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3 years ago

15

A person gets into an elevator at the lobby level of a hotel together with his 30-kg suitcase, and gets out at the 10th floor 35

m above. Determine the amount of energy consumed by the motor of the elevator that is now stored in the suitcase

1 answer:

jeka57 [31]3 years ago

8 0

Answer:

PE=10.3 Kj

Explanation:

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What is the length a rubberband was stretched if it has a spring constant of 5700N/m and is currently holding 8600J OF POTENTIAL

lozanna [386]

Answer:

$\displaystyle \Delta x=1.74\ m$

Explanation:

<u>Elastic Potential Energy </u>

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

$\displaystyle PE = \frac{1}{2}k(\Delta x)^2$

Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.

Solving for Δx:

$\displaystyle \Delta x=\swrt{\frac{2PE}{k}}$

Substituting:

$\displaystyle \Delta x=\sqrt{\frac{2*8600}{5700}}$

Calculating:

$\displaystyle \Delta x=\sqrt{3.0175}$

$\boxed{\displaystyle \Delta x=1.74\ m}$

6 0

3 years ago

If the rectangular barge is 3.0 m by 20.0 m and sits 0.70 m deep in the harbor, how deep will it sit in the river?

leva [86]

The harbour contains salt water while the river contains fresh water. So assuming that the densities of fresh water and salt water are:

density (salt water) = 1029 kg / m^3

density (fresh water) = 1000 kg / m^3

The amount of water (in mass) displaced by the barge should be equal in two waters.

mass displaced (salt water) = mass displaced (fresh water)

Since mass is also the product of density and volume, therefore:

<span>[density * volume]_salt water = [density * volume]_fresh water ---> 1</span>

First we calculate the amount of volume displaced in the harbour (salt water):

V = 3.0 m * 20.0 m * 0.70 m

V = 42 m^3 of salt water

Plugging in the values into equation 1:

1029 kg / m^3 * 42 m^3 = 1000 kg/m^3 * Volume fresh water

Volume fresh water displaced = 43.218 m^3

Therefore the depth of the barge in the river is:

43.218 m^3 = 3.0 m * 20.0 m * h

<span>h = 0.72 m (ANSWER)</span>

8 0

3 years ago

To Mr. H's disgust, a 450-g black crow is raiding the recently-filled bird feeder. As Mr. H runs out the back door with his broo

bagirrra123 [75]

Answer:

Actually it's 2.50 m/s, sorry

Explanation:

It is solved by using momentum conservation equation

combined mass of crow and feeder = 450+670=1120 gm

let the recoil speed of feeder be v m/s

Then applying momentum conservation we get;

1120×1.5 = 670×v

v= 2.50 m/s

the speed at which the feeder initially recoils backwards = 2.50 m/s

5 0

3 years ago

You are driving your car along a country road at a speed of 27.0 m/s. as you come over the crest of a hill, you notice a farm tr

Bezzdna [24]

speed of the car = 27 m/s

speed of truck ahead = 10 m/s

relative speed of car with respect to truck

$v_r = 27 - 10 = 17 m/s$

relative deceleration of car

$a_r = -7 m/s^2$

now the distance before they stop with respect to each other is given by

$v_f^2 - v_i^2 = 2 a d$

$0 - 17^2 = 2 *(-7)*d$

$d = 20.6 m$

so it will come at the same speed of truck after 20.6 m distance and hence it will not hit the truck as the distance of the truck is 25 m from car

Part b)

Distance traveled by car before it stops is given by

$v_f^2 - v_i^2 = 2 a s$

$0^2 - 27^2 = 2 * (-7)* s$

$s = 52.1 m$

so it will stop after it will cover total 52.1 m distance

Part c)

time taken by the car to stop

$v_f - v_i = at$

$0 - 27 = (-7) * t$

$t = 3.86 s$

now the distance covered by truck in same time

$d = 3.86 * 10 = 38.6 m$

now after the car will stop its distance from the truck is

$D = 25 + 38.6 - 52.1 = 11.5 m$

<em>so the distance between them is 11.5 m</em>

6 0

3 years ago

DanielleElmas [232]

A. it provides support

5 0

2 years ago