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Inga [223]
3 years ago
9

8) There is a bell at the top of a tower that is 87 m high. The bell weighs 78 N. What kind of energy does the bell have? Calcul

ate the energy.
Physics
1 answer:
taurus [48]3 years ago
7 0

Answer:

I think its potential energy

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How much power is needed to lift the 200-N object to a height of 4 m in 4 s?
Vitek1552 [10]

Answer: 2000 watts

Explanation:

Given that,

power = ?

Weight of object = 200-N

height = 4 m

Time = 4 s

Power is the rate of work done per unit time i.e Power is simply obtained by dividing work by time. Its unit is watts.

i.e Power = work / time

(since work = force x distance, and weight is the force acting on the object due to gravity)

Then, Power = (weight x distance) / time

Power = (200N x 4m) / 4s

Power = 8000Nm / 4s

Power = 2000 watts

Thus, 2000 watts of power is needed to lift the object.

3 0
4 years ago
A spring whose spring constant is 270 lbf/in has an initial force of 100 lbf acting on it. Determine the work, in Btu, required
cricket20 [7]

Answer:

0.02585 BTU

Explanation:

Given: Spring constant, k = 270 lbf/in

Initial force, f =100 lbf

Compression, x = 1 in

Work done can be calculated as follows:

W = \int {(f+kx)} \, dx \\W = fx + \frac{1}{2}kx^2\\W= (100 lbf)(1 in)+ \frac{1}{2}(270 lbf/in)(1 in)^2\\W= 100+135 = 235 lbf in\\W=235 \times 0.00011 BTU = 0.02585 BTU

6 0
4 years ago
What does the term Hubble time mean in cosmology, and what is the current best calculation for the Hubble time?
balu736 [363]

Hubble time in cosmology means the estimated age of the universe and the best calculation for it is T=1/H, where H is the Hubble constant
4 0
4 years ago
An airplane travels 2800 km at a speed of 700 km/h, decreases its speed to 500 km/h for the next 1500 km and travels the last 10
Dafna11 [192]

Answer:

he average speed for the airplane is 558 km/hr

Explanati

6 0
3 years ago
A circular loop with radius r is rotating with constant angular velocity ω in a uniform electric field with magnitude E. The axi
inn [45]

Answer:

\Phi_{E} = E\pi r^2 \omega t

Explanation:

The electric flux is defined as the multiple of electric field and the area that the electric field passes through, such that

\Phi_{E} = \vec{E}\vec{A}

When calculating the electric flux, the angle between the directions of electric field and the area becomes important, especially if the angle is changing with time.

The above formula can be rewritten as follows

\Phi_{E} = EA\cos(\theta)

where θ is the angle between the electric field and the area of the loop. Note that, the direction of the area of the loop is perpendicular to the plane of the loop.

If the loop is rotating with constant angular velocity ω, then the angle can be written as follows

\theta = \omega t

At t = 0, cos(0) = 1 and the electric flux through the loop is at its maximum value.

Therefore the electric flux can be written as a function of time

\Phi_{E} = E\pi r^2 \omega t

3 0
4 years ago
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