PLS HELP!!!- The moon rotates around the Earth every 28 days. What is the frequency of the moon’s revolution?

A hi-lo lifts an 25 N skid to top of a pallet rack. The pallet rack is 3.6 meters tall. The hi-lo takes 12 seconds to get the sk

mojhsa [17]

Answer:

7.5Watts

Explanation:

Given parameters:

Force of lift = 25N

Height = 3.6m

Time = 12s

Unknown:

Power output = ?

Solution:

Power is the rate at which work is done ;

Power = $\frac{force x height }{time}$

Power = $\frac{25 x 3.6}{12}$ = 7.5Watts

8 0

2 years ago

A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the

alex41 [277]

Answer:

(a) $v_1 = a_1t_1 = 1.76 t_1$

(b) It won't hit

(c) 110 m

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

$v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1$

(b) The velocity of the car before the driver begins braking is

$v_1 = 1.76*20 = 35.2m/s$

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

$a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2$

We can use the following equation of motion to calculate how far the car has travel since braking to stop

$s_2 = v_1t_2 + a_2t_2^2/2$

$s_2 = 35.2*5 - 7.04*5^2/2 = 88 m$

Also the distance from start to where the driver starts braking is

$s_1 = a_1t_1^2/2 = 1.76*20^2/2 = 352$

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

(c) The distance from the limb to where the car stops is 550 - 440 = 110 m

8 0

2 years ago

bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving +1.72 m/s what is the veloci

spin [16.1K]

Answer: V1 = 3.559 - 0.744V2

Explanation: Given that the

bumper car A has

Mass M1 = (281 kg) moving with

Velocity U1 = 2.82 m/s

bumper car B has

Mass M2 = (209 kg) moving with

Velocity U2 = 1.72 m/s

Where U1, U2 are the initial velocity of the two cars

Since the collision is elastic, we will use the formula below,

M1U1 + M2U2 = M1V1 + M2V2

Substitute the values into the formula

281×2.28 + 209×1.72 = 281V1 + 209V2

640.68 + 359.48 = 281V1 + 209V2

1000.16 = 281V1 + 209V2

Make V1 the subject of formula

281V1 = 1000.16 - 209V2

V1 = 1000.16/281 - 209V2/281

V1 = 3.559 - 0.744V2

Therefore, the velocity of car A which

is V1 after the collision will be expressed as V1 = 3.559 - 0.744V2

6 0

3 years ago

Uniform circular motion can best be described as A) movement with right-angle turns. B) the motion of an object orbiting at vary

murzikaleks [220]

D
The motion of an object in a circle at a constant speed.

5 0

3 years ago

A cement block accidentally falls from rest from the ledge of a 53.0-m-high building. When the block is 14.0 m above the ground,

katrin [286]

Answer:

0.405 seconds

Explanation:

Consider the amount of time it takes the block to fall from 53 m up to 14 m above the ground; then consider the amount of time it takes the block to fall from 53 m up to 2 m above the ground.

First, d = (1/2) gt^2 or t= ( 2 d / g)^1/2

= ( 2 × 39 / 9.8)^1/2 = 2.8212 seconds

Then, to fall from 53 down to 2 meters...

d = (1/2) gt^2 or t= ( 2 d / g)^1/2

= ( 2 * 51/ 9.8 )^1/2 = 3.2262 seconds

So the amount of time it takes for the block to fall from 14 m upto 2 m above the ground

3.2262 - 2.8212 = 0.405 seconds

this is how much time there is from when the man sees the block until it hits him. Not much time...

5 0

2 years ago