Bob has just finished climbing a sheer cliff above a beach, and wants to figure out how high he climbed. All he has to use, however, is a baseball, a stopwatch, and a friend on the beach below with a long measuring tape. Bob is a pitcher and he knows that the fastest he can throw the ball is about ????0=34.1 m/s.v0=34.1 m/s. Bob starts the stopwatch as he throws the ball (with no way to measure the ball's initial trajectory), and watches carefully. The ball rises and then falls, and after ????1=0.510 st1=0.510 s the ball is once again level with Bob. Bob cannot see well enough to time when the ball hits the ground. Bob's friend then measures that the ball landed ????=126 mx=126 m from the base of the cliff. How high up is Bob, if the ball started exactly 2 m above the edge of the cliff?

Answer:

[tex]0.496 kg m^2[/tex]

Explanation:

The torque exerted is given by

[tex]\tau = Fd[/tex]

where

[tex]F=2.00 \cdot 10^3 N[/tex] is the force applied

d = 3.10 cm = 0.031 m is the length of the lever arm

Substituting,

[tex]\tau=(2.00\cdot 10^3 N)(0.031 m)=62 Nm[/tex]

The equivalent of Newton's second law for rotational motion is:

[tex]\tau = I \alpha[/tex]

where

[tex]\tau = 62 Nm[/tex] is the net torque

I is the moment of inertia

[tex]\alpha = 125 rad/s^2[/tex] is the angular acceleration

Solving the equation for I, we find

[tex]I=\frac{\tau}{\alpha}=\frac{62 Nm}{125 rad/s^2}=0.496 kg m^2[/tex]

The photoelectric effect is a phenomenon that consists of the emission of electrons by certain metals when a beam of light impacts on its surface.

For this phenomenon to occur, certain conditions must be met, such as when the photon collides with the electron, in order to "pull it" from the metal, the photon must have a minimum energy equal to the ionization energy of the atom, so that the electron can leave the influence of the nucleus.  

This is achieved with the adequate intensity of the incident radiation, which is related to the number of photons that impact the metal.

This means:

Increasing the intensity of monochromatic light incident on a metal surface causes an increase in the rate of electron ejection due to the larger number of photons arriving on the surface per unit time. However, the kinetic energy of the ejected electrons doesn't change with changing light intensity.

The phenomenon described here is called the photoelectric effect, which occurs when monochromatic light is incident on a metal surface causing the ejection of electrons, also known as photoelectrons.

When the intensity of the light increases, the rate of electron ejection also increases. This is because the increased intensity of light means a larger number of photons are arriving on the surface per unit time, thus more electrons can gain energy from the photons and be ejected.

The kinetic energy of the ejected electrons, however, does not change with changing light intensity. The energy of an ejected electron equals to the energy of a single photon (which depends only on the light's frequency, not its intensity) minus the binding energy of the electron. This means, even though there are more electrons ejected per unit time due to increased intensity, they have the same kinetic energy as before.

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Answer: 0.1 m/s

Explanation:

m=1

s=10

1/10=0.1

The average velocity of the object is 0.1 m/s

From the question,

We are to determine the object's average velocity

Average velocity is defined as the change in position or displacement (∆x) divided by the time intervals (∆t)

That is,

[tex]Average\ velocity = \frac{x_{f}-x_{0} }{t_{f}-t_{0} }[/tex]

Where [tex]x_{f}[/tex] is the final position

[tex]x_{0}[/tex] is the initial position

[tex]t_{f}[/tex] is the final time

and [tex]t_{0}[/tex] is the initial time

From the question,

[tex]x_{f} = 5 \ m[/tex]

[tex]x_{0} = 0\ m[/tex]

[tex]t_{f} = 50 \ s[/tex]

[tex]t_{0} = 0 \ s[/tex]

Putting the parameters into the formula, we get

[tex]Average\ velocity = \frac{5-0}{50-0}[/tex]

[tex]Average\ velocity = \frac{5}{50}[/tex]

[tex]Average\ velocity = 0.1 \ m/s[/tex]

Hence, the average velocity of the object is 0.1 m/s

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Answer:

C) 21 m/s

Explanation:

The general formula of the Doppler effect is:

[tex]f'=(\frac{v+v_r}{v+v_s})f[/tex]

where

f' is the apparent frequency

f is the original frequency

v is the speed of the wave in the medium

[tex]v_r[/tex] is the velocity of the receiver, positive if the receiver is moving towards the source

[tex]v_s[/tex] is the velocity of the source, positive if the source is moving away from the receiver

Here we have

f = 440 Hz

f' = 415 Hz

v = 343 m/s

[tex]v_r = 0[/tex] (the observer is stationary)

[tex]v_s[/tex] is positive since we are considering when the train has passed the observer, so it is moving away from him

So we can rewrite the formula as

[tex]f'=(\frac{v}{v+v_s})f[/tex]

And solving for [tex]v_s[/tex], we find the speed of the train

[tex]v_s = v(\frac{f}{f'}-1)=(343 m/s)(\frac{440 Hz}{415 Hz}-1)=20.7 m/s \sim 21 m/s[/tex]

By applying Doppler's effect of a wave, the speed of this train is equal to 21 m/s.

Given the following data:

Maximum frequency = 440 Hz.

Apparent frequency = 415 Hz.

Speed of sound in air = 343 m/s.

Observer speed = 0 m/s (since his stationary).

Doppler effect can be defined as the change in frequency of a wave with respect to an observer that is in motion and moving relative to the source of the wave.

Mathematically, Doppler's effect of a wave is given by this formula:

[tex]F_o = \frac{V+V_r}{V+V_s}F[/tex]

Substituting the given parameters into the formula, we have;

[tex]415 = \frac{343+0}{343+V_s} \times 440\\\\142345+415V_s=150920\\\\415V_s=150920-142345\\\\415V_s=8575\\\\V_s=\frac{8575}{415} \\\\V_s=20.66[/tex]

Speed = 20.66 ≈ 21 m/s.

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Answer:

5 mg, [tex]5\cdot 10^{-3}g[/tex]

Explanation:

First of all, let's rewrite the mass in grams using scientific notation.

we have:

m = 0.005 g

To rewrite it in scientific notation, we must count by how many digits we have to move the dot on the right - in this case three. So in scientific notation is

[tex]m=5\cdot 10^{-3}g[/tex]

If  we want to convert into milligrams, we must remind that

1 g = 1000 mg

So we can use the proportion

[tex]1 g : 1000 mg = 0.005 g : x[/tex]

and we find

[tex]x=\frac{(1000 mg)(0.005 g)}{1 g}=5 mg[/tex]

To express the mass of the pill in grams, you can write it in scientific notation as 5.0 x 10^-3 g or in milligrams as 5 mg.

