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In-s [12.5K]
3 years ago
15

You are standing on a skateboard, initially at rest. A friend throws a very heavy ball towards you. You can either catch the bal

l or deflect the ball back towards your friend (such that it moves away from you with the same speed as it was originally thrown). What should you do in order to minimize your speed on the skateboard? Choose one of the following three answers.
A) Your final speed on the skateboard will be the same regardless whether you catch the ball or deflect the ball.

B) You should catch the hall.

B) You should deflect the ball

The answer is you should catch the ball to minimize speed. What I need is a math-based and THOROUGH explanation. Any formulas you use to prove, please show the ORIGNIAL equation with steps that you used to get to the final one. Please! will rate right away!
Physics
1 answer:
harina [27]3 years ago
7 0

To solve this problem we will apply the concept related to the moment, which describes the change in speed in proportion to the body mass. The momentum can be described under the general equation

p = mv

Where,

m = mass

v = Velocity

The change in momentum is the difference between the final moment and the initial moment

\Delta p = p_f - p_i

Where,

p_f = Final momentum

p_i = Initial momentum

After catching the ball, the ball comes to rest position and speed of the ball is zero.

p_f = 0

The change in momentum on catching the ball is

\Delta p = 0 - mv

\Delta p = -mv

The momentum during the catching of the ball is

p_i = mv

Now during deflection of the ball, we know that there is an initial momentum and a negative final momentum because it moves with the same speed but in the opposite direction, that is

p_f = -mv

The momentum is negative since during deflection the ball moves with same speed in opposite direction

\Delta p = -mv-mv

\Delta p = 2-mv

Therefore, the correct ansswer is B: catch the ball in order to minimize your speed on the skateboard

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A +1.0 nC charge is at x = 0 cm, a -1.0 nC charge is at x = 1.0 cm and a 4.0 nC at x= 2 cm. What is the electric potential energ
lesantik [10]

Answer:

- 2.7 x 10^-6 J

Explanation:

q1 = 1 nC  at x = 0 cm

q2 = - 1 nC at x = 1 cm

q3 = 4 nC at x = 2 cm

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U=\frac{Kq_{1}q_{2}}{r}

where r be the distance between the two charges

By use of superposition principle, the total energy of the system is given by

U = U_{1,2}+U_{2,3}+U_{3,1}

U=\frac{Kq_{1}q_{2}}{0.01}+\frac{Kq_{2}q_{3}}{0.01}+\frac{Kq_{3}q_{1}}{0.02}

U=-\frac{9\times10^{9}\times 1\times10^{-9}\times 1\times10^{-9}}}{0.01}-\frac{9\times10^{9}\times 1\times10^{-9}\times 4\times10^{-9}}}{0.01}+-\frac{9\times10^{9}\times 1\times10^{-9}\times 4\times10^{-9}}}{0.02}

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If a 20 kg green fish swimming at 2 m/s swallows a 1 kg orange fish at rest, in what direction, and how fast
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Explanation:

The question should be:

If a 20 kg green fish swimming at 2 m/s swallows a 1 kg orange fish at rest, in what direction, and how fast  will the green fish swim after eating the orange fish?

Ok, here we have conservation of momentum.

At the beginning, the total momentum is equal to the sum between the momentum of the green fish and the momentum of the orange fish.

Where the momentum is written as:

P = m*v

m = mass

v = velocity.

The momentum of the green fish is:

Pg = 20kg*2m/s = 40 kg*m/s.

The momentum of the orange fish is:

Po = 1kg*0m/s = 0

The total initial momentum is:

Pi = Pg + Po = 40 kg*m/s.

After the green fish eats the orange fish, we do not have an orange fish anymore, and the mass of the green fish will be equal to it's initial mass, plus the mass of the fish that it ate, this will be:

M = 20kg + 1kg = 21kg.

Then the momentum will be:

Pf = 21kg*V

Where V is the final velocity.

For conservation of momentum, the initial momentum is equal to the final momentum, then:

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40 kg*m/s = 21kg*V

(40/21) m/s = 1.9 m/s = V

The fish's final velocity is 1.9 m/s

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