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Tatiana [17]
3 years ago
8

The kinetic energy of a ball with a mass of 0.5 kg and a velocity of 10 m/s is

Physics
1 answer:
Bumek [7]3 years ago
7 0

Answer:

Answer is 25 kg m/s.

Explanation:

Given Data ;

mass = 0.5 kg

velocity = 10 m/s

Find ;

K.E = ?

Formula ;

KE = 1/2 mv 2

Solution:

KE = 1/2(0.5)(10)²(kg)(m/s)

     =25 kg m/s

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The engine of a 550kg car is producing a net force of 11,000N. What is its acceleration?
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During which phase of the moon do neap tides occur?
Fynjy0 [20]

Answer:

First Quarter and Third Quarter.

Explanation:

Tides are formed as a consequence of the differentiation of gravity due to the Moon across to the Earth sphere.

Since gravity variates with the distance:

F = G\frac{m1\cdot m2}{r^{2}} (1)

Where m1 and m2 are the masses of the two objects that are interacting and r is the distance between them.

For example, seeing the image below, point A is closer to the Moon than point b, and at the same time the center of mass of the Earth will feel more attracted to the Moon than point B. Therefore, that creates a tidal bulge in point A and point B.

When the Sun and the Moon are alight with respect to the Earth, then the Sun tidal force contributes to the tidal force of the Moon over the Earth. That makes the high tides even higher (spring tides).

               

However, when the Sun is not in the same line than the Moon (the Moon is at 90° with respect to the Sun), then the low tides are higher and the high tides are lower. That scenario is known as neap tides.

           

Therefore, that happens when the Moon is at First Quarter and Third Quarter.

4 0
3 years ago
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What is a device that minimizes the force on a microscopic object during a collision?
liraira [26]
An impact which stops a moving object must do enough work to take away its kinetic energy, so extending the distance moved during the collision reduces the impact force.
3 0
3 years ago
Which object is in static equilibrium
balu736 [363]

Answer:

The body is said to be in static equilibrium if the net force acting on a body at rest is zero.As the net force is zero,the body will not undergo motion.

Explanation:

3 0
2 years ago
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If a 1000-pound capsule weighs only 165 pounds on the moon, how much work is done in propelling this capsule out of the moon's g
Bas_tet [7]

Answer:

  W = 1,307 10⁶ J

Explanation:

Work is the product of force by distance, in this case it is the force of gravitational attraction between the moon (M) and the capsule (m₁)

              F = G m₁ M / r²

              W = ∫ F. dr

              W = G m₁ M ∫ dr / r²

we integrate

             W = G m₁ M (-1 / r)

                 

We evaluate between the limits, lower r = R_ Moon and r = ∞

           W = -G m₁ M (1 /∞ - 1 / R_moon)

            W = G m1 M / r_moon

Body weight is

             W = mg

             m = W / g

The mass is constant, so we can find it with the initial data

For the capsule

            m = 1000/32 = 165 / g_moon

            g_moom = 165 32/1000

            .g_moon = 5.28 ft / s²

I think it is easier to follow the exercise in SI system  

           W_capsule = 1000 pound (1 kg / 2.20 pounds)

           W_capsule = 454 N

           W = m_capsule g

           m_capsule = W / g

           m = 454 /9.8

           m_capsule = 46,327 kg

Let's calculate

          W = 6.67 10⁻¹¹ 46,327   7.36 10²² / 1.74 10⁶

          W = 1,307 10⁶ J

7 0
3 years ago
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