To express the mass of the pain-relieving pill in grams, we can either use scientific notation or a metric prefix. The mass of the pill is given as 0.005 g. To express it in scientific notation, we write it as 5.0 x 10-3 g. This means the mass is 5.0 times 10 raised to the power of -3 grams. Alternatively, we can express the mass in milligrams. Since 1 gram is equal to 1000 milligrams, the mass of the pill can be written as 5 mg.

The angle between the vector and the positive x axis is approximately 60°.

The vector points into the third quadrant, which means its x and y components are both negative. Let's assume that the magnitude of the vector is represented by |A| and the magnitude of the x component is represented by |Ax|. The given condition states that |A| = 2 * |Ax|. Using this information, we can find the angle between the vector and the positive x axis.

First, we need to find the values of |A| and |Ax|. Since |A| = 2 * |Ax|, we can substitute this into the Pythagorean theorem:

|A|² = |Ax|² + |Ay|²

Now let's substitute the given information into the equation:

(2 * |Ax|)² = |Ax|² + (|Ay|)²

Expanding and simplifying the equation:

4 * |Ax|² = |Ax|² + (|Ay|)²

3 * |Ax|² = (|Ay|)²

Now, we can take the square root of both sides:

√(3 * |Ax|²) = √((|Ay|)²)

√3 * |Ax| = |Ay|

Since both |Ax| and |Ay| are negative, we can ignore the negative sign. Therefore, |Ax| = -1 and |Ay| = -√3.

Now, we have all the information we need to find the angle between the vector and the positive x axis. We can use the formula: tan(A) = |Ay|/|Ax| = (-√3)/(-1) = √3

The angle A is the inverse tangent of √3: A = atan(√3) ≈ 60°

So, the angle between the vector and the positive x axis is approximately 60°.

Answer:

[tex]1.78\cdot 10^{-4} m[/tex]

Explanation:

The capacitance of the capacitor is given by:

[tex]C=\frac{Q}{V}[/tex]

where

[tex]Q=4.0\mu C=4.0\cdot 10^{-6} C[/tex] is the charge stored on the plates

V = 6.0 V is the potential difference across the capacitor

Substituting, we find

[tex]C=\frac{4.0\cdot 10^{-6} C}{6.0 V}=6.7\cdot 10^{-7} F[/tex]

The capacitance of a parallel-plate capacitor is also given by

[tex]C=k\frac{\epsilon_0 A}{d}[/tex]

where

k = 6.4 is the dielectric constant

[tex]\epsilon_0[/tex] is the vacuum permittivity

[tex]A=2.1 m^2[/tex] is the area of each plate

d is the separation between the plates

Solving for d, we find

[tex]d=\frac{k\epsilon_0 A}{C}=\frac{(6.4)(8.85\cdot 10^{-12} F/m)(2.1 m^2)}{6.7\cdot 10^{-7}F}=1.78\cdot 10^{-4} m[/tex]

(a) 3.5 Hz

The angular frequency in a spring-mass system is given by

[tex]\omega=\sqrt{\frac{k}{m}}[/tex]

where

k is the spring constant

m is the mass

Here in this problem we have

k = 160 N/m

m = 0.340 kg

So the angular frequency is

[tex]\omega=\sqrt{\frac{160 N/m}{0.340 kg}}=21.7 rad/s[/tex]

And the frequency of the motion instead is given by:

[tex]f=\frac{\omega}{2\pi}=\frac{21.7 rad/s}{2\pi}=3.5 Hz[/tex]

(b) 0.021 m

The block is oscillating up and down together with the upper end of the spring. The block will lose contact with the spring when the direction of motion of the spring changes: this occurs when the spring is at maximum displacement, so at

x = A

where A is the amplitude of the motion.

The maximum displacement is given by Hook's law:

[tex]F=kA[/tex]

where

F is the force applied initially to the spring, so it is equal to the weight of the block:

[tex]F=mg=(0.340 kg)(9.81 m/s^2)=3.34 N[/tex]

k = 160 N/m is the spring constant

Solving for A, we find

[tex]A=\frac{F}{k}=\frac{3.34 N}{160 N/m}=0.021 m[/tex]

a. The frequency (in Hz) of the motion is equal to 3.45 Hertz.

b. The amplitude at which the block will lose contact with the spring is 0.021 meters.

Given the following data:

a. To find the frequency (in Hz) of the motion:

Since the spring initiates simple harmonic motion, we would determine its angular frequency.

Mathematically, angular frequency of a spring is given by the formula:

[tex]\omega = \sqrt{\frac{k}{m} }[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]\omega = \sqrt{\frac{160}{0.340} }\\\\\omega = \sqrt{470.59}\\\\\omega = 21.69 \;rad/s[/tex]

Now, we can find the frequency of the motion by using the formula:

[tex]F = \frac{\omega}{2\pi} \\\\F = \frac{21.69}{2 \times 3.142} \\\\F = \frac{21.69}{6.284}[/tex]

Frequency, F = 3.45 Hz

b. To determine the amplitude at which the block will lose contact with the spring:

[tex]Force = kA\\\\mg = kA\\\\A = \frac{mg}{k}[/tex]

Where:

Substituting the parameters into the formula, we have;

[tex]A = \frac{0.340 \times 9.8}{160} \\\\A = \frac{3.332}{160}[/tex]

Amplitude, A = 0.021 meters

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Answer:

(a) There will be no water in the bucket by the time the bucket reaches the top of the well.

(b) The work done in this process will be approximately 1.98 × 10³ J.

Explanation:

How long will it take for the bucket to reach the top of the well?

[tex]\displaystyle t = \frac{s}{v} = \rm \frac{40.8}{0.8} = 51.0\;s[/tex].

How much water will leak out during that [tex]51.0[/tex] seconds?

[tex]\rm 51.0\;s \times 0.12\;kg\cdot s^{-1} = 6.12\;kg > 6\;kg[/tex].

That's more than all the water in the bucket. In other words, all [tex]\rm 6\;kg[/tex] of water in the bucket will have leaked out by the time the bucket reaches the top of the well. There will be no water in the bucket by the time the bucket reaches the top of the well.

How long will it take for all water to leak out of the bucket?

[tex]\displaystyle \rm \frac{6\; kg}{0.12\;kg \cdot s^{-1}} = 50\;s[/tex].

