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2 years ago

14

A 1.3 kg ball drops vertically onto a floor, hitting with a speed of 21 m/s. It rebounds with an initial speed of 9.8 m/s. (a) W

hat impulse acts on the ball during the contact? (b) If the ball is in contact with the floor for 0.0156 s, what is the magnitude of the average force on the floor from the ball?

1 answer:

Anarel [89]2 years ago

3 0

Answer:

a) Impulse(J)=40.04 kg.m/s

b)F=2566.66 N

Explanation:

Given that

$V_1=21 m/s,V_2=9.8m/s$

We know that impulse(J) impulse is a vector quantity

$J=P_2-P_1$

We know that P=mV

So $J=m(V_2-V_1)$

Now by putting the values

J=1.3(9.8-(-21))

So impulse(J)=40.04 kg.m/s

Force is given as

$F=\dfrac{dP}{dt}$

or we can say that

$F=m\left(\dfrac{v_2-V_1}{dt}\right)$

So $F=\dfrac{40.04}{0.0156}$

F=2566.66 N

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Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, $q_1=-3\ nC$

It is placed at a distance of 9 cm at x axis

Charge, $q_2=+4\ nC$

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

$\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0$

Here,

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So,

$\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}$

Squaring both sides,

$3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m$

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

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2 years ago

How much power is needed to lift the 200-N object to a height of 10 m in 4 s?

harkovskaia [24]

Answer:

500 watts

Explanation:

Recall that the definition of power is the amount of energy delivered per unit of time.

In our case, the energy delivered is potential energy which we can estimate as the product of the weight of the object times the distance it is lifted above ground:

200 N x 10 m = 2000 Nm

then the power is the quotient of this potential energy divided the time it took to lift the object to that position:

Power = 2000 / 4 Nm/s = 500 Nm/s = 500 watts

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Where in the motion is the magnitude of the force from the spring on the object zero? Where in the motion is the magnitude of th

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Answer:

1. The magnitude of the force from the spring on the object is zero on <em>Equilibrium.</em>

2. The magnitude of the force from the spring on the object is a maximum on <em>The top and bottom.</em>

3. The magnitude of the net force on the object is zero on <em>The Bottom.</em>

4. The magnitude of the force on the object is a maximum on <em>the Top.</em>

Explanation:

<em>1. Because the change in position delta X is zero.</em>

<em>2. Because of delta X.</em>

<em>3. Beacuse, the force of gravity and the force of the spring oppose each other to keep the block at rest, away from the equilibrium position.</em>

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3 years ago

Which of these processes are chemical reactions

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Explanation:

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3 years ago

g A cylinder of mass m is free to slide in a vertical tube. The kinetic friction force between the cylinder and the walls of the

sdas [7]

Answer:

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Explanation:

From the question we are told that

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3 years ago