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ludmilkaskok [199]
3 years ago
8

A plane is flying east when it drops some supplies to a designated target below. The supplies land after falling for 10 seconds.

How high above the ground was the airplane when the supplies were dropped?
120 meters

490 meters

980 meters

49 meters

As the supplies fell, they traveled 650 meters horizontally. How fast were the supplies and the airplane moving forward? Remember that it took 10 seconds for them to reach the ground.

65 m/s

6500 m/s

0.154 m/s

660 m/s
Physics
1 answer:
Ivanshal [37]3 years ago
4 0

Part 1)

here we know that supply took 10 s to reach the ground

so here we will have

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

y = \frac{1}{2}\times 9.8 \times 10^2

y = \frac{1}{2}\times 9.8 \times 100

y = 490 m

Part 2)

Here all the supply covered horizontal distance of 650 m in 10 s interval of time

so here we can say

speed = \frac{distance}{time}

v = \frac{650}{10}

v = 65 m/s


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Here we have three positively charged particles A,B and C, located at the following positions:

x_A = 0\\x_B = 10 m\\x_C = 20 m

The magnitudes of the three charges are:

q_A = q_B = q_C = 0.5 \mu C = 0.5\cdot 10^{-6}C

The force exerted by B on A is to the left (because the force between two positive charges is repulsive), and the force exerted by C on A is also to the left (also repulsive). Therefore, the net force on A is just the sum of the two forces exerted by charges B and C:

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brainly.com/question/8960054

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Dividing the above two equations we get

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