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Anna71 [15]
3 years ago
11

A scene in a movie has a stuntman falling through a floor onto a bed in the room below. The plan is to have the actor fall on hi

s back, but you have been hired to investigate the safety of this stunt. When you examine the mattress, you see that it effectively has a spring constant of 65144 N/m for the area likely to be impacted by the stuntman, but cannot depress more than 12.89 cm without injuring him. To approach this problem, consider a simplified version of the situation. A mass falls through a height of 3.32 m before landing on a spring of force constant 65144 N/m. Calculate the maximum mass that can fall on the mattress without exceeding the maximum compression distance.
=______________________ kg
Physics
1 answer:
tekilochka [14]3 years ago
6 0

Answer:

The maximum mass that can fall on the mattress without exceeding the maximum compression distance is 16.6 kg

Explanation:

Hi there!

Due to conservation of energy, the potential energy (PE) of the mass at a height of 3.32 m will be transformed into elastic potential energy (EPE) when it falls on the mattress:

PE = EPE

m · g · h = 1/2 k · x²

Where:

m = mass.

g = acceleration due to gravity.

h = height.

k = spring constant.

x = compression distance

The maximum compression distance is 0.1289 m, then, the maximum elastic potential energy will be the following:

EPE =1/2 k · x²

EPE = 1/2 · 65144 N/m · (0.1289 m)² = 541.2 J

Then, using the equation of gravitational potential energy:

PE = m · g · h =  541.2 J

m =  541.2 J/ g · h

m = 541.2 kg · m²/s² / (9.8 m/s² · 3.32 m)

m = 16.6 kg

The maximum mass that can fall on the mattress without exceeding the maximum compression distance is 16.6 kg.

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olga55 [171]

Answer:

11:1

Explanation:

At constant acceleration, an object's position is:

y = y₀ + v₀ t + ½ at²

Given y₀ = 0, v₀ = u, and a = -g:

y = u t − ½g t²

After 6 seconds, the ball reaches the maximum height (v = 0).

v = at + v₀

0 = (-g)(6) + u

u = 6g

Substituting:

y = 6g t − ½g t²

The displacement between t=0 and t=1 is:

Δy = [ 6g (1) − ½g (1)² ] − [ 6g (0) − ½g (0)² ]

Δy = 6g − ½g

Δy = 5½g

The displacement between t=6 and t=7 is:

Δy = [ 6g (7) − ½g (7)² ] − [ 6g (6) − ½g (6)² ]

Δy = (42g − 24½g) − (36g − 18g)

Δy = 17½g − 18g

Δy = -½g

So the ratio of the distances traveled is:

(5½g) / (½g)

11 / 1

The ratio is 11:1.

8 0
3 years ago
Explain the law of conservation of momentum.
WITCHER [35]

Answer:

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Explanation:

3 0
3 years ago
If two forces are in the same direction then do they cancel each other out
stellarik [79]
Im pretty sure that they do

3 0
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How do materials, such as nutrients and waste, pass through the cell membrane?
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3 years ago
Desperado, a roller coaster built in Nevada, has a mass of 800 kg. It also has a vertical drop of 225 feet down the first hill.
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Answer:

the work done by friction on the car is 524,582 J.

Explanation:

Given;

mass of the roller coaster, m = 800 kg

distance moved by the coaster, d = 225 ft = 68.58 m

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The time taken for the coaster to drop down the hill is calculated as;

t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\ \times \ 68.58}{9.8} } \\\\t = 3.74 \ s

The work done by friction on the car is calculated as;

W =F\ \times \ d = \frac{mv}{t} \times \ d\\\\W = \frac{800 \ \times \ 35.76  }{3.74} \times \ 68.58\\\\W = 524,582 \ J

Therefore, the work done by friction on the car is 524,582 J.

6 0
3 years ago
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