric field strength:
[tex]ES=\frac{q}{\epsilon_0}\\E=\frac{q}{S\epsilon_0}\\E=\frac{\sigma}{\epsilon_0}\\E=\frac{1.4*10^{-6}\frac{C}{m^2}}{8.85*10^{-12\frac{C^2}{N\cdot m^2}}}\\\\E=158192.09\frac{N}{C}=158.19\frac{kN}{C}[/tex]
The electric field strength just outside the surface of a conducting sphere with surface charge density of 1.4 μC/m^2 can be calculated using Gauss' Law. This gives an electric field strength of approximately 1.58 x 10^5 N/C.
Explanation:The electric field strength just outside the surface of a conducting sphere, with a surface charge density of 1.4 μC/m2, can be calculated using Gauss' Law, which states that the electric field is equal to the surface charge density divided by the permittivity of free space, ε0.
It is given by the formula:
E = σ/ε0
Where E is the electric field strength, σ is the charge density, and ε0 is the permittivity of free space. Using the given value for σ (1.4 x 10-6 C/m2) and the known value for ε0 (8.85 x 10-12 C2/N.m2), we find that:
E = (1.4 x 10-6) / (8.85 x 10-12)
This simplifies to approximately E = 1.58 x 105 N/C.
Therefore, the electric field strength just outside the surface of the conducting sphere is approximately 1.58 x 105 N/C.
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Two identical loudspeakers 2.0 m apart are emitting 1800 Hz sound waves into a room where the speed of sound is 340 m/s.
Is the point 4.0 m directly in front of one of the speakers, perpendicular to the plane of the speakers, a point of maximum constructive interference, perfect destructive interference, or something in between?
Answer:
a point of destructive interference.
Explanation:
the wavelength of the sound:
λ= v/f
v= velocity of sound =340 m/s
f= frequency of sound wave= 1800 Hz
L_1 = 4 m
then speaker is at the distance of
[tex]L_2 = sqrt(4^2+2^2)[/tex]
= 2√5 m
ΔL = L_2-L_1
x = ΔL/λ
Now, if this result is an integer, the waves will add up at the point. If it is nearly an integer + 0.5, the waves will have a destructive interference at the point. If it is neither of them , then point is "something in between".
[tex]x= \frac{2\sqrt{5}-4 }{\frac{340}{1800} } =2.4995[/tex]
which is close to 2.5, an integer + 0.5. So it's a point of destructive interference.
The result is within an integer value of +0.5, thus its a point of destructive interference.
The given parameters;
distance between the speakers, d = 2.0 mfrequency, f = 1800 Hzspeed of the sound, v = 340 m/sdistance below the speakers, c = 4 mThe resultant distance between the speakers is calculated as follows;
[tex]L = \sqrt{2^2 + 4^2} \\\\L = 4.47 \ m[/tex]
The wavelength of the sound wave is calculated as;
[tex]v = f\lambda\\\\\lambda = \frac{v}{f} \\\\\lambda = \frac{340}{1800} \\\\\lambda = 0.188 \ m[/tex]
Now, determine if the point is constructive interference, perfect destructive interference, or something in between?
[tex]x = \frac{\Delta L}{\lambda} \\\\x = \frac{4.47 - 4}{0.188} \\\\x = 2.5[/tex]
The result is within an integer value of +0.5, thus its a point of destructive interference.
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Driving along a crowded freeway, you notice that it takes a time tt to go from one mile marker to the next. When you increase your speed by 7.4 mi/hmi/h , the time to go one mile decreases by 15 ss . What was your original speed?
Answer:
38.6 mi/h
Explanation:
7.4 mi/h = 7.4mi/h * (1/60)hour/min * (1/60) min/s = 0.00206 mi/s
Let v (mi/s) be your original speed, then the time t it takes to go 1 mi/s is
t = 1/v
Since you increase v by 0.00206 mi/s, your time decreases by 15 s, this means
t - 15 = 1/(v+0.00206)
We can substitute t = 1/v to solve for v
[tex]\frac{1}{v} - 15 = \frac{1}{v + 0.00206}[/tex]
We can multiply both sides of the equation with v(v+0.00206)
v+0.00206 - 15v(v+0.00206) = v
[tex]-15v^2 - 0.0308v + 0.00206 = 0[/tex]
[tex]v= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
[tex]v= \frac{0.03083\pm \sqrt{(-0.03083)^2 - 4*(-15)*(0.00205)}}{2*(-15)}[/tex]
[tex]v= \frac{0.03083\pm0.35}{-30}[/tex]
v = -0.01278 or v = 0.01 0724 mi/s
Since v can only be positive we will pick v = 0.010724 mi/s or 0.010724*3600 = 38.6 mi/h
A hockey pick sliding along a frictional surface strikes a box at rest, after the collision the two objects stick together and move at the same final speed. Which of the following describes the change in momentum and energy of the puck during the collision?
