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3 years ago

12

0.5 kg air hockey puck is initially at rest. What will it’s kinetic energy be after a net force of .8 N acts on it for a distanc

e of 2 m

1 answer:

weqwewe [10]3 years ago

4 0

Answer:

1.6 J

Explanation:

Work = change in energy

W = ΔKE

Fd = KE

(0.8 N) (2 m) = KE

KE = 1.6 J

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3 years ago

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• A cable is suspended over a pulley. A 5.0 kg mass is attached to

Galina-37 [17]

Answer:

Tension= $\frac{20g}{3}$ (g=acceleration of gravity)

Explanation:

Given that,

A 5Kg and 10Kg are attached by a cable suspended over a pulley.

As 10Kg > 5Kg , the 10 kg mass accelerates down and the 5kg mass accelerates up, let it be a. Let the tension in the cable be T.

So, the equations of motion are

$10g-T = 10a$

$T-5g=5a$

Now adding them we get,

$5g=15a$

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3 years ago

if you run off the pavement, you should: turn the steering wheel quickly toward the road steer straight and slow down before att

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While driving on roads of rural areas, if our right wheel moves off the pavement, we should always hold the steering wheel firmly and then take our foot off the pedal, then apply brake lightly until we are moving at a low speed.

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brainly.com/question/28456065

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1 year ago

A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help"

Digiron [165]

Answer: a) The cliff is 532.05m high

b) Her speed just before hitting the ground is 102.12 m/s

Explanation: To solve This, I'll use a sketch diagram, attached to this solution,

In 3seconds, the teacher heard the echo of her initial scream back. We can obtain the distance the teacher had fallen at the end of 3 seconds using the equations of motion,

Y1 = ut + 0.5g(t^2)

Since she's falling under the influence of gravity, her initial velocity, u = 0m/s, g = 9.8m/s2, t = 3s

Y1, distance she fell through in 3 seconds = 0.5×9.8(3^2) = 44.1m

Let the total height of the cliff be (44.1 + x); where is the remaining height of cliff that the teacher will fall through.

Using the equations of motion again, we can obtain distance travelled by the sound waves in 3s. sound waves travel with a constant speed of 340m/s, no acceleration,

Y2 = ut + 0.5g(t^2) where g = 0, u = 340m/s, t = 3seconds

Y2 = 340 × 3 = 1020m

But in 3 secs, the sound waves would have travelled through the total height of the cliff (44.1 + x) and back to the teacher's current height, x. That is, 1020 = 44.1 + x + x

x = 487.95m

So, total height of cliff = 44.1 + 487.95 = 532.05m

b) the speed of the teacher just before she hits the ground.

Using the equations of motion again,

(V^2) = (U^2) + 2gs

Where v is the final velocity to be calculated

U is the initial velocity = 0m/s

g is acceleration due to gravity = 9.8m/s2

S is the total height she fell through, that is, the height of the cliff = 532.05m

(V^2) = 0 + 2×9.8×532.05 = 10428.18

V = √(10428.18) = 102.12m/s

QED!

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3 years ago

Pls help will give brainlist

ale4655 [162]

Potential energy

potential energy is the stored energy an object has, the bike is not moving and it just sitting on the hill, with the stored energy

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2 years ago