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Sergeu [11.5K]
2 years ago
13

Please help me, it's a simple equation

Physics
1 answer:
castortr0y [4]2 years ago
3 0

Because the elevator moves at a constant speed, it's in equilibrium and the net force acting on it is zero. Then the tension in the cable exactly equals the magnitude of the elevator's weight, which is

(3000 kg) (9.80 m/s²) = 29,400 N

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A 2,300-kg truck is traveling down a highway at 32 m/s. What is the kinetic energy of the truck?
kondor19780726 [428]

m = mass of the truck = 23 00 kg

v = speed of the truck down the highway = 32 m/s

K = kinetic energy of the truck = ?

kinetic energy of the truck down the highway is given as

K = (0.5) m v²

inserting the values

K = (0.5) (2300) (32)²

K = (0.5) (2300) (1024)

K = (1150) (1024)

K = 1177600 J

hence the kinetic energy of the truck comes out to be 1177600 J

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3 years ago
Give an example of a reputable website
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A stone dropped in a pond sends out a circular ripple whose radius increases at a constant rate of 4 ft/sec. After 12 seconds, h
Natali5045456 [20]

Answer:

After 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec

Explanation:

Area of a circle = πr²

where;

r is the circle radius

Differentiate the area with respect to time.

\frac{dA}{dt} = 2\pi r\frac{dr}{dt}

dr/dt = 4 ft/sec

after 12 seconds, the radius becomes = \frac{dr}{dt} X 12 = 4 \frac{ft}{sec}  X 12 sec = 48 ft

To obtain how rapidly is the area enclosed by the ripple increasing after 12 seconds, we calculate dA/dt

\frac{dA}{dt} = 2\pi r\frac{dr}{dt}

\frac{dA}{dt} = 2\pi (48)(4)

    dA/dt = 1206.528 ft²/sec

Therefore, after 12 seconds, the area enclosed by the ripple will be increasing rapidly at the rate of 1206.528 ft²/sec

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lorasvet [3.4K]

The answer to the blank is physical.

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3 years ago
A speed-time graph is shown below:
VladimirAG [237]

Answer: 0.5 m/s^{2}

Explanation:

Average acceleration a_{ave} is the variation of velocity \Delta V over a specified period of time \Delta t:

a_{ave}=\frac{\Delta V}{\Delta t}}

Where:

\Delta V=V_{f}-V_{o} being V_{o}=0 cm/s the initial velocity and V_{f}=4 cm/s the final velocity  (according to the information given from the described graph)

\Delta t=8 s

Then:

a_{ave}=\frac{4 cm/s -0 cm/s}{8 s}}

a_{ave}=0.5 m/s^{2}

5 0
3 years ago
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