In other words,

Express the mass [tex]m[/tex] of the bucket about time [tex]t[/tex] as a piecewise function:

[tex]\displaystyle m(t) = \left\{\begin{aligned} &6 - 0.12\;t,&&\;0\le t < 50\\& 2,&&\; 50\le t <51\end{aligned}[/tex].

Gravity on the bucket:

[tex]W(t) = m(t)\cdot g[/tex].

However, the bucket is moving at a constant velocity. There's no acceleration. By Newton's Second Law, the net force will be zero. Forces on the bucket are balanced. As a result, the size of the upward force shall be equal to that of gravity.

[tex]\displaystyle F(t) = W(t) = m(t)\cdot g[/tex].

The speed of the bucket [tex]v[/tex] is constant. Thus the power [tex]P[/tex] that pulls the bucket upward at time [tex]t[/tex] will be:

[tex]P(t) = F(t)\cdot v = m(t)\cdot v\cdot g[/tex].

Express work as a definite integral of power with respect to time:

[tex]\displaystyle \begin{aligned}W &= \int_{t_0}^{t_1}{P(t)\cdot dt} = \int_{t_0}^{t_1}{m(t)\cdot g \cdot v\cdot dt}\end{aligned}[/tex].

Both [tex]g[/tex] and [tex]v[/tex] here are constants. Factor them out:

[tex]\displaystyle \begin{aligned}W &= \int_{t_0}^{t_1}{m(t)\cdot g \cdot v\cdot dt} = v\cdot g\cdot \int_{t_0}^{t_1}{m(t)\cdot dt} \end{aligned}[/tex].

Integrate the piecewise function [tex]m(t)[/tex] piece-by-piece:

[tex]\displaystyle \begin{aligned}W &= v\cdot g\cdot \int_{t_0}^{t_1}{m(t)\cdot dt}\\ &=0.8\times 9.8 \left [\int_{0}^{50}(8-0.12\;t)\cdot dt + \int_{50}^{51}2\cdot dt\right]\\ &=0.8\times 9.8 \left [\left(8\;t - \frac{0.12}{2}\;t^{2}\right)\bigg|^{50}_{0} + (2\;t)\bigg|^{51}_{50}\right] \\&=0.8\times 9.8 \left [\left(8\times 50-\frac{0.12}{2}\times 50^{2}\right)+ (51\times 2 - 50\times 2)\right]\\&=\rm 1975.68\;J \end{aligned}[/tex].

Hence the work done pulling the bucket to the top of the well is approximately [tex]\rm 1.98\times 10^{3}\;J[/tex].

Answer:

0.169

Explanation:

There are three forces acting on the crate along the horizontal direction:

- The pushing force of the first worker, F1 = 450 N forward

- The pushing force of the second worker, F2 = 330 N forward

- The frictional force [tex]F_f[/tex] acting backward

The crate slides with constant speed, so its acceleration is zero: a = 0. This means that we can write Newton's second law as

[tex]\sum F = ma = 0\\F_1 + F_2 - F_f = 0[/tex]

The frictional force can be rewritten as

[tex]F_f = \mu mg[/tex]

where

[tex]\mu[/tex] is the coefficient of kinetic friction

m = 470 kg is the mass of the crate

g = 9.8 m/s^2 is the acceleration due to gravity

Substituting everything into the previous equation, we find:

[tex]F_1 + F_2 - \mu mg = 0\\\mu = \frac{F_1 + F_2}{mg}=\frac{450 N+330 N}{(470 kg)(9.8 m/s^2)}=0.169[/tex]

The coefficient of kinetic friction for the crate on this particular floor surface, given that the total force propelling the crate by the workers is balanced by the frictional force, is approximately 0.169.

In this problem, two workers are sliding a 470 kg crate across the floor. One pushes with a force of 450 N, and the other pulls with a force of 330 N, with both forces applied horizontally. Given that the crate slides at a constant speed, we can infer that the total force propelling the crate (450 N + 330 N) is balanced by the frictional force.

Since the crate moves at a constant pace, the force of friction equals the total force exerted by the workers, 780 N (450 N + 330 N). The frictional force is generally given as Ff = μkN, where μk is the coefficient of kinetic friction, and N is the normal force. For a crate on a horizontal surface, the normal force equals the weight of the crate, which is mass (m) times gravity (g), or 470 kg * 9.8 m/s², or approximately 4606 N.

So we can set up the equation as follows: 780 N = μk * 4606 N. Solving for μk gives us μk = 780 N / 4606 N = 0.169. Therefore, the coefficient of kinetic friction for the crate on this particular floor surface is approximately 0.169.

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Answer:

322 m

Explanation:

y = y₀ + v₀ t + ½ gt²

If we take down to be positive, then:

y = 0 + (5.10) (7.60) + ½ (9.8) (7.60)²

y = 322 m

The skyscraper is 322 m tall.

A) [tex]3.6\cdot 10^5 J[/tex]

The power used by an object is defined as

[tex]P=\frac{E}{t}[/tex]

where

E is the energy used

t is the time elapsed

In this problem, we have

P = 100 W is the power of the light bulb

t = 1 h = 3600 s is the time elapsed

Solving for E, we find the amount of energy used by the light bulb:

[tex]E=Pt = (100 W)(3600 s)=3.6\cdot 10^5 J[/tex]

B) [tex]1.1 \cdot 10^2 m/s[/tex]

The kinetic energy of an object is given by

[tex]K=\frac{1}{2}mv^2[/tex]

where

m is the mass

v is the speed of the object

In this problem, we have a man of mass

m = 65 kg

we want its kinetic energy to be

[tex]E=3.6\cdot 10^5 J[/tex]

Therefore, we can calculate its speed from the previous formula:

[tex]v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.6\cdot 10^5 J)}{65 kg}}=105.2 m/s = 1.1 \cdot 10^2 m/s[/tex]

A 100-watt light bulb uses 360,000 joules of energy per hour. A 65 kg person would need to run at approximately 105 m/s to have the same amount of kinetic energy.

Energy = Power × Time = 100 W × 3600 s = 360000 J

Therefore, a 100-watt light bulb uses 360,000 joules of energy per hour.

KE = ½ mv²

Where m is mass in kilograms and v is velocity in meters per second. Plugging in the values, we can solve for v:

360,000 J = ½ × 65 kg × v²

v² = (2 × 360,000 J) / 65 kg = 11076.92 m²/s²

v = √11076.92 m²/s²

v ≈ 105 m/s

Hence, a 65 kg person would need to run at approximately 105 meters per second to have 360,000 joules of kinetic energy, which is equivalent to the energy used by a 100-watt light bulb in one hour.