a. puck loses some but not all of its original momentum.
b. one cannot determine
c. puck conserves original momentum, but loses all mechanical energy
d. puck loses some momentum but conserves mechanical energy
e. puck loses conserves all momentum and mechanical energy
f. conserves momentum but loses some mechanical energy
Answer:
Explanation:
Option a is correct
If puck and pick constitute a system then the momentum of the system is conserved but not this may not be valid for the puck .
Option e is correct
If puck and pick is the system then momentum is conserved but because of the presence of friction, mechanical energy is not conserved.
Friction will cause the energy to dissipate in heat.
A 22.0-kg child is riding a playground merrygo-round that is rotating at 40.0 rev/min. What centripetal force is exerted if he is 1.25 m from its center?
Answer:
F=480.491 N
Explanation:
Given that
mass ,m = 22 kg
Angular speed ω = 40 rev/min
[tex]\omega=\dfrac{2\pi \times 40}{60}\ rad/s[/tex]
ω =4.18 rad/s
The radius r= 1.25 m
We know that centripetal force is given as
F=m ω² r
Now by putting the values in the above equation we get
[tex]F=22\times 4.18^2\times 1.25\ N[/tex]
F=480.491 N
Therefore the centripetal force on the child will be 480.491 N.
The rate constant of a reaction is 7.8 × 10−3 s−1 at 25°C, and the activation energy is 33.6 kJ/mol. What is k at 75°C? Enter your answer in scientific notation.
The rate constant at 75°C is calculated using the two-point form of the Arrhenius equation. The original conditions, the new temperature, and the activation energy are substituted into the equation and solved for the new rate constant, k2. The result is k2 = 0.048, or 4.8 x 10^-2 s^-1.
Explanation:For calculating the rate constant at a different temperature, we can use the Arrhenius equation: k = Ae^(-Ea/RT) where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant and T is the temperature in Kelvin.
To find the new temperature constant, we can transform the Arrhenius equation into the two-point form: ln(k2/k1) = (-Ea/R)(1/T2 - 1/T1).
Given:
k1 = 7.8 × 10−3 s−1, T1=25°C = 25 + 273 = 298K
Ea = 33.6 kJ/mol = 33,600 J/mol, R = 8.314 J/(mol·K)
T2 = 75°C = 75 + 273 = 348K
Substituting these values and solving for k2 (rate constant at 75°C), you get k2 = 0.048 or 4.8 x 10^-2 s^-1.
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A ray of light is incident on an air/water interface.
The ray makes an angle of θ1 = 32 degrees with respect to the normal of the surface. The index of the air is n1 = 1 while water is n2 = 1.33.
Choose an expression for the angle (relative to the normal to the surface) for the ray in the water, θ2.
a) θ2 = sin (θ1).n1/n2
b) θ2 = asin (n1/n2)
c) θ2 = asin (sin(θ1).n2/n1)
d) θ2 = asin (sin(θ1).n1/n2)
Answer:
[tex]\theta_2=sin^{-1}(\dfrac{n_1\ sin\theta_1}{n_2})[/tex]
Explanation:
Given that,
The ray makes an angle of 32 degrees with respect to the normal of the surface.
The refractive index of air, [tex]n_1=1[/tex]
The refractive index of water, [tex]n_2=1.33[/tex]
Snell's law is given by :
[tex]n_1\ sin\theta_1=n_2\ sin\theta_2[/tex]
[tex]sin\theta_2=\dfrac{n_1\ sin\theta_1}{n_2}[/tex]
[tex]\theta_2=sin^{-1}(\dfrac{n_1\ sin\theta_1}{n_2})[/tex]
So, option (4) is correct. Hence, this is the required solution.
The answer is option d.
The correct expression for the angle (relative to the normal to the surface) for the ray in the water is d) θ2 = asin (sin(θ1).n1/n2), based on Snell's law of refraction.
