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Paul [167]
3 years ago
14

A bolt falls off an airplane high above the ground. How far does the bolt have to fall before its speed reaches 100m/s (about 20

0 miles per hour)? You should ignore air friction in your calculation.
Physics
1 answer:
omeli [17]3 years ago
3 0
We can use the equation vf (the final velocity) =vi (the initial velocity) +at (aceleration times time)

We know the final velocity 100m/s, the initial velocity 0, and the acceleration (gravity) 9.8m/s^2. So, 100=0+9.8t. t=100/9.8
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Which of the following is not true of a conductor? A. It is used in electrical circuits. B. It allows easy passage of moving cha
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The Gulf Stream off the east coast of the United States can flow at a rapid 3.9 m/s to the north. A ship in this current has a c
Alex Ar [27]

Answer:

72.54 degree west of south

Explanation:

flow = 3.9 m/s north

speed = 11 m/s

to find out

point due west from the current position

solution

we know here water is flowing north and ship must go south at an equal rate so that the velocities cancel and the ship just goes west

so it become like triangle with 3.3 point down and the hypotenuse is 11

so by triangle

hypotenuse ×cos(angle) = adjacent side

11 ×cos(angle) = 3.3

cos(angle) = 0.3

angle = 72.54 degree west of south

3 0
3 years ago
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A(n) 7.7-kg object is sliding across the ice at 2.34 m/s in the positive x direction. An internal explosion occurs, splitting th
-BARSIC- [3]

Answer:

Average acceleration on first part of the chunk is given as

a_1 = 13.125 m/s^2

Average acceleration on second part of the chunk is given as

a_2 = -13.125 m/s^2

Explanation:

By momentum conservation along x direction we will have

mv_i = \frac{m}{2}v_1 + \frac{m}{2}v_2

so we have

v_1 + v_2 = 2v

v_1 + v_2 = 4.68

also by energy conservation

\frac{1}{2}(\frac{m}{2})v_1^2 + \frac{1}{2}(\frac{m}{2})v_2^2 - \frac{1}{2}mv^2 = 17 J

\frac{1}{4}m(v_1^2 + v_2^2) - \frac{1}{2}mv^2 = 17

(v_1^2 + v_2^2) - 2v^2 = \frac{4}{7.7}(17)

(4.68 - v_2)^2 + v_2^2 - 2v^2 = 8.83

21.9 + 2v_2^2 - 9.36 v_2 - 10.95 = 8.83

2v_2^2 - 9.36v_2 + 2.12 = 0

by solving above equation we will have

v_1 = 4.44 m/s

v_2 = 0.24 m/s

Average acceleration on first part of the chunk is given as

a_1 = \frac{4.44 - 2.34}{0.16}

a_1 = 13.125 m/s^2

Average acceleration on second part of the chunk is given as

a_2 = \frac{0.24 - 2.34}{0.16}

a_2 = -13.125 m/s^2

5 0
3 years ago
Which of the following is not a way in which humans directly impact algal bloom production? a. industrial pollution b. wastewate
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