This 200-kg horse ran the track at a speed of 5 m/s. What was the average kinetic energy?

Answer:

880J/kelvin

Explanation:

Q =MC ×change in t

c =C/m

C=Q/change in t

c= Q/ m× change in t

c = 26400 / 2.0 × 15

c = 880 J/kelvin

Answer: Kinetic Energy to Electrical.

Explanation: The magnet is rotated as a result of the spinning wheels, and this results in a powerful stream of electrons, therefore converting kinetic to electrical.

Answer:

here u go

Explanation:

Earth's rotation is the rotation of planet Earth around its own axis. Earth rotates eastward, in prograde motion. As viewed from the north pole star Polaris, Earth turns counterclockwise.

Answer:

The fraction of the object that is below the surface of the water is ¹⁷/₂₀

Explanation:

Given;

specific gravity of the object, γ = 0.850

Specific gravity is given as;

[tex]specific \ gravity = \frac{density \ of the \ object}{density \ of \ water}\\\\0.85= \frac{density \ of the \ object}{1000 \ kg/m^3} \\\\density \ of the \ object = 850 \ kg/m^3[/tex]

Fraction of the object's weight below the surface of water is calculated as;

[tex]= \frac{850}{1000} \ \times\ 100\%\\\\= 85 \% \\\\= \frac{17}{20}[/tex]

Therefore, the fraction of the object that is below the surface of the water is ¹⁷/₂₀

Answer:

The cantaloupe has a speed of 117.6 m/s

Explanation:

Free Fall Motion

It occurs when an object falls under the sole influence of gravity. Any object that is being acted upon solely by the force of gravity is said to be in a state of free fall. Free-falling objects do not face air resistance.

If an object is dropped from rest in a free-falling motion, it falls with a constant acceleration called the acceleration of gravity, which value is [tex]g = 9.8 m/s^2[/tex].

The final velocity of a free-falling object after a time t is given by:

vf=g.t

The cantaloupe has been dropped from rest. We are required to find the speed after t=12 seconds.

Calculate the final speed:

vf=9.8 * 12 = 117.6 m/s

The cantaloupe has a speed of 117.6 m/s

Answer:

The minimum coefficient of static friction required, µ = 0.10

Note. The question is incomplete. The complete question is given below:

While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll sideways, the police claim that the log could have slid out the back of the truck when accelerating from rest. The driver claims that the truck could not possibly accelerate at the level needed to achieve such an effect. Regardless, the police write a ticket anyway and now the driver court date is approaching.

The log has a mass of m = 929 kg; the truck has a mass of M = 8850 kg. According to the truck manufacturer, the truck can accelerate from 0 to 55 mph in 23.0 seconds, but this does not account for the additional mass of the log. Calculate the minimum coefficient of static friction μs needed to keep the log in the back of the truck.

Explanation:

First, velocity in mph is converted to m/s

1 mph = 0.447 m/s

55 mph ≈ 24.6 m/s

The acceleration of the empty truck is a = v/t = 24.6 / 23 = 1.07 m/s²

Force that can be generated by the truck, F = ma

F = 8850kg * 1.07 m/s² = 9469.5 N

However, with the added mass of the log on it, the acceleration of the truck will become;

a = F / m = 9469.5 N / (8550+929)kg = 0.97 m/s²

Frictional force between the log and the truck = 0.97 m/s² * 929 kg = 901.13 N

Normal reaction on the truck due to the weight of the log, R = mg

R = 929 kg * 9.8m/s² = 9104.2 N

Coefficient of static friction, µ = F/R

µ = 901.13/9104.2

µ = 0.098 ≈ 0.10

Therefore, the minimum static friction required is µ = 0.10