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Rasek [7]
3 years ago
7

What happens when the elevator is in free-fall, that is, what is the value for FN and what is the sensation experienced by the p

erson in the elevator? Also, discuss the meaning of FN<0.
Physics
1 answer:
Komok [63]3 years ago
7 0

Answer:

Weightlessness

Explanation:

When the elevator is in free fall, this can only occur when the cord of the elevator breaks.

The acceleration of the elevator will be equal to the acceleration due to gravity. That is:

a = g

Then, the body will experience what we called WEIGHTLESSNESS

Where the normal reaction N of the person will tend to zero. That is

N = 0

The value of FN < 0 because the person inside the lift will experience weightlessness.

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The distance traveled by the particle at the given time interval is 0.28 m.

<h3>Position of the particle at time, t = 0</h3>

The position of the particle at the given time is calculated as follows;

x = 2 sin2(t)

y = 2 cos2(t)

x(0) = 2 sin2(0) = 0

y(0) = 2 cos2(0) = 2(1) = 2

<h3>Position of the particle at time, t = 4</h3>

x = 2 sin2(t)

y = 2 cos2(t)

x(4) = 2 sin2(4) = 0.28

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<h3>Distance traveled by the particle at the given time interval</h3>

d = √[(x₄ - x₀)² + (y₄ - y₀)²]

d =  √[(0.28 - 0)² + (1.98 - 2)²]

d = 0.28 m

Thus, the distance traveled by the particle at the given time interval is 0.28 m.

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Now far is it to the moon?
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You are traveling in a car toward a hill at a speed of 36.4 mph. The car's horn emits sound waves of frequency 231 Hz, which mov
Marina CMI [18]

Answer:

<em>a. The frequency with which the waves strike the hill is 242.61 Hz</em>

<em>b. The frequency of the reflected sound wave is 254.23 Hz</em>

<em>c. The beat frequency produced by the direct and reflected sound is  </em>

<em>    11.62 Hz</em>

Explanation:

Part A

The car is the source of our sound, and the frequency of the sound wave it emits is given as 231 Hz. The speed of sound given can be used to determine the other frequencies, as expressed below;

f_{1} = f[\frac{v_{s} }{v_{s} -v} ] ..............................1

where f_{1} is the frequency of the wave as it strikes the hill;

f is the frequency of the produced by the horn of the car = 231 Hz;

v_{s} is the speed of sound = 340 m/s;

v is the speed of the car = 36.4 mph

Converting the speed of the car from mph to m/s we have ;

hint (1 mile = 1609 m, 1 hr = 3600 secs)

v = 36.4 mph *\frac{1609 m}{1 mile} *\frac{1 hr}{3600 secs}

v = 16.27 m/s

Substituting into equation 1 we have

f_{1} =  231 Hz (\frac{340 m/s}{340 m/s - 16.27 m/s})

f_{1}  = 242.61 Hz.

Therefore, the frequency which the wave strikes the hill is 242.61 Hz.

Part B

At this point, the hill is the stationary point while the driver is the observer moving towards the hill that is stationary. The frequency of the sound waves reflecting the driver can be obtained using equation 2;

f_{2} = f_{1} [\frac{v_{s}+v }{v_{s} } ]

where f_{2} is the frequency of the reflected sound;

f_{1}  is the frequency which the wave strikes the hill = 242.61 Hz;

v_{s} is the speed of sound = 340 m/s;

v is the speed of the car = 16.27 m/s.

Substituting our values into equation 1 we have;

f_{2} = 242.61 Hz [\frac{340 m/s+16.27 m/s }{340 m/s } ]

f_{2}  = 254.23 Hz.

Therefore, the frequency of the reflected sound is 254.23 Hz.

Part C

The beat frequency is the change in frequency between the frequency of the direct sound  and the reflected sound. This can be obtained as follows;

Δf = f_{2} -  f_{1}  

The parameters as specified in Part A and B;

Δf = 254.23 Hz - 242.61 Hz

Δf  = 11.62 Hz

Therefore the beat frequency produced by the direct and reflected sound is 11.62 Hz

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A circular loop of wire of cross-sectional area 0.12 m2 consists of 200 turns, each carrying 0.50 A. It is placed in a magnetic
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Answer:

0.52 Nm

Explanation:

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Angle between the normal of loop and the magnetic field = 90 - 30 = 60 degree

θ = 60°

Torque = N i A B Sinθ

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Torque = 0.52 Nm

4 0
3 years ago
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