(a) 7.18

The electric field within a parallel plate capacitor with dielectric is given by:

[tex]E=\frac{\sigma}{k \epsilon_0}[/tex] (1)

where

[tex]\sigma[/tex] is the surface charge density

k is the dielectric constant

[tex]\epsilon_0[/tex] is the vacuum permittivity

The area of the plates in this capacitor is

[tex]A=100 cm^2 = 100\cdot 10^{-4} m^2[/tex]

while the charge is

[tex]Q=8.9\cdot 10^{-7}C[/tex]

So the surface charge density is

[tex]\sigma = \frac{Q}{A}=\frac{8.9\cdot 10^{-7} C}{100\cdot 10^{-4} m^2}=8.9\cdot 10^{-5} C/m^2[/tex]

The electric field is

[tex]E=1.4\cdot 10^6 V/m[/tex]

So we can re-arrange eq.(1) to find k:

[tex]k=\frac{\sigma}{E \epsilon_0}=\frac{8.9\cdot 10^{-5} C/m^2}{(1.4\cdot 10^6 V/m)(8.85\cdot 10^{-12} F/m)}=7.18[/tex]

(b) [tex]7.66\cdot 10^{-7}C[/tex]

The surface charge density induced on each dielectric surface is given by

[tex]\sigma' = \sigma (1-\frac{1}{k})[/tex]

where

[tex]\sigma=8.9\cdot 10^{-5} C/m^2[/tex] is the initial charge density

k = 7.18 is the dielectric constant

Substituting,

[tex]\sigma' = (8.9\cdot 10^{-5} C/m^2) (1-\frac{1}{7.18})=7.66\cdot 10^{5} C/m^2[/tex]

And by multiplying by the area, we find the charge induced on each surface:

[tex]Q' = \sigma' A = (7.66\cdot 10^{-5} C/m^2)(100 \cdot 10^{-4}m^2)=7.66\cdot 10^{-7}C[/tex]

Final answer:

The dielectric constant cannot be determined without the electric field value without the dielectric (E0). The induced charge on each dielectric surface can be calculated once the electric field (E) is given, using the permittivity of free space and the area of the plates.

Explanation:

To solve the student's question regarding the dielectric constant and the induced charge on a dielectric material, we need to apply concepts from electrodynamics and the properties of capacitors.

Dielectric Constant (k)

The dielectric constant (k) is given by the ratio of the electric field without the dielectric (E0) to the electric field with the dielectric (E). Here, you provided us with the electric field within the dielectric material (E) which is 1.4 × 106 V/m, but have not given us the electric field without the dielectric (E0). To find k, we would normally use the following equation:

k = E0 / E

Without knowing E0, we cannot calculate the dielectric constant directly.

Induced Charge (Qi)

We can calculate the induced charge on each surface of the dielectric (Qi) using the relation between electric field (E), surface charge density (σ), and the permittivity of free space (ε0).

E = σ / ε0

From that, the induced surface charge density is:

σi = E × ε0

Where the permittivity of free space (ε0) is approximately 8.85 × 10-12 F/m. After calculating σi, we can find Qi by multiplying σi with the plate area (A), which you've given as 100 cm2 or 0.01 m2.

Explanation:

Photons have momentum, this was proved by he American physicist Arthur H. Compton after his experiments related to the scattering of photons from electrons (Compton Effect or Compton Shift). In addition, energy and momentum are conserved in the process.

In this context, the Compton Shift [tex]\Delta \lambda[/tex] in wavelength when the photons are scattered is given by the following equation:

[tex]\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)[/tex]     (1)

Where:

[tex]\lambda_{c}=2.43(10)^{-12} m[/tex] is a constant whose value is given by [tex]\frac{h}{m_{e}.c}[/tex], being [tex]h=4.136(10)^{-15}eV.s[/tex] the Planck constant, [tex]m_{e}[/tex] the mass of the electron and [tex]c=3(10)^{8}m/s[/tex] the speed of light in vacuum.

[tex]\theta=30\°[/tex] the angle between incident phhoton and the scatered photon.

We are told the scattered X-rays (photons) are detected at [tex]30\°[/tex]:

[tex]\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos(30\°))[/tex]   (2)

[tex]\Delta \lambda=\lambda' - \lambda_{o}=3.2502(10)^{-13}m[/tex]   (3)

Now, the initial energy [tex]E_{o}=400keV=400(10)^{3}eV[/tex] of the photon is given by:

 [tex]E_{o}=\frac{h.c}{\lambda_{o}}[/tex]    (4)

From this equation (4) we can find the value of [tex]\lambda_{o}[/tex]:

[tex]\lambda_{o}=\frac{h.c}{E_{o}}[/tex]    (5)

[tex]\lambda_{o}=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{400(10)^{3}eV}[/tex]    

[tex]\lambda_{o}=3.102(10)^{-12}m[/tex]    (6)

Knowing the value of [tex]\Delta \lambda[/tex] and [tex]\lambda_{o}[/tex], let's find [tex]\lambda'[/tex]:

[tex]\Delta \lambda=\lambda' - \lambda_{o}[/tex]

Then:

[tex]\lambda'=\Delta \lambda+\lambda_{o}[/tex]  (7)

[tex]\lambda'=3.2502(10)^{-13}m+3.102(10)^{-12}m[/tex]  

[tex]\lambda'=3.427(10)^{-12}m[/tex]  (8)

Knowing the wavelength of the scattered photon [tex]\lambda'[/tex]  , we can find its energy [tex]E'[/tex] :

[tex]E'=\frac{h.c}{\lambda'}[/tex]    (9)

[tex]E'=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{3.427(10)^{-12}m}[/tex]    

[tex]E'=362.063keV[/tex]    (10) This is the energy of the scattered photon

So, if we want to know the energy of the recoiling electron [tex]E_{e}[/tex], we have to calculate all the energy lost by the photon, which is:

[tex]E_{e}=E_{o}-E'[/tex]  (11)

[tex]E_{e}=400keV-362.063keV[/tex]  

Finally we obtain the energy of the recoiling electron:

[tex]E_{e}=37.937keV[/tex]  

The energy of the recoiling electron after the Compton scattering is 38 KeV.

