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jarptica [38.1K]

2 years ago

11

The International Space Station has a mass of 419,455 kg. When the station

1 answer:

Ludmilka [50]2 years ago

4 0

Answer is B
When you use the formula of gravitational force provided by the earth to the ISS from Newton’s universal law of gravitation, (Fg=GMeM/Rt^2), then add all the given data, you will get your answer, but Rt=average distance from the surface of the earth to the ISS + the radius of the earth, since gravitational force is a central force

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Water is being boiled in an open kettle that has a 0.52-cm-thick circular aluminum bottom with a radius of 12.0 cm. If the water

tangare [24]

Answer:

$T_b=107.3784\ ^{\circ}C$

Explanation:

Given:

thickness of the base of the kettle, $dx=0.52\ cm=5.2\times 10^{-3}\ m$

radius of the base of the kettle, $r=0.12\ m$

temperature of the top surface of the kettle base, $T_t=100^{\circ}C$

rate of heat transfer through the kettle to boil water, $\dot Q=0.409\ kg.min^{-1}$

We have the latent heat vaporization of water, $L=2260\times 10^3\ J.kg^{-1}$

and thermal conductivity of aluminium, $k=240\ W.m^{-1}.K^{-1}$

<u>So, the heat rate:</u>

$\dot Q=\frac{0.409\times 2260000}{60}$

$\dot Q=15405.67\ W$

<u>From the Fourier's law of conduction we have:</u>

$\dot Q=k.A.\frac{dT}{dx}$

$\dot Q=k\times \pi.r^2\times \frac{T_b-T_t}{5.2\times 10^{-3}}$

where:

$A=$ area of the surface through which conduction occurs

$T_b=$ temperature of the bottom surface

$15405.67=240\times \pi\times 0.12^2\times \frac{T_b-100}{5.2\times 10^{-3}}$

$T_b=107.3784\ ^{\circ}C$ is the temperature of the bottom of the base surface of the kettle.

6 0

3 years ago

You are trying to climb a castle wall so, from the ground, you throw a hook with a rope attached to it at 24.1 m/s at an angle o

Serhud [2]

Answer:

The value is $h = 13.2 \ m$

Explanation:

From the question we are told that

The speed of the rope with hook is $u = 24 .1 \ m/s$

The angle is $\theta = 65.0^o$

The speed at which it hits top of the wall is $v = 16.3 m/s$

Generally from kinematic equation we have that

$v_y^2 = u_y ^2 + * 2 (-g)* h$

Here h is the height of the wall so

$[16.3 sin (65)]^2 = [24.1 sin (65)] ^2+ 2 (-9.8)* h$

=> $h = 13.2 \ m$

4 0

2 years ago

What is the answer to my question

Andrew [12]

Well that depends...what is your question?

7 0

2 years ago

Read 2 more answers

In the reaction 2Ca + O2 → 2CaO for every 2 Ca you will need how much O2?

Serjik [45]

Answer:

1

Explanation: that is the ratio

5 0

2 years ago

What speed must a 600 kg car have in order to have the same momentum as a 1200 kg truck traveling at a velocity of 10 m/s to the

yKpoI14uk [10]

Answer:

20 m/s

Explanation:

4 0

2 years ago