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LenaWriter [7]
4 years ago
10

Two helium-filled balloons are released simultaneously at points A and B on the x axis in an earth-based reference frame. Balloo

n A is to the left of balloon B. Which one of the following statements is true for an observer moving in the +x direction?
A. The observer sees balloon B released before balloon A.
B. The observer always sees the balloons released simultaneously.
C. The observer could see either balloon released first depending on her speed and the distance between A and B.
D. The observer sees balloon A released before balloon B.
Physics
1 answer:
Vedmedyk [2.9K]4 years ago
5 0

Answer:

The correct answer is C

Explanation:

In this exercise you are asked to analyze the following situation: for a fixed observer, the two balloons are launched at the same time.

If the observer is mobile moving towards the positive side of axis ax we have several

possibilities when starting the movement

* the observer is to the left of the two balloons

* the observer is between the two balloons

* the observer is to the right of the two baloomls

The time it takes for the signal to arrive to know which ball goes first is

             v = d / t

             t = d / v

If the signal goes at the speed of light, the speed is a constant and the time will depend only on the distance, so we see that the trigger changes depending on the relative position between a given ball and the observer.

Consequently, it will be seen which comes out first, depending on the relative position with the observer.

           

The correct answer is C

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Answer:

<em>1.228 x </em>10^{-6}<em> mm </em>

<em></em>

Explanation:

diameter of aluminium bar D = 40 mm  

diameter of hole d = 30 mm

compressive Load F = 180 kN = 180 x 10^{3} N

modulus of elasticity E = 85 GN/m^2  = 85 x 10^{9} Pa

length of bar L = 600 mm

length of hole = 100 mm

true length of bar = 600 - 100 = 500 mm

area of the bar A = \frac{\pi D^{2} }{4} =  \frac{3.142* 40^{2} }{4} = 1256.8 mm^2

area of hole a = \frac{\pi(D^{2} - d^{2}) }{4} = \frac{3.142*(40^{2} - 30^{2})}{4} = 549.85 mm^2

Total contraction of the bar = \frac{F*L}{AE} + \frac{Fl}{aE}

total contraction = \frac{F}{E} * (\frac{L}{A} +\frac{l}{a})

==> \frac{180*10^{3}}{85*10^{9}} *( \frac{500}{1256.8} + \frac{100}{549.85}) = <em>1.228 x </em>10^{-6}<em> mm </em>

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A nuclear fission reactor produces electricity. What is the role of the fuel, the control rods, water, and generator in this pro
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v = 15.45 m/s

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As per mechanical energy conservation we can say that here since friction is present in the barrel so we will have

Work done by friction force = Loss in mechanical energy

so we will have

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here we know that

W_f = F_f . d

W_f = 40 \times 4

W_f = 160 J

Initial compression in the spring is given as

F = kx

4400 = 1100 x

x = 4 m

now from above equation

W_f = (\frac{1}{2}kx^2 + 0) - (mgh + \frac{1}{2}mv^2)

160 = (\frac{1}{2}1100(4^2) + 0) - (60 \times 9.8\times 2.50 + \frac{1}{2}(60)v^2)

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3 years ago
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Answer:

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For small disk, I_small = ½MR²

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I_A = 4MR²

We are also told that System B consists of one of the larger disks and a number of the smaller disks. Thus;

I_B = I_large + n(I_small)

Where n is the number of smaller disks.

I_B = 2MR² + n(½MR²)

I_B = MR²(2 + n/2)

We are told that the moment of inertia for system A equals the moment of inertia for system B. Thus;

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