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KengaRu [80]
3 years ago
14

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 a researcher has been hired to investigate the safety of this stunt. When the researcher examines the mattress, she sees that it effectively has a spring constant of 77144 N/m for the area likely to be impacted by the stuntman, but it cannot depress more than 13.11 cm without injuring him. To approach this problem, consider a simplified version of the situation. A mass falls through a height of 4.12 m before landing on a spring of force constant 77144 N/m. Calculate the maximum mass that can fall on the mattress without exceeding the maximum compression distance.
Physics
1 answer:
Alina [70]3 years ago
4 0

Answer:

131 kg

Explanation:

Data:

Given,

spring constant, k = 65 144 N/m

Height of the fall = 3.32 m

Spring compression distance, x = 13. 55 × 10⁻²m

acceleration due to gravity, g = 9.81 ms⁻²

Now, during the fall, the gravitational potential energy is translated to the kinetic energy.

The kinetic energy is then later absorbed by the spring which absorbs it as the spring potential energy or simply (SPE).

Thus, the equation becomes:

Potential energy = Spring potential energy

mh (h + x) = \frac{1}{2} kx^{2} \\m(9.81 * (3.32 + 0.1311) = \frac{1}{2}*65 144 * (0.1311)^{2}  \\4.268 m = 559.82\\           m = 131 kg

maximum mass = 131 kg

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During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to come near, the clown turns due east and runs 19.8 m to exit the arena. The magnitude of the clown’s displacement is 27 m.

<u>Explanation: </u>

As the clown is running in the north direction for about 7.7 m and then he turns 49.9 degrees east of north. In the east of north, he covers a distance of 6.4 m and then turns east to exit the arena after covering a distance of 19.8 m. Let’s have a simple diagram to easily understand the problem.

In first step, the clown runs 7.7 m in north direction, so the image will be  as in fig 1. Then he takes a direction of north east and covers a distance of 6.4 m, so the image will be modified as in fig 2. Then after the bull comes, he turns east and runs 19.8 m to exit the arena, so the image will be as in figure 3.

So, the extension of North line and the East line at a point shown as the dotted line in the above image, forms the total displacement as the hypotenuse of a right angled triangle. The extended dotted lines is nothing but the horizontal and vertical components of the angle 49.9 degree.

By using Pythagoras theorem, the total displacement can be found as

\text { Total displacement }=\sqrt{(o p p)^{2}+(a d j)^{2}}

\text { Distance covered by the clown in east direction }=(6.4 \times \cos 49.9)+19.8=23.9 \mathrm{m}

Similarly, the adjacent side of this imaginary triangle is the distance covered by the clown in the North direction.

\text { Distance covered by the clown in north direction }=6.4 \sin 49.9+7.7=12.6 \mathrm{m}

Thus, the total displacement covered by the clown is

\text { Total displacement }=\sqrt{(23.9)^{2}+(12.6)^{2}}=\sqrt{571.21+158.76}=\sqrt{729.97}=27 \mathrm{m}

Thus, the total displacement by the clown is 27 m.

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3 years ago
At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2(g)+O2(g)→2SO3(g) At equilibrium, the partial
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Answer : The partial pressure of SO_3 is, 67.009 atm

Solution :  Given,

Partial pressure of SO_2 at equilibrium = 30.6 atm

Partial pressure of O_2 at equilibrium = 13.9 atm

Equilibrium constant = K_p=0.345

The given balanced equilibrium reaction is,

2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)

The expression of K_p will be,

K_p=\frac{(p_{SO_3})^2}{(p_{SO_2})^2\times (p_{O_2})}

Now put all the values of partial pressure, we get

0.345=\frac{(p_{SO_3})^2}{(30.6)^2\times (13.9)}

p_{SO_3}=67.009atm

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3 years ago
(a) What is the intensity in W/m2 of a laser beam used to burn away cancerous tissue that, when 90.0% absorbed, puts 500 J of en
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Answer:

4.42 x 10⁷ W/m²

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A = energy absorbed = 500 J

η = efficiency = 0.90

E = Total energy

Total energy is given as

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E = 555.55 J

t = time = 4.00 s

Power of the beam is given as

P = E /t

P = 555.55/4.00

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Area of the circular spot is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

Intensity of the beam is given as

I = P /A

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I = 4.42 x 10⁷ W/m²

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