PHYSICS COLLEGE

While on a hike, a pair of friends get caught in a thunderstorm. Four seconds after seeing the flash of a distant lightning strike, they hear the thunder. How far away was this lightning strike in miles? Note: sound, in air, travels at 340 m/s.

Answers

Answer 1

Answer:

Answer:

1360 m

Explanation:

Time taken for the thunder to travel the distance to the hikers = 4 seconds

Speed of the thunder = 340 m/s

Speed of light = 3×10⁸ m/s

It can be seen that the speed of light is substantially faster than the speed of sound. This is the reason why there is a delay in seeing the lightning and hearing the thunder.

Distance = Speed × Time

$\text{Distance}=340* 4\n\Rightarrow \text{Distance}=1360\ m$

Hence, the lightning strike was 1360 m away from the hikers

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Calculate the linear momentum per photon,energy per photon, and the energy per mole of photons for radiation of wavelength; (a) 600 nm (red), (b) 550 nm (yellow), (c) 400 nm (violet), (d) 200 nm (ultraviolet), (e) 150 pm (X-ray), and (f ) 1.0 cm (microwave).

Answers

Answer:

The detailed explanations is attached below

Explanation:

What is applied is the De brogile equation and the equation showing a relationship between Energy, speed of light and wavelength.

The explanation is as attached below.

In order to work well, a square antenna must intercept a flux of at least 0.040 N⋅m2/C when it is perpendicular to a uniform electric field of magnitude 5.0 N/C.

Answers

Answer:

L > 0.08944 m or L > 8.9 cm

Explanation:

Given:

- Flux intercepted by antenna Ф = 0.04 N.m^2 / C

- The uniform electric field E = 5.0 N/C

Find:

- What is the minimum side length of the antenna L ?

Solution:

- We can apply Gauss Law on the antenna surface as follows:

Ф = $\int\limits^S {E} \, dA$

- Since electric field is constant we can pull it out of integral. The surface at hand is a square. Hence,

Ф = E.(L)^2

L = sqrt (Ф / E)

L > sqrt (0.04 / 5.0)

L > 0.08944 m

Final answer:

The area of a square antenna needed to intercept a flux of 0.040 N⋅m2/C in a uniform electric field of magnitude 5.0 N/C is 0.008 m². Consequently, each side of the antenna must be about 0.089 meters (or 8.9 cm) long.

Explanation:

The question pertains to the relationship between electric field and flux. The electric flux through an area is defined as the electric field multiplied by the area through which it passes, oriented perpendicularly to the field.

We are given that the electric field (E) is 5.0 N/C and the flux Φ must be 0.040 N⋅m2/C.

Hence, to intercept this amount of flux, the antenna must have an area (A) such that A = Φ / E.

That is, A = 0.040 N⋅m2/C / 5.0 N/C = 0.008 m².

Since the antenna is square, each side will have a length of √(0.008) ≈ 0.089 meters (or 8.9 cm).

Learn more about Electric Flux here:

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The universe is filled with photons left over from the Big Bang that today have an average energy of about 2 × 10−4 eV (corresponding to a temperature of 2.7 K). As derived in lecture, the number of available energy states per unit volume for photons is ????(????)????????

Answers

Answer:

The number of available energy states per unit volume is $4.01*10^(48)$

Explanation:

Given that,

Average energy $E=2*10^(-4)\ eV$

Photon = $4*10^(-5)\ eV$

We need to calculate the number of available energy states per unit volume

Using formula of energy

$g(\epsilon)d\epsilon=(8\pi E^2dE)/((hc)^3)$

Where, E = energy

h = Planck constant

c = speed of light

Put the value into the formula

$g(\epsilon)d\epsilon=(8*\pi*2*10^(-4)*4*10^(-5)*1.6*10^(-19))/((6.67*10^(-34)*3*10^(8))^3)$

$g(\epsilon)d\epsilon=4.01*10^(48)$

Hence, The number of available energy states per unit volume is $4.01*10^(48)$

Where would the normal force exerted on the rover when it rests on the surface of the planet be greater

Answers

Answer:

Normal force exerted on the rover would be greater at a point on the surface of the planet where the weight of the rover is experienced to be greater.

Explanation:

Since weight is a vector quantity, it can vary with position. Weight is the amount of force the planet exerts on the rover centered towards the planet.

Such a force is the result of gravitational pull and is quantified as:

$F=G* (M.m)/(R^2)$

and $M=\rho* (4\pi.r^3)/(3)$

where:

R = distance between the center of mass of the two bodies (here planet & rover)

G = universal gravitational constant

M = mass of the planet

m = mass of the rover

This gravitational pull varies from place to place on the planet because the planet may not be perfectly spherical so the distance R varies from place to place and also the density of the planet may not be uniform hence there is variation in weight.

