Describe the motion of an object has an acceleration of 0 m/s2

Physics

1 answer:

SIZIF [17.4K]3 years ago

8 0

Answer:

<h2>The object is either traveling at a constant speed or is not moving.</h2><h2></h2>

Explanation:

<h2> It is not possible for the position of an object to be changing and the acceleration to be zero. Acceleration is the change in velocity divided by the change in time.</h2>

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If 3,600 J of work is done in 3.0 s, what is the power?

Firlakuza [10]

Answer:

1,200 watts

Explanation:

1 watt = 1 Joule (J) of work / second

So, 3600 Joules of work / 3 seconds is:

3600 J / 3 seconds = 1,200 watts

8 0

3 years ago

In a perfectly elastic collision between two perfectly rigid objects

ipn [44]

Both the total momentum and the total kinetic energy are conserved

Explanation:

- In a collision between two or more objects, if there are no external forces acting on the system (isolated system), the total momentum of the objects is always conserved. This is called principle of conservation of momentum, and can be written as follows:

$mu+MU = mv+MV$

where

m, M are the masses of the two objects

u, U are the initial velocities of the two objects

v, V are the final velocities of the two objects

- The total kinetic energy, however, is not always conserved. In fact, we have two types of collision:

1) In a perfectly elastic collision, the total kinetic energy of the objects is conserved. This means that we can write the following equation:

$\frac{1}{2}mu^2 + \frac{1}{2}MU^2 = \frac{1}{2}mv^2+\frac{1}{2}MV^2$

2) In an inelastic collision, the total kinetic energy of the object is NOT conserved. This means that part of the total kinetic energy is "lost", converted into other forms of energy (mainly thermal energy, due to the presence of frictional forces within the system). The most extreme case is called perfectly inelastic collision, in which the two objects stick together after the collision, and there is the maximum loss of kinetic energy.

Learn more about collisions:

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7 0

3 years ago

A girl rolls a basketball across a basketball court. The ball slowly decelerates at a rate of -0.20m/s^2. If the initial velocit

Dafna1 [17]

By equation of motion :

$2as=v^2-u^2\\\\s=\dfrac{v^2-u^2}{2a}\\\\s=\dfrac{0^2-2^2}{2\times( -0.2)}\\\\s=10\ m$