Explanation:The question addresses the refraction of light, specifically the change in angle as light moves from air to water. According to Snell's law, which is used to calculate the angle of refraction, the correct expression in your options is d) θ2 = asin (sin(θ1).n1/n2). Here's a step by step process:
First, it's important to understand that light changes direction when it moves from one medium to another, a process known as refraction.Snell's law mathematically expresses this change and is written as n1*sin(θ1) = n2*sin(θ2). In your case, you want to find the angle θ2. So, rearranging Snell's law to solve for θ2 gives you θ2 = asin(n1*sin(θ1)/n2).Learn more about Refraction:https://brainly.com/question/2459833?referrer=searchResults
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What is the net charge of the Earth if the magnitude of its electric field near the terrestrial surface is 1.08 ✕ 102 N/C? Assume the Earth is a sphere of radius 6.40 ✕ 106 m.
To solve this problem we will apply the concepts related to the electric field based on the laws of Coulomb. Said electric field is equivalent to the product between the Coulomb constant and the rate of change of the charge and the squared distance. Mathematically this is,
[tex]E = \frac{kq}{r^2}[/tex]
Here,
k = Coulomb's constant
q = Charge
r = Distance
Replacing we have that
[tex]E = \frac{kq}{r^2}[/tex]
[tex]1.08*10^2 = \frac{(9*10^{9})q}{(6.4*10^{6})^2}[/tex]
Solving for q,
[tex]q = 491520 C[/tex]
Therefore the net charge of the Earth under the previous condition is 491520 C
An object starts from rest and accelerates at a rate of 2 rad/s^2 until it reaches an angular speed of 24 rad/s. The object then accelerates at a rate of -3 rad/s2 until it stops. Through what angular displacement (in rad) does the object move between when it starts moving and when it stops?
Answer:
Total angular displacement will be 240 radian
Explanation:
In first case object starts from rest so initial angular speed [tex]\omega _i=0rad/sec[/tex]
Angular acceleration is given [tex]\alpha =2rad/sec^2[/tex]
Final angular speed[tex]\omega _f=24rad/sec[/tex]
From third equation of motion [tex]\omega _f^2=\omega _i^2+2\alpha \Theta[/tex]
So [tex]24^2=0^2+2\times 2\times \Theta[/tex]
[tex]\Theta =144[/tex] radian
Now in second case as the objects finally stops
So final velocity [tex]\omega _f=0rad/sec[/tex]
Angular acceleration [tex]\alpha =-3rad/sec^2[/tex]
So [tex]0^2=24^2-2\times 3\times \Theta[/tex]
[tex]\Theta =96[/tex] radian
So total angular displacement will be 96+144 = 240 radian
To find the angular displacement, we can analyze the motion of the object in two phases: the first phase of acceleration and the second phase of deceleration. We can calculate the time and angular displacement in each phase using the formulas for angular speed and angular displacement. Finally, we can add the angular displacements from both phases to find the total angular displacement.
Explanation:To find the angular displacement, we need to analyze the motion of the object in two phases: the first phase of acceleration and the second phase of deceleration. In the first phase, the object starts from rest and accelerates at a rate of 2 rad/s^2 until it reaches an angular speed of 24 rad/s. We can use the formula:
Final Angular Speed = Initial Angular Speed + Angular Acceleration * Time
Using this formula, we can find the time taken in the first phase. Then, we can calculate the angular displacement during the first phase using the formula:
Angular Displacement = Initial Angular Speed * Time + 0.5 * Angular Acceleration * Time^2
In the second phase, the object decelerates at a rate of -3 rad/s^2 until it stops. We can use the same formulas to find the time and angular displacement in the second phase. Finally, we can add the angular displacements from both phases to get the total angular displacement.
The total angular displacement will be the sum of the angular displacements in the two phases.
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The driver of a car wishes to pass a truck that is traveling at a constant speed of 20.0 m/s (about 45 mi/h). Initially, the car is also traveling at 20.0 m/s, and its front bumper is 24.0 m behind the truck’s rear bumper. The car accelerates at a constant 0.600 m/s2, then pulls back into the truck’s lane when the rear of the car is 26.0 m ahead of the front of the truck. The car is 4.5 m long, and the truck is 21.0 m long. (a) How much time is required for the car to pass the truck? (b) What distance does the car travel during this time? (c) What is the final speed of the car?
