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inn [45]
3 years ago
7

Use Newton's Law of Gravitation to compute the work W required to propel a 800 kg satellite out of the earth's gravitational fie

ld. You may assume that the earth's mass is 5.98✕1024 kg and is concentrated at its center. Take the radius of the earth to be 6.37✕106 m and G = 6.67✕10-11 Nm2/kg2.
Physics
1 answer:
Leni [432]3 years ago
8 0

Answer:

5.0\cdot 10^{10}J

Explanation:

The work W required to make an object escape from a gravitational field is given by

W=m(V_{\infty}-V)

where

m is the mass of the object

V is the gravitational potential at the initial position of the object

V_{\infty}=0 is the potential at infinity

In this problem, we have:

m = 800 kg is the mass of the satellite

The gravitational potential at the Earth's surface is given by

V=-\frac{GM}{R}

where

G = 6.67\cdot 10^{-11} Nm^2/kg^2 is the gravitational constant

M=5.98\cdot 10^{24} kg is the mass of the Earth

R=6.37\cdot 10^6 m is the Earth's radius

Substittuing into the initial equation, we find:

W=-mV=\frac{GMm}{r}=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})(800)}{6.37\cdot 10^6}=5.0\cdot 10^{10}J

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valentinak56 [21]

Answer:

15193.62 m/s

Explanation:

t = Time taken = 6.5 hours

u = Initial velocity = 0 (Assumed)

m = Mass of rocket = 1380 kg

F = Thrust force = 896 N

v = Final velocity

a = Acceleration of the rocket

Force

F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{896}{1380}\\\Rightarrow a=0.6493\ m/s^2

Equation of motion

v=u+at\\\Rightarrow v=0+0.6493\times 6.5\times 60\times 60\\\Rightarrow v=15193.62\ m/s

The velocity of the rocket after 6.5 hours of thrust is 15193.62 m/s

5 0
3 years ago
Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma
Lesechka [4]

Answer:

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

\omega_f = Final angular velocity

\omega_i = Initial angular velocity

\alpha = Angular acceleration

Mass of inertia is given by

I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2

Angular acceleration is given by

\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}

Equation of rotational motion

\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454

The coefficient of friction is 0.54454

At r = 0.25 m

\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N

The force needed to stop the wheel is 104.00902 N

5 0
3 years ago
Give two examples of pure substances and two examples of mixtures​
nikdorinn [45]

Answer:

1 kerosene is the pure substance, salt and water is the mixture substance

7 0
2 years ago
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1 thing you learned about your culture that you did not know
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Answer: That There are diffrent names for my skin color

Explanation:

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2 years ago
A book rest on a table which it's face having a sides 30cm by 25cm . if it exerts apressure of 200pa then determine the mass of
Simora [160]

Answer:

Approximately 1.5\; \rm kg. (Assuming that this table is level, and that the gravitational field strength is g = 9.8\; \rm N \cdot kg^{-1}.

Explanation:

Convert the dimensions of this book to standard units:

\displaystyle 30\; \rm cm = 30\; \rm cm \times \frac{1\; \rm m}{100\; \rm cm}  = 0.30\; \rm m.

\displaystyle 25\; \rm cm = 25\; \rm cm \times \frac{1\; \rm m}{100\; \rm cm}  = 0.25\; \rm m.

Calculate the surface area of this book:

0.30\; \rm m  \times 0.25\; \rm m = 0.075\; \rm m^{2}.

Pressure is the ratio between normal force and the area over which this force is applied.

\displaystyle \text{Pressure} = \frac{\text{normal Force}}{\text{contact Area}}.

Equivalently:

(\text{normal Force}) = \text{Pressure} \cdot (\text{contact Area}).

In this question, \text{Pressure} = 200\; \rm Pa = 200\; \rm N \cdot m^{-2}.

It was found that (\text{contact Area}) = 0.075\; \rm m^{2} (assuming that the entire side of this book is in contact with the table.

Hence:

\begin{aligned}& (\text{normal Force}) \\ &= \text{Pressure} \cdot (\text{contact Area}) \\ &= 200\; \rm N \cdot m^{-2} \times 0.075\; \rm m^{2} \\ &= 15\; \rm N \end{aligned}.

If that the table is level, this normal force would be equal to the weight of this book:

\text{weight} = 15\; \rm N.

Assuming that the gravitational field strength is g = 9.8\; \rm N \cdot kg^{-1}. The mass of this book would be:

\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{15\; \rm N}{9.8\; \rm N \cdot kg^{-1}}\approx 1.5\; \rm kg\end{aligned}.

4 0
2 years ago
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