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2 years ago

A 0.2-kg ball moving at 2.0 m/s perpendicular to a wall rebounds from the wall at 1.5 m/s. The impulse given to the ball is:

Physics

1 answer:

kiruha [24]2 years ago

8 0

Answer:

J = 0.7 N-s away from wall

Explanation:

Given that,

Mass of the ball, m = 0.2 kg

Initial speed of the ball, u = 2 m/s

Final speed of the ball, v = -1.5 m/s (as it rebounds)

Impulse is equal to the change in momentum. So,

$J=m(v-u)\\\\J=0.2\times (-1.5-2)\\\\J=-0.7\ N-m$

So, the impulse given to the ball is 0.7 N-m and it is away from the wall. Hence, the correct option is (c).

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A ball of mass is released from rest at a height of 30 how fast is it going when it hits the ground

elena55 [62]

Answer:

If the height is in metres, the speed is 24.25m/s

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3 years ago

If you wanted to know the location of a vehicle that ran out of gas after taking a zigzag route through the city, which quantity

Papessa [141]

Explanation:

the vehicles displacement, since displacement deals with position

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2 years ago

Q. No. 9 A body falls freely from the top of a tower and during the last second of its fall, it falls through 25m. Find the heig

HACTEHA [7]

Answer:

45.6m

Explanation:

The equation for the position y of an object in free fall is:

$y=-\frac{1}{2} gt^2+v_0t+y_0$

With the given values in the question the equation has one unknown v₀:

$v_0=\frac{y-y_0}{t}+\frac{1}{2}gt$

Solving for t=1:

1) $v_0=y-y_0+\frac{g}{2}$

To find the hight of the tower you can use the concept of energy conservation:

The energy of the body 1 sec before it hits the ground:

2) $E=\frac{1}{2}m{v_0}^2+mgy_0$

If h is the height of the tower, the energy on top of the tower:

3) $E=mgh$

Combining equation 2 and 3 and solving for h:

4) $h=\frac{{v_0}^2}{2g}+y_0$

Combining equation 1 and 4:

$h=\frac{{(y-y_0+\frac{g}{2}})^2}{2g}+y_0$

4 0

3 years ago

A 14gram ovarian tumor is treated using a sodium phosphate in which the phosphorus atoms are the radioactive phosphorus 32 isoto

Nata [24]

I don’t know sorry ;khbadkhb didhwbck( khwdicdwbihwd

6 0

3 years ago

If the resultant of two velocity vectors of equal magnitude is also of the same magnitude, then which statement must be correct?

Tamiku [17]

The correct option is C) The angle between the vectors is 120°.

Why?

We can solve the problem and find the correct option using the Law of Cosine.

Let A and B, the given two sides and R the resultant (sum),

Then,

$R=A=B$

So, using the law of cosines, we have:

$R^{2}=A^{2}+B^{2}+2ABCos(\alpha)\\ \\A^{2}=A^{2}+A^{2}+2*A*A*Cos(\alpha)\\\\0=A^{2}+2*A^{2}*Cos(\alpha)\\\\Cos(\alpha)=-\frac{A^{2}}{2*A^{2}}=-\frac{1}{2}\\\\\alpha =Cos(-\frac{1}{2})^{-1}=120\°$

Hence, we have that the angle between the vectors is 120°. The correct option is C) The angle between the vectors is 120°

Have a nice day!

4 0

3 years ago