Answer:
a) 15.864s
b) 392.78m
c) 29.52 m/s
Explanation:
The total distance (relative to the truck) that the (front bumper of the) car travels from 24m behind the truck's rear bumper to in front of the car is
distance from car's front bumper to the truck's rear bumper + distance from truck's rear bumper to truck's front bumper (truck's length) + distance from truck's front bumper to car's rear bumper's + distance from the car's rear bumper to the car's front bumper (car's length)
= 24 + 21 + 26 + 4.5 = 75.5 m
As they start at the same speed, we can draw the following equation of motion for the car distance relative to the truck
[tex]s = at^2/2[/tex]
[tex]75.5 = 0.6t^2/2[/tex]
[tex]t^2 = 251.67[/tex]
[tex]t = \sqrt{251.67} = 15.864s[/tex]
b) The actual distance relative to Earth that the car has traveled during this time is the distance car traveled relative to the truck plus distance truck traveled relative to Earth within this time
= 75.5 + 20*15.864 = 392.78 m
c) final speed of the car is the initial speed plus the change in speed
[tex]v = v_0 + \Delta v = v_0 + at = 20 + 15.864*0.6 = 29.52 m/s[/tex]
To pass the truck, it takes the car 33.3 seconds to accelerate and overtake the truck. The car travels a distance of 333 meters during this time. The final speed of the car is 1.80 m/s.
Explanation:u is the initial velocity of the car and a is the acceleration. Since the car is initially traveling at the same speed as the truck (20.0 m/s) and accelerates at a constant rate of 0.600 m/s², the equation becomes: t = (0 - 20) / -0.600. Solving for t gives us t = 33.3 seconds. To find the distance traveled by the car during this time, we can use the equation: s = ut + (1/2)at², where s is the distance, u is the initial velocity, t is the time, and a is the acceleration. Plugging in the values, we get: s = 20(33.3) + (1/2)(-0.600)(33.3)². Solving for s gives us s = 333 meters. To find the final speed of the car, we can use the equation: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Plugging in the values, we get: v = 20 + (-0.600)(33.3). Solving for v gives us v = 1.80 m/s.
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A particle has a velocity of v→(t)=5.0ti^+t2j^−2.0t3k^m/s.
(a) What is the acceleration function?
(b) What is the acceleration vector at t = 2.0 s? Find its magnitude and direction.
Answer:
a)[tex]a=5 i+2t j - 6\ t^2k[/tex]
b)[tex]a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2[/tex]
Explanation:
Given that
v(t) = 5 t i + t² j - 2 t³ k
We know that acceleration a is given as
[tex]a=\dfrac{dv}{dt}[/tex]
[tex]\dfrac{dv}{dt}=5 i+2t j - 6\ t^2k[/tex]
[tex]a=5 i+2t j - 6\ t^2k[/tex]
Therefore the acceleration function a will be
[tex]a=5 i+2t j - 6\ t^2k[/tex]
The acceleration at t = 2 s
a= 5 i + 2 x 2 j - 6 x 2² k m/s²
a=5 i + 4 j -24 k m/s²
The magnitude of the acceleration will be
[tex]a=\sqrt{5^2+4^2+24^2}\ m/s^2[/tex]
a= 24.83 m/s²
The direction of the acceleration a is given as
[tex]a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2[/tex]
a)[tex]a=5 i+2t j - 6\ t^2k[/tex]
b)[tex]a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2[/tex]
At the point of fission, a nucleus of 235U that has 92 protons is divided into two smaller spheres, each of which has 46 protons and a radius of 5.9 × 10−15 m. What is the magnitude of the repulsive force pushing these two spheres apart? The value of the Coulomb constant is 8.98755 × 109 N · m2 /C 2 .
Answer:
Force = 3481.1 N.
Explanation:
Below is an attachment containing the solution.
If the pressure of a substance is increased during a boiling process, will the temperature also increase, or will it remain constant? Why?
Answer:
on increasing pressure, temperature will also increase.
Explanation:
Considering the ideal gas equation as:
[tex]PV=nRT[/tex]
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Thus, at constant volume and number of moles, Pressure of the gas is directly proportional to the temperature of the gas.
P ∝ T
Also,
Also, using Gay-Lussac's law,
[tex]\frac {P_1}{T_1}=\frac {P_2}{T_2}[/tex]
Thus, on increasing pressure, temperature will also increase.
A 0.0575 kg ice cube at −30.0°C is placed in 0.557 kg of 35.0°C water in a very well insulated container, like the kind we used in class. The heat of fusion of water is 3.33 x 105 J/kg, the specific heat of ice is 2090 J/(kg · K), and the specific heat of water is 4190 J/(kg · K). The system comes to equilibrium after all of the ice has melted. What is the final temperature of the system?
