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Ghella [55]
3 years ago
6

Three observers watch a train pull away from a station toward the right of the platform. Observer A is in one of the train’s car

s; Observer B is on the station platform; and Observer C is on another train traveling in the opposite direction along a parallel track. How would the observer in each frame of reference describe the motion of the train? According to Newton’s laws of motion, how would the motion in the station platform’s frame of reference change if the conductor applied the brakes to the train? Does universal gravitation affect the train? Explain why or why not.
Physics
1 answer:
juin [17]3 years ago
5 0

Observer A is moving inside the train

so here observer A will not be able to see the change in position of train as he is standing in the same reference frame

So here as per observer A the train will remain at rest and its not moving at all

Observer B is standing on the platform so here it is a stationary reference frame which is outside the moving body

So here observer B will see the actual motion of train which is moving in forward direction away from the platform

Observer C is inside other train which is moving in opposite direction on parallel track. So as per observer C the train is coming nearer to him at faster speed then the actual speed because they are moving in opposite direction

So the distance between them will decrease at faster rate

Now as per Newton's II law

F = ma

Now if train apply the brakes the net force on it will be opposite to its motion

So we can say

- F = ma

a = \frac{-F}{m}

so here acceleration negative will show that train will get slower and its distance with respect to us is now increasing with less rate

It is not affected by the gravity  because the gravity will cause the weight of train and this weight is always counterbalanced by normal force on the train

So there is no effect on train motion



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Gala2k [10]

37° Celsius is equal to 98.6° Fahrenheit

6 0
3 years ago
a ultrasonic wave at 8x10^4 Hz is emitted into a vien where the speed of sound in blood is 1570 m/s. the wave reflects off the r
Aneli [31]

Answer: 0.392 m/s

Explanation:

The Doppler shift equation is:

f'=\frac{V+V_{o}}{V-V_{s}} f

Where:

f=8(10)^{4} Hz is the actual frequency of the sound wave

f'=8.002(10)^{4} Hz is the "observed" frequency

V=1570 m/s is the speed of sound

V_{o}=0 m/s is the velocity of the observer, which is stationary

V_{s} is the velocity of the source, which are the red blood cells

Isolating V_{s}:

V_{s}=\frac{V(f'-f)}{f'}

V_{s}=\frac{1570 m/s(8.002(10)^{4} Hz-8(10)^{4} Hz)}{8.002(10)^{4} Hz}

Finally:

V_{s}=0.392 m/s

3 0
3 years ago
You drop a pencil from your desk, which is 1 meter above the floor. How long does it take for the pencil to hit the floor? How f
vova2212 [387]

Answer:

1. 0.45 s.

2. 4.41 m/s

Explanation:

From the question given above, the following data were obtained:

Height (h) = 1 m

Time (t) =?

Velocity (v) =?

1. Determination of the time taken for the pencil to hit the floor.

Height (h) = 1 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

1 = ½ × 9.8 × t²

1 = 4.9 × t²

Divide both side by 4.8

t² = 1/4.9

Take the square root of both side

t = √(1/4.9)

t = 0.45 s.

Thus, it will take 0.45 s for the pencil to hit the floor.

2. Determination of the velocity with which the pencil hit the floor.

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) = 0.45 s.

Final velocity (v) =?

v = u + gt

v = 0 + (9.8 × 0.45)

v = 0 + 4.41

v = 4.41 m/s

Thus, the pencil hit the floor with a velocity of 4.41 m/s

6 0
3 years ago
Estimate the smallest possible period of a satellite in a circular orbit around earth. (mass and radius of earth is 5.98 x 1024
myrzilka [38]
Given: Mass of earth Me = 5.98 x 10²⁴ Kg

           Radius of earth r = 6.37 x 10⁶ m

            G = 6.67 x 10⁻¹¹ N.m²/Kg²

Required: Smallest possible period T = ?

Formula: F = ma;  F = GMeMsat/r²     Centripetal acceleration ac = V²/r

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equate T from all equation.

F = ma

GMeMsat/r² = Msat4π²/rT²    

GMe = 4π²r³/T²

T² = 4π²r³/GMe  

T² = 39.48(6.37 x 10⁶ m)³/6.67 x 10⁻¹¹ N.m²/Kg²)(5.98 x 10²⁴ Kg)

T² = 1.02 x 10²² m³/3.99 x 10¹⁴ m³/s²

T² = 25,563,909.77 s²

T = 5,056.08 seconds    or around   1.4 Hour

3 0
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kifflom [539]

Answer:

U² = 142.86 N

U¹ = 357.14 N

Explanation:

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W(2\ m) - U^2(7\ m) = 0

where,

W = weight of boy = 500 N

U² = reaction ay B = ?

Therefore,

(500\ N)(2\ m)-(U^2)(7\ m)=0\\U^2=\frac{1000\ Nm}{7\ m}\\

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Now, taking summation of forces on the plank. Taking upward direction as positive, for equilibrium position:

W-U^1-U^2=0\\500\ N - 142.86\ N = U^1\\

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3 0
3 years ago
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