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Effectus [21]
3 years ago
10

A 2-kg ball is moving with a speed of 4 m/s, and a 4-kg ball is moving with a speed of 2 m/s. What can you conclude about the of

kinetic energies of the two balls?
Physics
1 answer:
Mila [183]3 years ago
4 0
<h2>Kinetic energy of mass 4 kg ball is less than kinetic energy of mass 2 kg ball</h2>

Explanation:

Kinetic energy = 0.5 x Mass x Velocity²

For ball of mass 2 kg

           Mass, m = 2 kg

           Velocity, v = 4 m/s

           Kinetic energy = 0.5 x 2 x 4² = 16 J

For ball of mass 4 kg

           Mass, m = 4 kg

           Velocity, v = 2 m/s

           Kinetic energy = 0.5 x 4 x 2² = 8 J

Kinetic energy of mass 4 kg ball is less than kinetic energy of mass 2 kg ball

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A motorcyclist heading east through a small town accelerate at constant 4.0meter per seconds square after he leaves the limits.
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A) The position at t = 2.0 sec is 43.0 m east

B) The position is 55 m east

Explanation:

A)

In order to solve the problem, we take the east direction as positive direction.

We know that:

- at t = 0, the motorcyclist is at a position of x_0 = 5.0 m

- at t = 0, the initial velocity of the motorcyclist is v_0 = 15.0 m east

- The acceleration of the motorcyclist is constant and it is a=4.0 m/s^2

Since the motion is a uniformly accelerated motion, the position of the motorcylist is given by the expression

x(t)=x_0 + v_0t + \frac{1}{2}at^2

where t is the time.

Substituting t = 2.0 s, we find the position:

x(2.0)=(5.0)+(15)(2.0)+\frac{1}{2}(4.0)(2.0)^2=43 m

B)

The velocity of the motoryclist can be found by calculating the derivative of the position. Therefore, it is:

v(t)=x'(t)=v_0 + at

where:

v_0=15.0 m/s is the initial velocity

a=4.0 m/s^2 is the acceleration

We want to find the time t at which the velocity is

v = 25 m/s

Solving the equation for t,

t=\frac{v-v_0}{a}=\frac{25-15}{4}=2.5 s

And therefore, the position at t = 2.5 s is:

x(2.5s)=5.0+(15.0)(2.5)+\frac{1}{2}(4)(2.5)^2=55 m

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

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