The wavelength of the photons after the Compton scattering is calculated as follows;

λ = λ₀ + λc(1 - cosθ)

[tex]\lambda' = \frac{hc}{E_0} \ + \ 2.43 \times 10^{-12}(1 - cos\theta)\\\\\lambda' = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{400,000 \times 1.6 \times 10^{-19}} \ + \ 2.43 \times 10^{-12}(1 - cos30)\\\\\lambda ' = 3.108 \times 10^{-12} \ + \ 3.26 \times 10^{-13}\\\\\lambda ' = 3.434 \times 10^{-12} \ m[/tex]

The energy of photons after the Compton scattering is calculated as follows;

[tex]E ' = \frac{hc}{\lambda '} \\\\E ' = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{3.434 \times 10^{-12} } \\\\E ' = 5.79\times 10^{-14} \ J\\\\E' = \frac{5.79\times 10^{-14} \ J}{1.6\times 10^{-19} J/eV} = 362,000 \ eV = 362 \ KeV[/tex]

The energy of the recoiling electron is the change in the energy of the photons.

E = 400 KeV - 362 KeV

E = 38 KeV

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Answer:

9.60 m/s

Explanation:

The escape speed of an object from the surface of a planet/asteroid is given by:

[tex]v=\sqrt{\frac{2GM}{R}}[/tex]

where

G is the gravitational constant

M is the mass of the planet/asteroid

R is the radius of the planet/asteroid

In this problem we have

[tex]\rho = 2.02\cdot 10^6 g/m^3[/tex] is the density of the asteroid

[tex]V=3.09\cdot 10^{12}m^3[/tex] is the volume

So the mass of the asteroid is

[tex]M=\rho V=(2.02\cdot 10^6 g/m^3)(3.09\cdot 10^{12} m^3)=6.24\cdot 10^{18} g=6.24\cdot 10^{15} kg[/tex]

The asteroid is approximately spherical, so its volume can be written as

[tex]V=\frac{4}{3}\pi R^3[/tex]

where R is the radius. Solving for R,

[tex]R=\sqrt[3]{\frac{3V}{4\pi}}=\sqrt[3]{\frac{3(3.09\cdot 10^{12} m^3)}{4\pi}}=9036 m[/tex]

Substituting M and R inside the formula of the escape speed, we find:

[tex]v=\sqrt{\frac{2(6.67\cdot 10^{-11})(6.24\cdot 10^{15})}{(9036)}}=9.60 m/s[/tex]

Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth to study the earliest stages of star formation, before a star begins to emit visible light, the wavelength is 2.325 x 10¹ um.

The wavelength can be calculated by the information given:

- Frequency f = 25.9 THz = [tex]\(25.9 \times 10^{12}\)[/tex] Hz

- Speed of light c = [tex]\(3.00 \times 10^8\)[/tex] m/s

We'll use the formula [tex]\(\lambda = \frac{c}{f}\)[/tex] to calculate the wavelength ([tex]\(\lambda\)[/tex]).

Substitute the values:

[tex]\[\lambda = \dfrac{3.00 \times 10^8 \, \text{m/s}}{25.9 \times 10^{12} \, \text{Hz}}.\][/tex]

Calculate the value of [tex]\(\lambda\)[/tex]:

[tex]\[\lambda = 1.157 \times 10^{-5} \, \text{m}.\][/tex]

Now, convert the wavelength to micrometers:

[tex]\[\lambda = 1.157 \times 10^{-5} \, \text{m} \times \dfrac{10^6 \, \mu m}{1 \, \text{m}}.\][/tex]

Calculate the value in micrometers:

[tex]\[\lambda \approx 2.325 \, \mu m.\][/tex]

Thus, the wavelength is 2.325 x 10¹ um or 2.325μm.

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The wavelength of the infrared radiation emitted from young stars with a frequency of 25.9 THz can be calculated using the formula relating the speed of light, wavelength, and frequency (c = λv). By rearranging the formula to λ = c/v and substituting the values, we find that the wavelength is approximately 11.6 micrometers.

To calculate the wavelength of the infrared radiation, we must use the formula for the relation between the speed of light, wavelength, and frequency. This formula is c = λv, where c is the speed of light, λ (lambda) is the wavelength, and v is the frequency.

The speed of light is a constant and is approximately 3.00 x 108 m/s. The frequency provided is 25.9 THz or 25.9 x 1012 Hz.

To find the wavelength, we rearrange the formula to λ = c/v. We substitute the given values to get λ = (3.00 x 108 m/s) / (25.9 x 1012 Hz). After performing the calculation, the wavelength comes out to approximately 1.16 x 10-5 meters, or 11.6 micrometers when rounded to three significant figures.

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Answer:

E - B - D - A - C

Explanation:

The magnitude of the emf induced in the loop of wire is given by Faraday-Newmann-Lenz

[tex]\epsilon=\frac{\Delta \Phi_B}{\Delta t}[/tex] (1)

where

[tex]\Delta \Phi_B[/tex] is the variation of magnetic flux

[tex]\Delta t[/tex] is the time interval

Rewriting the flux as product between magnetic field strength (B) and area enclosed by the coil (A):

[tex]\Phi_B = BA[/tex]

and since the area of the coil does not change, the variation of flux can be rewritten as

[tex]\Delta \Phi_B = \Delta B A[/tex]

So (1) becomes

[tex]\epsilon=\frac{\Delta B}{\Delta t}A[/tex]

Which means that the induced emf is proportional to the rate of change of the magnetic field, [tex]\frac{\Delta B}{\Delta t}[/tex]. So we just need to calculate this quantity for each scenario, and rank them from greatest to latest.

We have:

A) [tex]\frac{\Delta B}{\Delta t}=\frac{1 T - 0T}{6 s}=0.167 T/s[/tex]

B) [tex]\frac{\Delta B}{\Delta t}=\frac{4 T - 1T}{2 s}=1.500 T/s[/tex]

C) [tex]\frac{\Delta B}{\Delta t}=\frac{4 T - 4T}{60 s}=0 T/s[/tex]

D) [tex]\frac{\Delta B}{\Delta t}=\frac{3 T - 4T}{4 s}=-0.250 T/s[/tex]

E) [tex]\frac{\Delta B}{\Delta t}=\frac{0 T - 3T}{1 s}=-3.000 T/s[/tex]

So, from greatest to least magnitude, we have:

E - B - D - A - C

The plane of a loop of wire is perpendicular to a magnetic field. Rank, from greatest to least, the magnitudes of the loop's induced emf for each situation will be E - B - D - A-C.