Weight is basically a force that a mass on the surface of the planet experiences.

According to Newton's third law the there is an equal and opposite reaction force on the body (here rover) which is the normal force.

A model airplane with a mass of 0.741kg is tethered by a wire so that it flies in a circle 30.9 m in radius. The airplane engine provides anet thrust of 0.795 N perpendicular tothe tethering wire.(a) Find the torque the net thrust producesabout the center of the circle.
N·m

(b) Find the angular acceleration of the airplane when it is inlevel flight.
rad/s2

(c) Find the linear acceleration of the airplane tangent to itsflight path.
m/s2

Answers

(a) 24.6 Nm

The torque produced by the net thrust about the center of the circle is given by:

$\tau = F r$

where

F is the magnitude of the thrust

r is the radius of the wire

Here we have

F = 0.795 N

r = 30.9 m

Therefore, the torque produced is

$\tau = (0.795 N)(30.9 m)=24.6 N m$

(b) $0.035 rad/s^2$

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

$\tau = I \alpha$

where

$\tau$ is the torque

I is the moment of inertia

$\alpha$ is the angular acceleration

If we consider the airplane as a point mass with mass m = 0.741 kg, then its moment of inertia is

$I=mr^2 = (0.741 kg)(30.9 m)^2=707.5 kg m^2$

And so we can solve the previous equation to find the angular acceleration:

$\alpha = (\tau)/(I)=(24.6 Nm)/(707.5 kg m^2)=0.035 rad/s^2$

(c) $1.08 m/s^2$

The linear acceleration (tangential acceleration) in a rotational motion is given by

$a=\alpha r$

where in this problem we have

$\alpha = 0.035 rad/s^2$ is the angular acceleration

r = 30.9 m is the radius

Substituting the values, we find

$a=(0.035 rad/s^2)(30.9 m)=1.08 m/s^2$

A 10 kg block moving at 10 m/s in a direction 45 degrees above the horizontal. When it has fallen to a point that is 10 m below the initial point measured vertically (without air friction), the block's kinetic energy is closest to

Answers

The block's kinetic energy is closest to 1500 Joules.

Kinetic energy :

The energy is always conserved.

So that, the total kinetic energy will be sum of initial potential energy and kinetic energy during falling.

Given that, mass(m)=10kg, v=10m/s, h=10m,g=10m/s^2

K.E=(1/2)mv^2 + mgh

K.E=(1/2)*10*100 + (10*10*10)

K.E=500 + 1000=1500Joule

The block's kinetic energy is closest to 1500 Joules.

Learn more about the kinetic energy here:

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Answer:

Kinetic energy = 1500 J

Explanation:

The computation of the block's kinetic energy is shown below:

As we know that

Conservation of energy is

PE_i + KE_i = PE_f + KE_f

where,

Initial Potential energy = PE_i = m gh = 10kg× 10m/s^2 × 10m = 1000 J

Initial Kinetic energy = KE_i = (0.5) m V^2 = (0.5) (10 kg) (10 m/s)^2 = 500 J

Final potential energy = PE_f = mgh = 0

As h = 0 which is at reference line

PE_i + KE_i = PE_f + KE_f

Now put these valeus to the above formulas

1000 J + 500 J = 0 + KE_f

After solving this

Kinetic energy = 1500 J

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The equations for single-slit and multiple-slit interference both contain the variable θ. For the multiple-slit case, the angle is: a. the angular location of the first order minimum in the diffraction pattern. Which means at this point the light experiences constructive interference. b. the angular location of the first order minimum in the diffraction pattern. Which means at this point the light experiences destructive interference. c. the angular location of bright interference maxima in the pattern. Which means at this point the light experiences constructive interference. d. the angular location of bright interference maxima in the pattern. Which means at this point the light experiences destructive interference.

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A long copper cylindrical shell of inner radius 5 cm and outer radius 8 cm surrounds concentrically a charged long aluminum rod of radius 1 cm with a charge density of 7 pC/m. All charges on the aluminum rod reside at its surface. The inner surface of the copper shell has exactly opposite charge to that of the aluminum rod while the outer surface of the copper shell has the same charge as the aluminum rod. Find the magnitude and direction of the electric field at points that are at the following distances from the center of the aluminum rod: (a) 0.5 cm,(b) 1.5 cm,(c) 2.5 cm,(d) 3.5 cm,(e) 7 cm.