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3 years ago

9

An elevator was a full mass of 2400 kg has a power rating on 16.5 horsepower how much time will it take the elevator to lift a f

ull load 12 meters if there is 1000n of frictional force

1 answer:

ehidna [41]3 years ago

6 0

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The kinetic energy of an object depends on what?

liraira [26]

It depends on "Potential Energy", the amount energy it could have, the amount depending on certain circumstances, like height or force. This was how traditional and some modern rollercoasters work. As the "conveyer belt" pulls you up, the higher you go, the more potential energy you have. Once you are falling down the hill, you are experiencing "Kinetic Energy". Hope it makes sence.

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2 years ago

Read 2 more answers

NEED HELP NOW DUE TODAY Which of these statements is most likely correct about Newton's law on gravity? (2 points)

yKpoI14uk [10]

_______________________a

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3 years ago

Certain rifles can fire a bullet with a speed of 950 m/s just as it leaves the muzzle (this speed is called the muzzle velocity)

TiliK225 [7]

Answer:

a) By $v^2 = u^2 + 2as$ => a= 70291.70.

(b)By $v = u + at$ => t= 1.58 ms.

(c)By $v^2 = u^2 - 2gh$ => H = 46045.92 m.

Explanation:

a) By $v^2 = u^2 + 2as$

$(950)^2 = 0 + 2 \times a \times 0.75\\a = 601666.67 m/s^2\\a/g = 688858.70/9.8 = 70291.70$

(b)By $v = u + at$

$950 = 0 + 601666.67 \times t\\t = 1.58 x 10^-3 sec = 1.58 ms$

(c)By $v^2 = u^2 - 2gh$

$0 = (950)^2 - 2 \times9.8 \times H\\H = 46045.92 m$

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2 years ago

Two vehicles approach a right angle intersection and then suddenly collide. After the collision, they become entangled. If their

geniusboy [140]

Answer:

a. 11 m/s at 76° with respect to the original direction of the lighter car.

Explanation:

In this exercise, since both cars make a right angle, let's assume that the lighter car only has a horizontal velocity component (vx) and that the heavier one only has a vertical velocity component (vy). The final velocities for both components for the system can be determined as:

$m_{1} v_{x1}+m_{2}v_{x2}=(m_{1}+m_{2})v_{fx}\\m_{1} v_{y1}+m_{2}v_{y2} =(m_{1}+m_{2})v_{fy}$

Assume that the lighter car has a 1kg mass and that the heavier car has a 4 kg mass.

$1*13 + 4*0 = (1+4) v_{fx}\\v_{fx}=\frac{13}{5} =2.6\\1*0 + 4*13 = (1+4) v_{fy}\\v_{fy}=\frac{13*4}{5} =10.4\\$

The magnitude of the final velocity of the wreck can be found as:

$v_{f}^{2}= v_{fx}^{2}+ v_{fy}^{2}\\v_{f}=\sqrt[]{2.6^{2} + 10.4^{2}} \\v_{f}= 10.72$

The final velocity has an intensity of roughly 11 m/s

As for the angle, it can be determined in respect to the lighter car (x axis) as follows:

$\theta = cos^{-1}(\frac{v_{fx} }{v_{f}} )\\\theta = cos^{-1}(\frac{2.6}{10.7} )\\\theta = 76^{o}$

Therefore, the wreck has a velocity with an intensity of 11 m/s at 76° with respect to the original direction of the lighter car.

5 0

3 years ago

I need help with #10

kherson [118]

Conduction. Because they are connected as they tranfer energy

3 0

3 years ago