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KonstantinChe [14]
3 years ago
7

I need help ASAP.

Physics
2 answers:
Zina [86]3 years ago
7 0

Answer:

1. 2.23

2. 0.90

3. 0.28

aivan3 [116]3 years ago
4 0

1. Average time for the first 0.25 m: 2.23 s

Explanation:

The average time that it takes for the car to travel the first 0.25 m is given by the average of the first three measures, so:

t=\frac{2.24s+2.21s+2.23s}{3}=2.227 s=2.23 s

2. Average time to travel between 0.25 m and 0.50 m: 0.90 s

Explanation:

First of all, we need to calculate the time the car takes to travel between 0.25 m and 0.50 m for each trial:

t1 = 3.16 s - 2.24 s = 0.92 s

t2 = 3.08 s - 2.21 s = 0.87 s

t3 = 3.15 s - 2.23 s =  0.92 s

So, the average time is

t=\frac{0.92 s + 0.87 s + 0.92 s}{3}=0.903 s=0.90 s

3. Velocity in the second 0.25 m section: 0.28 m/s

Explanation:

The average velocity in the second 0.25 m section is equal to the ratio between the distance covered (0.25 m) and the average time taken (0.90 s):

v=\frac{d}{t}=\frac{0.25 m}{0.90 s}=0.28 m/s

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Hans Langseth's beard measured 5.33 m in 1927. Consider two charges, q1 = 2.42 nC and an unspecified charge, q2, are separated 5
schepotkina [342]

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-7.89 * 10^(-9) C

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Parameters given:

q1 = 2.42 nC = 2.42 * 10^(-9) C

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F = (k*q1*q3)/1.9² + (k*q2*q3)/3.43²

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q2 = -7.89 * 10^(-9) C

3 0
3 years ago
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