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Shkiper50 [21]
4 years ago
13

Mechanical energy that has been "lost" to friction has actually been

Physics
2 answers:
umka21 [38]4 years ago
5 0
The answer to the question is true
Alex Ar [27]4 years ago
5 0

Answer:

it false, APEX, i just did this one

Explanation:

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Obesity does not lead to addiction. Obesity is a weight condition characterized by a body.

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How deep is the event horizon
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The supermassive black holes that the Event Horizon Telescope is observing are far larger; Sagittarius A*, at the center of the Milky Way, is about 4.3 million times the mass of our sun and has a diameter of about 7.9 million miles (12.7 million km), while M87 at the heart of the Virgo A galaxy is about 6 billion solar ..

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A head-on, elastic collision between two particles with equal initial speed v leaves the more massive particle (mass m1) at rest
ZanzabumX [31]
<span>1/3 The key thing to remember about an elastic collision is that it preserves both momentum and kinetic energy. For this problem I will assume the more massive particle has a mass of 1 and that the initial velocities are 1 and -1. The ratio of the masses will be represented by the less massive particle and will have the value "r" The equation for kinetic energy is E = 1/2MV^2. So the energy for the system prior to collision is 0.5r(-1)^2 + 0.5(1)^2 = 0.5r + 0.5 The energy after the collision is 0.5rv^2 Setting the two equations equal to each other 0.5r + 0.5 = 0.5rv^2 r + 1 = rv^2 (r + 1)/r = v^2 sqrt((r + 1)/r) = v The momentum prior to collision is -1r + 1 Momentum after collision is rv Setting the equations equal to each other rv = -1r + 1 rv +1r = 1 r(v+1) = 1 Now we have 2 equations with 2 unknowns. sqrt((r + 1)/r) = v r(v+1) = 1 Substitute the value v in the 2nd equation with sqrt((r+1)/r) and solve for r. r(sqrt((r + 1)/r)+1) = 1 r*sqrt((r + 1)/r) + r = 1 r*sqrt(1+1/r) + r = 1 r*sqrt(1+1/r) = 1 - r r^2*(1+1/r) = 1 - 2r + r^2 r^2 + r = 1 - 2r + r^2 r = 1 - 2r 3r = 1 r = 1/3 So the less massive particle is 1/3 the mass of the more massive particle.</span>
8 0
3 years ago
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On his way off to college, Russell drags his suitcase 19 m from the door of his house to the car at a constant speed with a hori
Mashcka [7]

Answer:

The work done on the suitcase is, W = 1691 J

Explanation:

Given data,

The force on the suitcase is, F  = 89 N

The distance Russell dragged the suitcase, S = 19 m

The work done on the suitcase by Russell is equal to the work done on the suitcase to overcome the friction

The work done on the suitcase by Russell is given by the formula

                          W = F · S

Substituting the given values,

                           W = 89 N x 19 m

                           W = 1691 J

Hence, the work done on the suitcase is, W = 1691 J

8 0
3 years ago
A moving bumper car hits the back bumper of a stationary bumper car. The momentum of the stationary car increases. Which happens
kolbaska11 [484]
It decreases because it gave its momentum to the other car.
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3 years ago
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