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Drupady [299]
3 years ago
12

Three students have been studying relative motion and decide to do an experiment to demonstrate their knowledge. The experiment

plan calls for Jane to drive her pickup in a straight line across the parking lot at a constant speed of 12.6 m/s. Fred is in the back of the truck and throws a baseball backward and upward at an angle ? out the back of the truck. Sue observes the flight of the ball while standing nearby in the parking lot.
(a) If Fred can throw the ball 30.0 m/s, at what angle relative to the horizontal should he throw the ball in order for Sue to see the ball travel vertically upward? (Enter your answer to at least one decimal place.)
�

(b) If Fred throws the ball at this angle, how high does Sue observe it to travel above the level at which it was thrown?
Physics
1 answer:
miskamm [114]3 years ago
7 0

Answer with Explanation:

We are given that

Constant speed of Jane=12.6 m/s

a.When Fred can throw the ball 30  m/s

We have to find the angle relative to the horizontal when he throw the ball in order for Sue to see the ball travel vertically upward.

Let \theta be the angle .

Therefore,

30 cos\theta=12.6

cos\theta=\frac{12.6}{30}=0.42

\theta=cos^{-1}(0.42)=65.165^{\circ}

b.We have to find the height to which ball reach.

v^2-v^2_0=2aS

S=\frac{v^2-v^2_0}{2a}=\frac{0-(30 sin65.165)^2}{2(-9.81)}

S=37.78 m

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In a cup game if the teams have the same score at the end of the match, 30 minutes of ------- are played.
Alina [70]

Answer:

second lag

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If in a cup game, a specified time limit is assigned to both teams to score high. If both teams are unable to score or if score of both the teams is equal then there is another second lag played where each team tries to score high. Even if in second lag both teams fail to score higher than other the last third lag is played or else game is declared draw.

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For exercise, an athlete lifts a barbell that weighs 400 N from the ground to a height of 2.0 m in a time of 1.6 s. Assume the e
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Answer:

Explanation:

(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)

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ΔK = 0

ΔUg  is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .

ΔUg = positive

ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .

ΔUs = 0

ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence

ΔEth = negative .

b )

work done by athlete

= 400 x 2 = 800 J

energy output = 800 J

c )

It is 25% of metabolic energy output of his body

so metalic energy output of body

= 4x 800 J .

3200 J

power = energy output / time

= 3200 / 1.6

= 2000 W .

d )

1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .

2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .

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Read 2 more answers
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mafiozo [28]

Answer:

d) 0 V

Explanation:

It can be showed that the potential due to a point charge q, to a distance d from the charge, can be expressed as follows:

V = \frac{k*q}{r}

where k = \frac{1}{4*\pi*\epsilon0} = 9e9 N*m2/C2

As the potential is an scalar, and is linear with the charge, we can apply the superposition principle, which means that we can find the potential due to one of the charges, as if the other were not present.

By symmetry, all four charges are at the same distance from the center, so we can write the total potential, as follows:

V = \frac{k}{d} ( q1 + q2 + q3 + q4) (1)

where d, is the semi-diagonal of the square, that we can find applying Pythagorean theorem, as follows:

d = \sqrt{\frac{L^{2}}{4} + \frac{L^{2}}{4} } = L*\frac{\sqrt{2}}{2}

Replacing by the values in (1) we have:

V = \frac{9e9N*m2/C2}{\frac{L}{2}*\sqrt{2} }* ( +3q -q + 2q + -4q)  = 0 V

which is equal to the option d).

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3 years ago
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