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3 years ago

11

The mass of a moving object increases, but its speed stays the same. What happens to the kinetic energy of the object as a resul

t? A.It decreases. B. It increases. C. It remains unchanged. D. It fluctuates up and down.

2 answers:

Nuetrik [128]3 years ago

8 0

Answer:

B. It increases.

Explanation:

ozzi3 years ago

5 0

The correct answer is
B It increases.

In fact, the kinetic energy of a moving object is given by
$K= \frac{1}{2}mv^2$
where m is the mass of the object and v is its speed. We see that the kinetic energy is proportional to the mass and proportional to the square of the speed: in this problem, the speed of the object remains the same, while its mass increases, therefore the kinetic energy will increase as well.

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2 years ago

What is the moment of inertia of an object that rolls without slipping down a 3.5-m- high incline starting from rest, and has a

Daniel [21]

Answer:

I = 0.287 MR²

Explanation:

given,

height of the object = 3.5 m

initial velocity = 0 m/s

final velocity = 7.3 m/s

moment of inertia = ?

Using total conservation of mechanical energy

change in potential energy will be equal to change in KE (rotational) and KE(transnational)

PE = KE(transnational) + KE (rotational)

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2$

v = r ω

$mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\dfrac{Iv^2}{r^2}$

$I = \dfrac{m(2gh - v^2)r^2}{v^2}$

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3 0

3 years ago

Consider two objects (Object 1 and Object 2) moving in the same direction on a frictionless surface. Object 1 moves with speed v

Semenov [28]

Answer:

A)Object 1 has the greater magnitude of its momentum.

B)The objects 2 have the greater kinetic energy.

Explanation:

For object 1 :

v₁ = v ,m₁ = 2 m

For object 2 :

$v_2=2\sqrt{v}$ ,m₂=m

We know that linear momentum given as

P = M V

M=Mass , V=Velocity

For object 1 :

P₁ =m₁ v₁

P₁ =2 m v

For object 2

$P_2=m_2v_2$

$P_2=2m\sqrt {v}$

We can say that object 1 have more momentum.

The kinetic energy

$KE_1=\dfrac{1}{2}m_1v_1^2$

$KE_1=\dfrac{1}{2}\times 2m\times v^2$

$KE_1=mv^2$

$KE_2=\dfrac{1}{2}m_2v_2$

$KE_2=\dfrac{1}{2}\times m\times 4v^2$

$KE_2=2mv^2$

Therefore both the object 2 have higher kinetic energy.

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3 years ago