Answer:
t= 22.9ºC
Explanation:
Assuming no heat exchange outside the container, before reaching to a condition of thermal equilibrium, defined by a common final temperature, the body at a higher temperature (water at 35ºC) must give heat to the body at a lower temperature (the ice), as follows:
Qw = c*m*Δt = 4190 (J/kg.ºC)*0.557 kg*(35ºC-t) (1)
This heat must be the same gained by the ice, which must traverse three phases before arriving at a final common temperature t:
1) The heat needed to reach in solid state to 0º, as ice:
Qi =ci*m*(0ºC-(-30ºC) = 0.0575kg*2090(J/kg.ºC)*30ºC = 3605.25 J
2) The heat needed to melt all the ice, at 0ºC:
Qf = cfw*m = 3.33*10⁵ J/kg*0.0575 kg = 19147.5 J
3) Finally, the heat gained by the mass of ice (in liquid state) in order to climb from 0º to a final common temperature t:
Qiw = c*m*Δt = 4190 (J/kg.ºC)*0.0575 kg*(t-0ºC)
So, the total heat gained by the ice is as follows:
Qti = Qi + Qf + Qiw
⇒Qti = 3605.25 J + 19147.5 J + 240.9*t = 22753 J + 240.9*t (2)
As (1) and (2) must be equal each other, we have:
22753 J + 240.9*t = 4190 (J/kg.ºC)*0.557 kg*(35ºC-t)
⇒ 22753 J + 240.9*t = 81684 J -2334*t
⇒ 2575*t = 81684 J- 22753 J = 58931 J
⇒ [tex]t= \frac{58931J}{2575 J/C} = 22.9C[/tex]
⇒ t = 22.9º C
The magnitude of the velocity vector of the car is ∣∣v→∣∣ = 78 ft/s. If the vector v→ forms an angle θ = 0.09 rad with the horizontal direction, determine the Cartesian representation of v→ relative to the (iˆ, jˆ) component system.
Answer:
[tex]\vec{v} = (77.68~{\rm ft/s})\^i + (7.01~{\rm ft/s})\^j[/tex]
Explanation:
The x- and y- components of the velocity vector can be written as following:
[tex]\vec{v}_x = ||\vec{v}||\cos(\theta)\^i[/tex]
[tex]\vec{v}_y = ||\vec{v}||\sin(\theta)\^j[/tex]
Since the angle θ and the magnitude of the velocity is given, the vector representation can be written as follows:
[tex]\vec{v} = 78\cos(0.09)\^i + 78\sin(0.09)\^j\\\vec{v} = (77.68~{\rm ft/s})\^i + (7.01~{\rm ft/s})\^j[/tex]
A square plate of copper with 55.0 cm sides has no net charge and is placed in a region of uniform electric field of 82.0 kN/C directed perpendicularly to the plate.a. Find the charge density of each face of the plate.b. Find the total charge on each face.
Answer
given,
Side of copper plate, L = 55 cm
Electric field, E = 82 kN/C
a) Charge density,σ = ?
using expression of charge density
σ = E x ε₀
ε₀ is Permittivity of free space = 8.85 x 10⁻¹² C²/Nm²
now,
σ = 82 x 10³ x 8.85 x 10⁻¹²
σ = 725.7 x 10⁻⁹ C/m²
σ = 725.7 nC/m²
change density on the plates are 725.7 nC/m² and -725.7 nC/m²
b) Total change on each faces
Q = σ A
Q = 725.7 x 10⁻⁹ x 0.55²
Q = 219.52 nC
Hence, charges on the faces of the plate are 219.52 nC and -219.52 nC
If the center atom has three groups of electrons around it, what type of electron geometry is present?
Answer:
Trigonal planar
Explanation:
Trigonal planar - it is referred to the molecular shape of atom in which three bonds exist around any central atom. As there is no lone pair at the center hence all three atoms have taken the form of a triangle. All three atom lies at same plane and known as peripheral atoms
The angle between all the three atoms is 120 degree
A trapezoidal channel with 6.0 ft bed width, 3 ft water depth and 1:1 side slope, carries a discharge of 250 ft3/s. Determine whether the flow is supercritical or subcritical.