The number of magnetic flux lines on a unit area passing perpendicular to the given line direction is known as induced magnetic field strength .it is denoted by B.

The magnitude of the  induced emf in the loop of wire is;

[tex]\rm \epsilon = \frac{\triangle \phi }{ \triangle t} \\\\[/tex]

The megnetic flux is given by;

[tex]\phi_B= \triangle BA[/tex]

[tex]\rm \epsilon = \frac{ \triangle BA }{ \triangle t} \\\\[/tex]

The above expression shows that the induced emf is proportional to the rate of change of the magnetic field,

[tex]\rm \frac{ \triangle B }{ \triangle t} = \frac{ 1T-0T }{ 6} =0.167\ T/sec[/tex]

[tex]\rm \frac{ \triangle B }{ \triangle t} = \frac{ 4T-1T }{ 2} =1.500\ T/sec[/tex]

[tex]\rm \frac{ \triangle B }{ \triangle t} = \frac{ 4T-4T }{ 60} =0\ T/sec[/tex]

[tex]\rm \frac{ \triangle B }{ \triangle t} = \frac{ 3T-4T }{ 4} =-0.25 \ T/sec[/tex]

[tex]\rm \frac{ \triangle B }{ \triangle t} = \frac{ 0T-3T }{ 1} =-3 \ T/sec[/tex]

From the above, it is observed that Rank, from greatest to least, the magnitudes of the loop's induced emf for each situation will be E - B - D - A-C.

Hence the order will be E - B - D - A-C.

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I believe it should be

Answer:

B

Explanation:

In a DC current, the current supplied is steady motion (a straight horizontal line in a graph) and this is what makes it had to transform. The alternating current takes a sine wave motion in a graph. This means is voltage varies from zero to peak and reverses polarity. The rate at which it achieves this is its frequency in Hertz.

Final answer:

DC current is a unidirectional flow of electric charge, unlike AC current which periodically reverses direction. The true statement about DC current is that there can only be one voltage supplied in a DC circuit (option B).

Explanation:

The question concerns the characteristics of DC current. Direct current, or DC, is the flow of electric charge in only one direction, which is characteristic of systems with a constant voltage source, such as a battery. Unlike alternating current, or AC, which periodically reverses direction, DC's flow is unidirectional. The given statements regarding DC current must be assessed for their correctness. To clarify:

Given this information, we can conclude that the true statement regarding DC current is B. There can only be one voltage supplied in a DC circuit.

If the line on the graph is horizontal, the car is at the same position no matter what time you look at it.  That means that the car is sitting motionless in one place, and not moving.  The car's speed, velocity, and acceleration are all zero, although its resale value may be decreasing.

Answer:

Q: remains the same

C: decreases

ΔV: increases

Explanation:

When the capacitor is disconnected from the battery, and the wire connected to the plates are not touching anything else, it means that the charge cannot flow out from the capacitor: so, the charge stored on the plates of the capacitor, Q, will not change, regardless of the distance between the plates.

The capacitance of a parallel plate is given by

[tex]C=\frac{\epsilon_0 A}{d}[/tex]

where A is the area of the plates and d the separation between the plates. As we see from the formula, C is inversely proportional to d: so, if the plates are pulled apart to a larger separation, it means that d increases, and so C decreases.

Finally, the voltage across the capacitor is given by

[tex]\Delta V=\frac{Q}{C}[/tex]

and since we said that Q does not change while C decreases, it means that [tex]\Delta V[/tex] increases, since [tex]\Delta V[/tex] is inversely proportional to C.

The angular speed of the wheel when pedaling is approximately [tex]\(3.73 \, \text{s}^{-1}\)[/tex].

Given the information about the radii and the rotations, we can analyze the relationships between the angular speeds of the sprockets and the wheel.

The angular speed of the pedal sprocket can be found using the information that it rotates at one revolution every t = 1.1 seconds.

The formula for angular speed omega is:

[tex]\[ \omega = \frac{{2\pi \text{ (revolutions)}}}{{\text{time}}} \][/tex]

For the pedal sprocket:

[tex]\[ \omega_{\text{pedal}} = \frac{{2\pi}}{{1.1 \, \text{s}}} \][/tex]

Now, knowing the relationship between the angular speeds of the wheel sprocket and the pedal sprocket:

[tex]\[\omega_{\text{wheel sprocket}} = \omega_{\text{pedal}}\][/tex]

The linear speed of a point on the wheel (where the chain makes contact) is given by the product of the angular speed and the radius:

[tex]\[ v = \omega \times r \][/tex]

Since the chain rotates without slipping, the linear speed of the chain where it contacts the wheel is equal to the linear speed of the wheel.

For the wheel sprocket:

[tex]\[ \omega_{\text{wheel sprocket}} \times r_{\text{wheel sprocket}} = \omega_{\text{wheel}} \times r_{\text{wheel}} \][/tex]

Given:

- [tex]\( r_{\text{pedal}} = 9.5 \, \text{cm} \)[/tex]

- [tex]\( r_{\text{wheel sprocket}} = 6.5 \, \text{cm} \)[/tex]

- [tex]\( R_{\text{wheel}} = 65 \, \text{cm} \)[/tex]

We have:

[tex]\[ \omega_{\text{pedal}} = \frac{{2\pi}}{{1.1 \, \text{s}}} \][/tex]

[tex]\[ \omega_{\text{wheel sprocket}} = \omega_{\text{pedal}} = \frac{{2\pi}}{{1.1 \, \text{s}}} \][/tex]

[tex]\[ \omega_{\text{wheel sprocket}} \times r_{\text{wheel sprocket}} = \omega_{\text{wheel}} \times R_{\text{wheel}} \][/tex]

Let's solve for the angular speed of the wheel.

[tex]\[ \omega_{\text{wheel sprocket}} \times r_{\text{wheel sprocket}} = \omega_{\text{wheel}} \times R_{\text{wheel}} \][/tex]

Given:

- [tex]\( r_{\text{wheel sprocket}} = 6.5 \, \text{cm} \)[/tex]

- [tex]\( R_{\text{wheel}} = 65 \, \text{cm} \)[/tex]

We found earlier that [tex]\( \omega_{\text{wheel sprocket}} = \frac{2\pi}{1.1 \, \text{s}} \)[/tex]. Substituting the known values into the equation:

[tex]\[ \frac{2\pi}{1.1 \, \text{s}} \times 6.5 \, \text{cm} = \omega_{\text{wheel}} \times 65 \, \text{cm} \][/tex]

Now, solve for [tex]\( \omega_{\text{wheel}} \)[/tex], the angular speed of the wheel:

[tex]\[ \omega_{\text{wheel}} = \frac{\frac{2\pi}{1.1 \, \text{s}} \times 6.5 \, \text{cm}}{65 \, \text{cm}} \][/tex]

[tex]\[ \omega_{\text{wheel}} = \frac{2\pi \times 6.5}{1.1 \times 65} \, \text{s}^{-1} \][/tex]

[tex]\[ \omega_{\text{wheel}} = \frac{13 \pi}{11} \, \text{s}^{-1} \][/tex]

[tex]\[ \omega_{\text{wheel}} \approx 3.73 \, \text{s}^{-1} \][/tex]

Therefore, the angular speed of the wheel when pedaling is approximately [tex]\(3.73 \, \text{s}^{-1}\)[/tex].

