3 years ago

I need help on the data section of the circuit design lab on Edg.

Physics

1 answer:

Arte-miy333 [17]3 years ago

8 0

I hope it's not too late, but here you go

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A bullet with mass 1.0kg and velocity 180 m/s is brought to rest in 0.02 s by a sandbag.assuming constant acceleration in the sa

kotykmax [81]

Hello
The bullet is moving by uniformly accelerated motion.
The initial velocity is $v_i=180~m/s$ , the final velocity is $v_i=0~m/s$ , and the total time of the motion is $\Delta t=0.02~s$ , so the acceleration is given by
$a= \frac{v_f-v_i}{\Delta t} = -9000~m/s^2$
where the negative sign means that is a deceleration.
Therefore we can calculate the total distance covered by the bullet in its motion using
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3 years ago

A 2500 kg car traveling to the north is slowed down uniformly from an initial velocity of 20 m/s by a 5620 N braking force actin

Dennis_Churaev [7]

Answer:

the distance traveled by the car is 42.98 m.

Explanation:

Given;

mass of the car, m = 2500 kg

initial velocity of the car, u = 20 m/s

the braking force applied to the car, f = 5620 N

time of motion of the car, t = 2.5 s

The decelaration of the car is calculated as follows;

-F = ma

a = -F/m

a = -5620 / 2500

a = -2.248 m/s²

The distance traveled by the car is calculated as follows;

s = ut + ¹/₂at²

s = (20 x 2.5) + 0.5(-2.248)(2.5²)

s = 50 - 7.025

s = 42.98 m

Therefore, the distance traveled by the car is 42.98 m.

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3 years ago

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25% i believe because if were talking 50 percent half it would be 25.

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