Answer
given,
width of trapezoidal channel, b = 6 ft
depth of water, d = 3 ft
discharge,Q = 250 ft³/s
now, we have to calculate Froude number
[tex]F_r = \dfrac{Q}{A\sqrt{gD}}[/tex]
Where D is the hydraulic radius
[tex]D = \dfrac{A}{P}[/tex]
[tex]F_r = \dfrac{Q}{A\sqrt{g\times \dfrac{A}{P}}}[/tex]
P is the width of the channel
P = b + 2 z d
P = 6 + 2 x 1 x 3
P = 13 ft
A = d(b + z d) = 3 (6 + 3) = 27 ft²
g = 32.2 ft/s²
now,
[tex]F_r = \dfrac{Q}{A\sqrt{g\times \dfrac{A}{P}}}[/tex]
[tex]F_r = \dfrac{250}{27\sqrt{32.2\times \dfrac{27}{13}}}[/tex]
F_r = 1.087
F_r > 1
Froude number is greater than 1 so, the flow is Super critical flow.
A driver has a reaction time of 0.50s , and the maximum deceleration of her car is 6.0m/s2 . She is driving at 20m/s when suddenly she sees an obstacle in the road 50m in front of her.
Can she stop the car in time to avoid the collision?
Answer:
given,
reaction time.t_r = 0.50 s
deceleration of the car = 6 m/s²
initial speed,v = 20 m/s
distance at which the car stop = ?
distance travel by the car in reaction time
d= v x t_r
d = 20 x 0.5 = 10 m
using equation of motion
distance travel to stop the car
v² = u² + 2 a s
0² = 20² - 2 x 6 x s
12 s = 400
s = 33.33 m
Total distance travel by the car
D = 10 + 33.33
D = 43.33 m
Hence, the car stops before to avoid collision.
Canada geese migrate essentially along a north–south direction for well over a thousand kilo-meters in some cases, traveling at speeds up to about 100 km/h. If one goose is flying at 100 km/h relative to the air but a 40-km/h wind is blowing from west to east, (a) at what angle relative to the north–south direction should this bird head to travel directly southward relative to the ground? (b) How long will it take the goose to cover a ground distance of 500 km from north to south? (Note: Even on cloudy nights, many birds can navigate by using the earth’s magnetic field to fix the north–south direction.)
Answer:
a) 66.4 relative to the west in the south-west direction
b) 5.455 hours
Explanation:
a)If the wind is blowing east-ward at a speed of 40km/h, then the west component of the geese velocity must be 40km/h in order to counter balance it. Geese should be flying south-west at an angle of
[tex]cos(\alpha) = 40 / 100 = 0.4[/tex]
[tex]\alpha = cos^{-1}(0.4) = 1.16 rad = 180\frac{1.16}{\pi} = 66.4^0[/tex] relative to the West
b) The south-component of the geese velocity is
[tex]100sin(\alpha) = 100sin(66.4^0) = 91.65 km/h[/tex]
The time it would take for the geese to cover 500km at this rate is
t = 500 / 91.65 = 5.455 hours
How much taller (in m) does the Eiffel Tower become at the end of a day when the temperature has increased by 17°C? Its original height is 324 m and you can assume it is made of steel.
Answer:
324.066096 m.
Explanation:
Given that
height of the tower ,h= 324 m
The increase in temperature ,ΔT = 17°C
We know that coefficient of thermal expansion for steel ,α= 12 x 10⁻⁶ C⁻¹
The increase in height is given as
Δ h = α h ΔT
Now by putting the values in the above equation we get
Δ h= 12 x 10⁻⁶ x 324 x 17 m
Δ h=66096 x 10⁻⁶ m
Δ h=0.066096 m
Therefore the height of the tower become 324.066096 m.
Due to thermal expansion, the Eiffel Tower, made of steel, would increase in height by approximately 0.066 meters or 6.6 cm over a day when the temperature increases by 17°C.
Explanation:The height of the Eiffel Tower increases due to the phenomenon of thermal expansion, which is an increase in volume, including height, in response to an increase in temperature. The amount of expansion can be calculated using this formula: ΔL = α * L_original * ΔT. Let's apply the given values:
'α' the coefficient of linear expansion for steel is approximately 0.000012 per degree Celsius. 'L_original' is the original length in meters, which is 324 m. 'ΔT' is the change in temperature, which is 17°C.
So ΔL = 0.000012 * 324 * 17, which equates to approximately 0.066048 m. Therefore, the Eiffel Tower would increase in height by about 0.066 meters (or 6.6 cm) over the course of a day when the temperature increases by 17°C.
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