All categories

3 years ago

A mass of 2000 kg is raised 5.0 m in 10 seconds. What is the power output in horsepower to raise the object?

Physics

1 answer:

Lapatulllka [165]3 years ago

3 0

Answer:

The power output in horsepower to raise the object is, P = 13.14 Hp

Explanation:

Given data,

The mass of the object, m = 2000 kg

The object is raised to a height, h = 5.0 m

The time duration of the work, t = 10 s

The force acting on the mass,

F = mg

= 2000 x 9.8

19600 N

The work done on the object is,

W = F x h

= 19600 x 5

= 98000 J

The rate of doing work is defined as the power. It is given by the formula,

P = W / t

= 98000 /10

= 9800 Watts

1 Hp = 745.7 Watts

Therefore,

9800 watts = 13.14 Hp

Hence, the power output in horsepower to raise the object is, P = 13.14 Hp

You might be interested in

To qualify to run in the 2005 Boston Marathon, a distance of 26.2 miles, an 18-year-old woman had to have completed another mara

Korvikt [17]

Answer:

7,14545 mph and 3,1936 m/s

Explanation:

The average speed is calculated by dividing the displacement over time, then it is 26,2 miles/(3 2/3 hours), here 3 (2/3) hours is a mixed number, that represents 11/3 hours or 3,66 hours. Then the average speed is 7,14545 mph, now to turn this into meters per second, we notice as mentioned that 1 mile =1609 meters and 1 hour=3600 seconds. Then 7,14545 miles/hour* (1 hour/3600 seconds) * (1609 meters/1 mile)=3,1936 m/s

3 0

3 years ago

What is E(r)E(r)E(r), the radial component of the electric field between the rod and cylindrical shell as a function of the dist

andriy [413]

Answer:

E(r) = λ/2πrε0

Explanation:

If we consider an infinitely long line of charge with the charge per unit length being λ, we can take advantage of the cylindrical symmetry of this situation.

By symmetry, i mean that the electric fields all point radially away from the line of charge and thus there is no component parallel to the line of charge.

Niw, let's use a cylinder (with an arbitrary radius (r) and length (l)) centred on the line of charge as our Gaussian surface.

Doing that will mean that the electric field would be perpendicular to the curved surface of the cylinder. Therefore, the angle between the electric field and area vector is equal to zero and cos θ = cos 0 = 1

Now, the top and bottom surfaces of the cylinder will lie parallel to the electric field. Therefore, the angle between the area vector and the electric field would be 90° and cos θ = cos 90 = 0

Now, we know that according to Gauss Law,

Electric Flux, Φ = E•dA

Thus,

Total Φ = Φ_curved + Φ_top + Φ_bottom

Thus,

Φ = ∫E•dA cos 0 + ∫E•dA cos 90° + ∫E•dA cos 90°

We now have ;

Φ = ∫E . dA × 1

Since we are dealing with the radial component, the curved surface would be equidistant from the line of charge and the electric field in the surface will be the same magnitude throughout.

Thus,

Φ = ∫E•dA = E∫dA = E•2πrl

The net charge enclosed by the surface is given by:

q_net = λl

So using gauss theorem, we have;

Φ = E•2πrl = q_net/εo = λl/εo

E•2πrl = λl/ε0

Making E the subject, we obtain ;

E = λ/2πrε0

4 0

3 years ago

ILI

Nikolay [14]

The original kinetic energy will be 0 J and the final kinetic energy will be 7500 J and the amount of work utilized will be similar to the final kinetic energy i.e., 7500 J.

<u>Explanation:</u>

As it is known that the kinetic energy is defined as the energy exhibited by the moving objects. So the kinetic energy is equal to the product of mass and square of the velocity attained by the car. Thus,

$\text {Kinetic energy}=\frac{1}{2} m v^{2}$

So the initial kinetic energy will be the energy exerted by the car at the initial state when the initial velocity is zero. Thus the initial kinetic energy will be zero.

The final kinetic energy is

$\text {Kinetic energy}=\frac{1}{2} m v^{2}=\frac{1}{2} \times 600 \times 5 \times 5$ = 7500 J