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1 year ago

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suppose you need to design a parachute system to help a remote camera land safely at the bottom of a cave. the camera will drop

from a height of one story. it has a mass of about 500 g. a carton of eggs inside a box will model the camera. for your test to be successful, none of the eggs can break when the box lands. you may use string, plastic bags, paper, and any fabrics you have to make a model parachute.

1 answer:

Marina86 [1]1 year ago

3 0

The answer is hang glider.

<h3>What is the model?</h3>

In high school, I created a project for a similar experiment that was accepted.

It matters what you carry the egg in because it is an egg, so something cushioned.

Use the fabric and cut the paper bag to make the kind of parachute I mentioned before.

When people go sky divining, they use a specific kind of parachute because the design is supposed to slow the fall and give them a chance to make a safe landing.

To learn more about hang glider, refer

brainly.com/question/2374318

#SPJ4

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$\theta_2 - \theta_1 = 168.46 - 11.53$

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022 (part 1 of 4) 10.0 points A ball is thrown vertically upward with a speed of 24.5 m/s. How high does it rise? The accelerati

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1)

Answer:

Part 1)

H = 30.6 m

Part 2)

t = 2.5 s

Part 3)

t = 2.5 s

Part 4)

$v_f = 24.5 m/s$

Explanation:

Part 1)

initial speed of the ball upwards

$v_i = 24.5 m/s$

so maximum height of the ball is given by

$H = \frac{v_i^2}{2g}$

$H = \frac{24.5^2}{2(9.80)}$

$H = 30.6 m$

Part 2)

As we know that final speed will be zero at maximum height

so we will have

$v_f - v_i = at$

$0 - 24.5 = (-9.8)t$

$t = 2.5 s$

Part 3)

Since the time of ascent of ball is same as time of decent of the ball

so here ball will same time to hit the ground back

so here it is given as

t = 2.5 s

Part 4)

since the acceleration due to earth will be same during its return path as well as the time of the motion is also same

so here its final speed will be same as that of initial speed

so we have

$v_f = 24.5 m/s$

2)

Answer:

a = 9.76 m/s/s

Explanation:

As we know that the object is released from rest

so the displacement of the object in vertical direction is given as

$y = \frac{1}{2}at^2$

$4.88 = \frac{1}{2}a(1^2)$

$a = 9.76 m/s^2$

3)

Answer:

v = 29.7 m/s

Explanation:

acceleration of the rocket is given as

$a = 90 m/s^2$

time taken by the rocket

t = 0.33 min

final speed of the rocket is given as

$v_f = v_i + at$

$v_f = 0 + (90)(0.33)$

$v_f = 29.7 m/s$

4)

Answer:

Part 1)

y = 25.95 m

Part 2)

d = 6.72 m

Explanation:

Part 1)

As it took t = 2.3 s to hit the water surface

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$y = \frac{1}{2}gt^2$

$y = \frac{1}{2}(9.81)(2.3^2)$

$y = 25.95 m$

Part 2)

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$d = v_x t$

$d = 2.92 \times 2.3$

$d = 6.72 m$

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