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labwork [276]
4 years ago
8

The gravitational acceleration is 9.81 m/s2 here on Earth at sea level. What is the gravitational acceleration at a height of 35

0 km above the surface of the Earth, where the International Space Station (ISS) flies? (The mass of the Earth is 5.97×1024 kg, and the radius of the Earth is 6370 km.)
Physics
1 answer:
azamat4 years ago
5 0

To solve this problem it is necessary to apply the definition of severity of Newtonian laws in which it is specified that gravity is defined by

g= \frac{GM}{R^2}

Where

G= Gravitational Constant

M = Mass of Earth

R= Radius from center of the planet

According to the information we need to find the gravity 350km more than the radius of Earth, then

g_{ss} = \frac{GM}{R+h^2}

g_{ss} = \frac{6.67*10^{-11}*5.972*10^{24}}{(6371*10^3+350*10^3)^2}

g_{ss} = 8.82m/s^2

Therefore the gravitational acceleration at 350km is 8.82m/s^2

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Which statements describe nuclear reactions? Check all that apply.
Alexandra [31]

Explanation:

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7 0
3 years ago
Read 2 more answers
A T-shirt cannon can shoot a 0.085 kg T-shirt at nearly 30 m/s. The T-shirt cannon has a mass of 33 kg. If the initial net momen
IgorLugansk [536]

Answer:

Approximately 0.077\; {\rm m\cdot s^{-1}} (assuming that external forces on the cannon are negligible.)

Explanation:

If an object of mass m is moving at a velocity of v, the momentum p of that object would be p = m\, v.

Momentum of the t-shirt:

\begin{aligned} p(\text{t-shirt}) &= m(\text{t-shirt}) \, v(\text{t-shirt}) \\ &= 0.085\; {\rm kg} \times 30\; {\rm m \cdot s^{-1}} \\ &= 2.55 \; {\rm kg \cdot m \cdot s^{-1}} \end{aligned}.

If there is no external force (gravity, friction, etc.) on this cannon, the total momentum of this system should be conserved. In other words, if p(\text{cannon}) denote the momentum of this cannon:

p(\text{t-shirt}) + p(\text{cannon}) = 0.

p(\text{cannon}) = -p(\text{t-shirt}) = -2.55\; {\rm kg \cdot m \cdot s^{-1}}.

Rewrite p = m\, v to obtain v = (p / m). Since the mass of this cannon is m(\text{cannon}) = 33\; {\rm kg}, the velocity of this cannon would be:

\begin{aligned} v(\text{cannon}) &= \frac{p(\text{cannon})}{m(\text{cannon})} \\ &= \frac{-2.55\; {\rm kg \cdot m \cdot s^{-1}}}{33\; {\rm kg}} \\ &\approx 0.077\; {\rm m \cdot s^{-1}}\end{aligned}.

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How is it possible for two objects to have the same momentum, but different velocities?
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Answer: momentum has the same direction as that of velocity but when 2 bodies with the same linear momentum & different velocities it has different masses because a vector quantity is represented by a cross product of mass and velocity of object .

8 0
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