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Approximately 8.35 rad/s is the angular speed of the wheel sprocket in radians per second.

We'll first find the angular speed of the pedal sprocket, then use that to find the angular speed of the wheel sprocket and the bicycle wheel.

Given:

- Radius of the pedal sprocket, [tex]\( r_p = 9.5 \, \text{cm} \)[/tex]

- Radius of the wheel sprocket, [tex]\( r_w = 6.5 \, \text{cm} \)[/tex]

- Radius of the bicycle wheel,  R = 65 cm

- Time taken for one revolution of the pedal, t = 1.1 s

1. Calculate the angular speed of the pedal sprocket [tex]\( \omega_p \)[/tex]:

 [tex]\[ \omega_p = \frac{2\pi}{1.1 \, \text{s}} \][/tex]

 [tex]\[ \omega_p \approx 5.71 \, \text{rad/s} \][/tex]

2. Calculate the linear speed of a point on the pedal sprocket:

[tex]\[ v_p = r_p \cdot \omega_p \][/tex]

[tex]\[ v_p = 9.5 \, \text{cm} \cdot 5.71 \, \text{rad/s} \][/tex]

[tex]\[ v_p \approx 54.295 \, \text{cm/s} \][/tex]

3. Find the angular speed of the wheel sprocket and the bicycle wheel:

[tex]\[ \omega_w = \frac{v_w}{r_w} = \frac{v_p}{r_w} \][/tex]

[tex]\[ \omega_w = \frac{54.295 \, \text{cm/s}}{6.5 \, \text{cm}} \][/tex]

[tex]\[ \omega_w \approx 8.35 \, \text{rad/s} \][/tex]

The angular speed of the wheel sprocket and the bicycle wheel is approximately 8.35 rad/s.

Complete Question:

A bicyclist notes that the pedal sprocket has a radius of rp = 9.5 cm while the wheel sprocket has a radius of rw = 6.5 cm. The two sprockets are connected by a chain which rotates without slipping. The bicycle wheel has a radius R = 65 cm. When pedaling the cyclist notes that the pedal rotates at one revolution every t = 1.1 s. When pedaling, the wheel sprocket and the wheel move at the same angular speed.

Calculate the angular speed of the wheel sprocket in radians per second.

Answer:

22m/s

Explanation:

To find the velocity we employ the equation of free fall: v²=u²+2gh

where u is initial velocity, g is acceleration due to gravity h is the height, v is the velocity the moment it hits the ground, taking the direction towards gravity as positive.

Substituting for the values in the question we get:

v²=2×9.8m/s²×25m

v²=490m²/s²

v=22.14m/s which can be approximated to 22m/s

Answer:

The atoms are isotopes of each other

Explanation:

- Two atoms are said to be isotopes of each other if they have same atomic number Z (which corresponds to the number of protons) but different mass number A (which corresponds to the sum of protons and neutrons in the nucleus).

For atom A, we have:

Z = 10 (10 protons)

A = 10+12 = 22 (10 protons + 12 neutrons)

For atom B, we have:

Z = 10 (10 protons)

A = 10+10 = 20 (10 protons + 10 neutrons)

The two atoms have same atomic number Z but different mass number A, so they are isotopes of each other.

Atom A and Atom B are isotopes of the same element because they have an equal number of protons (10 each), but differ in the number of neutrons (12 in Atom A and 10 in Atom B).

The atoms identified as A and B are isotopes of each other. Isotopes are variations of a chemical element, which means they have the same number of protons (and thus belong to the same element) but differ in the number of neutrons. In this case, both atoms A and B share the same number of protons (10) which identifies the element, but differ in the number of neutrons (12 in atom A and 10 in atom B).

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Explanation:

Let's begin by explaining that the Work [tex]W[/tex] done by a Force [tex]F[/tex] refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a distance [tex]y[/tex].  When the applied force is constant and the direction of the force and the direction of the movement are parallel, the equation to calculate it is:  

[tex]W=F.y[/tex] (1)

In the case of this rectangular aquarium (see figure attached), we know its initial volume [tex]V_{i}[/tex] (which is the same volume occupied by water, because the tank is full) is:

[tex]V_{i}=4m.1m.1m=4m^{3}[/tex] (2)

In addition, we know the force needed to pump a small amount of water is:

[tex]F=m_{water}.g[/tex] (3)

On the other hand, we know the density of the water [tex]\rho_{water}[/tex] is given by the following equation:

[tex]\rho_{water}=\frac{m_{water}}{V}[/tex] (4)

Where [tex]m_{water}[/tex] is the mass of water and [tex]V=4m.1m.dy[/tex] is the volume of a thin "sheet" of water.

Finding [tex]m_{water}[/tex]:

[tex]m_{water}=\rho_{water}.V[/tex]  (5)

Substituting (5) in (3):

[tex]F=\rho_{water}.V.g[/tex]   (6)

And substituting (6) in (1):

[tex]W=\rho_{water}.V.g.y[/tex] (7)

Now, we are asked to find the work needed to pump half of the water out of the aquarium. So, if the aquarium is [tex]1m[/tex] deep, the half is [tex]0.5m[/tex]:

[tex]W=(1000\frac{kg}{m^{3}})(4m.1m.dy)(9.8\frac{m}{s^{2}})y[/tex] (8)

[tex]W=39200\frac{kg.m^{6}}{s^{2}}ydy[/tex] (9)

Well, in order to solve this, we have to write the definite integral from y=0 mto y=0.5m:

[tex]W=39200\frac{kg.m^{6}}{s^{2}}\int\limits^{0.5}_0 {y} \, dy[/tex] (10)

Knowing [tex]\int\limits^b_a {f(y)} \, dy= F(b)-F(a)[/tex]:

[tex]W=39200\frac{kg.m^{6}}{s^{2}}(\frac{(0.5m)^{2}}{2}-\frac{(0m)^{2}}{2})[/tex] (11)

[tex]W=4900kg\frac{m^{2}}{s^{2}}[/tex] (12)

[tex]W=4900J[/tex] >>>>This is the work needed to pump half of the water out of the aquarium

Work is the product of force and displacement. The amount of work needed to pump half of the water out of the aquarium is 4900 J.

