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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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Answer:

The force on q₁ due to q₂ is (0.00973i + 0.02798j) N

Explanation:

F₂₁ = \frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|}

Where;

F₂₁ is the vector force on q₁ due to q₂

K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²

r₂₁ is the unit vector

|r₂₁| is the magnitude of the unit vector

|q₁| is the absolute charge on point charge one

|q₂| is the absolute charge on point charge two

r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)

|r₂₁| = \sqrt{(4^2)+(11.5^2)} = \sqrt{148.25}

(|r₂₁|)² = 148.25

F_2_1=\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|} = \frac{8.99X10^9(14X10^{-6})(60X10^{-6})}{148.25}.\frac{(4i + 11.5j)}{\sqrt{148.25} }

      = 0.050938(0.19107i + 0.54933j) N

      = (0.00973i + 0.02798j) N

Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N

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Answer:

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V is the final velocity

u is the initial velocity

Given

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Substitute

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Change in momentum = 0.45×10

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Answer:

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