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AlexFokin [52]
3 years ago
7

Conservation equations are powerful equations, because of how versatile and flexible they are to different physical situations.

Conservation of mass is a simple one that can even be applied to fluids with care. The conservation of momentum also applies to any closed system, such as collisions. We've examined the nuances of energy conservation in our labs with total mechanical energy. Does kinetic energy remain conserved in all collisions too?
a) true
b) false
Physics
1 answer:
Temka [501]3 years ago
8 0

Answer:

b) false

Explanation:

Since in the given situation it is mentioned that the mass conversation would be simple and same applied to the fluids with care also the conservation of momentum would be applied to any type of closed system like collisions

But as we know that

In the inelastic collisions, the total kinetic energy would not be remain conserved

So the given statement is false

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Your shadow follows you everywhere. How is it formed? Is it always the same size or shape? Do you always have one shadow?
Yanka [14]
Formed from an object blocking the light to a certain area.

Is not always the same size or shape, it varies depending on the position of the light.

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8 0
2 years ago
In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 1.40×1016 kg and a ra
Arturiano [62]

A) 8.11 m/s

For a satellite orbiting around an asteroid, the centripetal force is provided by the gravitational attraction between the satellite and the asteroid:

m\frac{v^2}{(R+h)}=\frac{GMm}{(R+h)^2}

where

m is the satellite's mass

v is the speed

R is the radius of the asteroide

h is the altitude of the satellite

G is the gravitational constant

M is the mass of the asteroid

Solving the equation for v, we find

v=\sqrt{\frac{GM}{R+h}}

where:

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}

M=1.40\cdot 10^{16}kg

R=8.20 km=8200 m

h=6.00 km = 6000 m

Substituting into the formula,

v=\sqrt{\frac{(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=8.11 m/s

B) 11.47 m/s

The escape speed of an object from the surface of a planet/asteroid is given by

v=\sqrt{\frac{2GM}{R+h}}

where:

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}

M=1.40\cdot 10^{16}kg

R=8.20 km=8200 m

h=6.00 km = 6000 m

Substituting into the formula, we find:

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=11.47 m/s

5 0
3 years ago
An object needs a force of 152 Newtrons to move 8 meters. How much work is required?
GuDViN [60]

Answer:

960J

Explanation:

Given parameters:

Force  = 120N

Distance = 8m

Unknown:

Work required  = ?

Solution:

The work done by a body is the force applied to move a body in a specific direction.

 Work done  = Force x distance

Insert the parameters and solve;

  Work done  = 120  x  8  = 960J

6 0
2 years ago
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