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Oxana [17]
3 years ago
11

The International Space Station has a mass of 1.8 × 105 kg. A 70.0-kg astronaut inside the station pushes off one wall of the st

ation so she accelerates at 1.50 m/s2. What is the magnitude of the acceleration of the space station as the astronaut is pushing off the wall? Give your answer relative to an observer who is space walking and therefore does not accelerate with the space station due to the push.
Physics
1 answer:
Aleonysh [2.5K]3 years ago
3 0

Answer:

a = 5.83 \times 10^{-4} m/s^2

Explanation:

Since the system is in international space station

so here we can say that net force on the system is zero here

so Force by the astronaut on the space station = Force due to space station on boy

so here we know that

mass of boy = 70 kg

acceleration of boy = 1.50 m/s^2

now we know that

F = ma

F = 70(1.50) = 105 N

now for the space station will be same as above force

F = ma

105 = 1.8 \times 10^5 (a)

a = \frac{105}{1.8 \times 10^5}

a = 5.83 \times 10^{-4} m/s^2

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A playground merry-go-round of radius 2.00 m has a moment of inertia I = 275 kg · m2 and is rotating about a frictionless vertic
ryzh [129]

Answer:

0.2932 rad/s

Explanation:

r = Radius = 2 m

I_i = Initial angular momentum = 275\ kgm^2

\omega_i = Initial angular velocity = 14 rev/min

I_f = Final angular momentum

\omega_f = Final angular velocity

Here the angular momentum of the system is conserved

I_i\omega_i=I_f\omega_f\\\Rightarrow 275\times 14\times \dfrac{2\pi}{60}=(275+275(2)^2)\omega_f\\\Rightarrow \omega_f=\dfrac{275\times 14\times \dfrac{2\pi}{60}}{275+275(2)^2}\\\Rightarrow \omega_f=0.2932\ rad/s

The final angular velocity is 0.2932 rad/s

7 0
3 years ago
As the mass of an object increases so does its gravitational force. <br> a. True <br> b. False
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Answer:

False

Explanation:

Gravity force is constant.

There some places on Earth that gravity has a variation, but in general, it is the same everywhere.

If you analyze the equation for the weight, which is the action of the gravity to mass, you'll see that W=mg, where m is the mass and g, is gravity.

If you increase the mass, what you are increasing is weight and not gravity.

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As charges move in a closed loop, they
mamaluj [8]

As charges move in a closed loop, they gain as much energy as they lose.

<h3>What is principle of conservation of energy?</h3>
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  • The energy gained by the particles or charges in a closed system is equal to the energy lost  by the charges.

Thus, we can conclude the following based on principles of conservation of energy;

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Determine the velocity that a car should have while traveling around a frictionless curve of radius 100m and that is banked 20 d
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Answer:

v=18.89\frac{m}{s}

Explanation:

From the free-body diagram for the car, we have that the normal force has a vertical component and a horizontal component, and this component act as the centripetal force on the car:

\sum F_x:Nsin(20^\circ)=ma_c(1)\\\sum F_y:Ncos(20^\circ)=mg(2)

Solving N from (2) and replacing in (1):

N=\frac{mg}{cos(20^\circ)}\\(\frac{mg}{cos(20^\circ)})sin(20^\circ)=ma_c\\gtan(20^\circ)=a_c

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