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Semenov [28]
3 years ago
7

A football quarterback runs 15.0 m straight down the playing field in 2.50 s. he is then hit and pushed 3.00 m straight backward

in 1.75 s. he breaks the tackle and runs straight forward another 21.0 m in 5.20 s. calculate his average velocity (a) for each of the three intervals and (b) for the entire motion.

Physics
1 answer:
My name is Ann [436]3 years ago
3 0
Define
t = time (s),
dt = change in time (s)
 d =distance (m).

The scenario is as follows:
t = 0 s: Begin. Quarterback runs forward.
d = 0

t = 2.5 s: Quarterback is tackled and pushed backward.
d = 15 m

t = 2.5+1.75 = 4.25 s: Quarterback resumes the run forward.
d = 15 - 3 = 12 m

t = 4.25+5.2 = 9.45 s: Quarerback runs another 21 m forward.
d = 12+21 = 33 m

The graph showing the scenario is shown below.

Calculations:
Interval 1:
Average velocty = 15/2.5 = 6.0 m/s

Interval 2:
Average velocity = (12 - 15)/(1.75) = - 1.7 m/s

Interval 3:
Average velocity = (33 - 12)/5.2 = 4.0 m/s

Entire motion:
Average velocity = (33 - 0)/9.45 = 3.5 m/s

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A particular interaction force does work wint inside a system. the potential energy of the interaction is u. which equation rela
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Molodets [167]

Answer:

The current through the inductor at the end of 2.60s is 9.7 mA.

Explanation:

Given;

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initial current in the inductor, I₀ = 1.5 mA

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The emf of the inductor is given by;

V = L\frac{di}{dt} \\\\V = \frac{L(I_1-I_o)}{dt} \\\\L(I_1-I_o) = V*dt\\\\I_1-I_o = \frac{V*dt}{L}\\\\I_1 =  \frac{V*dt}{L} + I_o\\\\I_1 = \frac{41*10^{-3}*2.6}{13} +1.5*10^{-3}\\\\I_1 = 8.2*10^{-3} + 1.5*10^{-3}\\\\I_1 = 9.7 *10^{-3} \ A\\\\ I_1 = 9.7 \ mA

Therefore, the current through the inductor at the end of 2.60 s is 9.7 mA.

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