Which of the following is an example of kinetic mechanical energy?

Physics

1 answer:

nata0808 [166]2 years ago

8 0

Answer:

Explanation:

Kinetic energy must be moving. Potential energy has the ability to move but is not doing so at the moment.

A is likely the answer. But there's lots involved in that kind of motion.

B If the ball is elevated, it implies it is not moving yet. It has potential energy.

C Again, the spring is compressed. It will push something when it moves, but it is not moving yet.

D The load gun's bullet is not moving. It's still potential energy.

E. The mouse trap is set, but it is not moving. When the mouse eats the bait then it's potential energy will transform into kinetic energy.

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a stone is thrown by a person from the top of the building, which is 200m tall. at the same time, another stone is thrown with v

solniwko [45]

Answer:

The time after which the two stones meet is tₓ = 4 s

Explanation:

Given data,

The height of the building, h = 200 m

The velocity of the stone thrown from foot of the building, U = 50 m/s

Using the II equation of motion

S = ut + ½ gt²

Let tₓ be the time where the two stones meet and x be the distance covered from the top of the building

The equation for the stone dropped from top of the building becomes

x = 0 + ½ gtₓ²

The equation for the stone thrown from the base becomes

S - x = U tₓ - ½ gtₓ² (∵ the motion of the stone is in opposite direction)

Adding these two equations,

x + (S - x) = U tₓ

S = U tₓ

200 = 50 tₓ

∴ tₓ = 4 s

Hence, the time after which the two stones meet is tₓ = 4 s

6 0

3 years ago

An airplane pilot wishes to fly due west. A wind of 80.0 km/h (about 50 mi/h) is blowing toward the south. (a) If the airspeed o

Vesnalui [34]

Answer:

a) 75.5 degree relative to the North in north-west direction

b) 309.84 km/h

Explanation:

a)If the pilot wants to fly due west while there's wind of 80km/h due south. The north-component of the airplane velocity relative to the air must be equal to the wind speed to the south, 80km/h in order to counter balance it

So the pilot should head to the West-North direction at an angle of

$cos(\alpha) = 80/320 = 0.25$

$\alpha = cos^{-1}(0.25) = 1.32 rad = 180\frac{1.32}{\pi} = 75.5^0$ relative to the North-bound.

b) As the North component of the airplane velocity cancel out the wind south-bound speed. The speed of the plane over the ground would be the West component of the airplane velocity, which is

$320sin(\alpha) = 320sin(75.5^0) = 309.84 km/h$