Work done can be defined as the amount of force needed to displace an object from one location to another.

[tex]W = F \cdot ds[/tex]

Given to us

Volume of the aquarium, V = 4 x 1 x s = 4m³

Height of the aquarium, s = 1 m

Acceleration due to gravity, g = 9.8 m/s²

Density of water, ρ = 1000 kg/m³

We know about work done, it is given as,

[tex]W = F \cdot ds[/tex]

We also know that force can be written as,

[tex]F = m \cdot a[/tex]

Also, the mass can be written as,

[tex]m = \rho \times V[/tex]

Therefore, work can be written as,

[tex]W = F\cdot ds\\\\W = m \cdot a \cdot ds\\\\W = (\rho \times V \times a )ds\\\\W = \int (\rho \times V \times a )ds[/tex]

As we need to pump half of the water, therefore, from below the tank to half the distance

[tex]W = \int_0^{\frac{1}{2}} (\rho \times V \times a )ds\\\\W = \int_0^{\frac{1}{2}} (\rho \times (4 \times 1 \times s) \times a )ds\\\\W = (\rho \times 4 \times a )[s^2]_0^{\frac{1}{2}\\\\[/tex]

Substitute all the values,

[tex]W = 1000\times 4 \times g \times[(\dfrac{0.5}{2}^2) -0^2]\\\\W = 1000\times 4 \times 9.8 \times 0.125\\\\W = 4900\ J[/tex]

Hence, the amount of work needed to pump half of the water out of the aquarium is 4900 J.

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Answer:

A) 0.50 mV

Explanation:

In this problem, we can think the wings of the bird as a metal rod moving across a magnetic field. So, and emf will be induced into the wings of the bird, according to the formula:

[tex]\epsilon = BvL sin \theta[/tex]

where

[tex]B=5\cdot 10^{-5} T[/tex] is the strength of the magnetic field

v = 13 m/s is the speed of the bird

L = 1.2 m is the wingspan of the bird

[tex]\theta=40^{\circ}[/tex] is the angle between the direction of motion and the direction of the magnetic field

Substituting numbers into the formula, we find

[tex]\epsilon = (5.0\cdot 10^{-5} T)(13 m/s)(1.2 m) sin 40^{\circ}=0.00050 V = 0.50 mV[/tex]

Answer:

In an elastic collision, the momentum is conserved and the mechanical energy is conserved too.

Explanation:

There are two types of collisions:

- Elastic collision: in an elastic collision, the total momentum before and after the collision is conserved; also, the total mechanical energy before and after the collision is conserved.

- Inelastic collision: in an inelastic collision, the total momentum before and after the colllision is conserved, while the total mechanical energy is not conserved (in fact, part of the energy is converted into other forms of energy such that thermal energy, due to the presence of frictional forces)

Answer:

B) 4/5

Explanation:

The magnitude of the electric force between the two spheres is given by

[tex]F=k\frac{q_1 q_2}{r^2}[/tex]

where

k is the Coulombs' constant

q1 and q2 are the charges on the two spheres

r is the distance between the two spheres

Initially, we have

[tex]q_1 = 5.0\mu C=5.0\cdot 10^{-6}C\\q_2 = 1.0 \mu C=1.0 \cdot 10^{-6}C\\r = L[/tex]

So the force is

[tex]F_1=k\frac{(5.0\cdot 10^{-6}C)(1.0\cdot 10^{-6}C)}{L^2}=(5.0 \cdot 10^{-12} C^2)\frac{k}{L^2}[/tex]

Later, the two spheres are brought together so they are in contact: this means that the total charge will redistribute equally on the two spheres (because they are identical).

The total charge is

[tex]Q=q_1 + q_2 = +4.0 \mu C=4.0\cdot 10^{-6}C[/tex]

So each sphere will have a charge of

[tex]q=\frac{Q}{2}=2.0\cdot 10^{-6} C[/tex]

So, the new force will be

[tex]F_2=k\frac{(2.0\cdot 10^{-6}C)(120\cdot 10^{-6}C)}{L^2}=(4.0 \cdot 10^{-12} C^2)\frac{k}{L^2}[/tex]

And so the ratio of the two forces is

[tex]\frac{F_2}{F_1}=\frac{4.0\cdot 10^{-12} C}{5.0\cdot 10^{-12} C}=\frac{4}{5}[/tex]

After two identical conducting spheres carrying different charges come into contact and separate, their total charge is distributed evenly. Using Coulomb's Law, the ratio of the forces after and before contact is 4/5. The answer is B) 4/5.

When two identical conducting spheres come into contact, their total charge is redistributed evenly between them. In this case, we have one sphere with a charge of +5.0 μC and another with –1.0 μC, totaling +4.0 μC for both spheres. After contact and redistribution, each sphere will have half of the total charge, which is +2.0 μC per sphere.

To find the ratio of the magnitudes of the electric forces before and after contact, we use Coulomb's Law, which states that the electric force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between their centers:F = k * |q1 * q2| / L2

Where F is the force, k is Coulomb's constant, q1 and q2 are the charges, and L is the separation distance. Before contact, the force is: Fbefore = k * |(+5.0 μC) * (–1.0 μC)| / L2 = k * 5.0 μC / L2

After contact, the force is: Fafter = k * |(+2.0 μC) * (+2.0 μC)| / L2 = k * 4.0 μC / L2

The ratio of Fafter to Fbefore is therefore (4.0/5.0 μC), which simplifies to 4/5. The answer to the question is option B) 4/5.

Explanation:

When a swimmer is underwater and a light pasess from air to water, there is a change in the medium and its index of refraction, hence the light is refracted. This means it changes its direction.

Nevertheless, in this process the refracted ray of light does not change its frequency [tex]f[/tex], because frequency is:

[tex]f=\frac{1}{T}[/tex]

Where [tex]T[/tex] is the period of the wave and this remains unchanged, hence the frequency, as well.

So, as the frequency does not change and the color of light depends on frequency, the color of the light remains the same.

10.8 seconds is the correct answ