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3 years ago

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A 6.0x10-2kg hollow racquetball with an initial speed of 18.6 m/s collides with a backboard. It rebounds with a speed of 4.6 m/s

. Calculate the total impulse that the wall imparts on the ball on the ball.

1 answer:

trapecia [35]3 years ago

6 0

Answer:

The total impulse that the wall imparts on the ball on the ball is 1.392 kg-m/s.

Explanation:

Given that,

Mass of the racquetball, $m=6\times 10^{-2}\ kg$

Initial speed of the ball, u = 18.6 m/s

Final speed of the ball, v = -4.6 m/s (the ball rebounds so it will be negative)

We need to find the total impulse that the wall imparts on the ball on the ball. We know that the change in momentum of an object. It is given by :

$J=m(v-u)$

$J=6\times 10^{-2}\times (-4.6-18.6)$

J = -1.392 kg-m/s

So, the total impulse that the wall imparts on the ball on the ball is 1.392 kg-m/s. Hence, this is the required solution.

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Which force is greater the earth’s pull on the moon or the moon’s pull on the earth

Vitek1552 [10]

Answer:

The earth's pull on the moon

Explanation:

Earth exerts a gravitational pull on the moon 80 times stronger than the moon's pull on the Earth.

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4 years ago

A train slows down as it rounds a sharp horizontal turn, going from 94.0 km/h to 46.0 km/h in the 17.0 s that it takes to round

Svetllana [295]

Answer:

1.41 m/s^2

Explanation:

First of all, let's convert the two speeds from km/h to m/s:

$u = 94.0 km/h \cdot \frac{1000 m/km}{3600 s/h} = 26.1 m/s$

$v=46.0 km/h \cdot \frac{1000 m/km}{3600 s/h}=12.8 m/s$

Now we find the centripetal acceleration which is given by

$a_c=\frac{v^2}{r}$

where

v = 12.8 m/s is the speed

r = 140 m is the radius of the curve

Substituting values, we find

$a_c=\frac{(12.8 m/s)^2}{140 m}=1.17 m/s^2$

we also have a tangential acceleration, which is given by

$a_t = \frac{v-u}{t}$

where

t = 17.0 s

Substituting values,

$a_t=\frac{12.8 m/s-26.1 m/s}{17.0 s}=-0.78 m/s^2$

The two components of the acceleration are perpendicular to each other, so we can find the resultant acceleration by using Pythagorean theorem:

$a=\sqrt{a_c^2+a_t^2}=\sqrt{(1.17 m/s^2)+(-0.78 m/s^2)}=1.41 m/s^2$

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3 years ago

Read 2 more answers

An electric wall clock has a second hand 15 cm long. at the tip of his hand, what is the magnitude of the velocity?

Mila [183]

The velocity of the tip of the second hand is 0.0158 m/s

Explanation:

First of all, we need to calculate the angular velocity of the second hand.

We know that the second hand completes one full circle in

T = 60 seconds

Therefore, its angular velocity is:

$\omega = \frac{2\pi}{T}=\frac{2\pi}{(60)}=0.105 rad/s$

Now we can calculate the velocity of a point on the tip of the hand by using the formula

$v=\omega r$

where

$\omega=0.105 rad/s$ is the angular velocity

r = 15 cm = 0.15 m is the radius of the circle (the distance of the point from the centre of rotation)

Substituting,

$v=(0.105)(0.15)=0.0158 m/s$

Learn more about angular motion here:

brainly.com/question/9575487

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#LearnwithBrainly

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3 years ago

Calculate the energy used by a radio of power 30W in 1 minute.

Alekssandra [29.7K]

Answer:

<u>1.8kJ</u>

Explanation:

Formula :

<u>Energy used = Power x time</u>

<u />

===============================================================

Given :

⇒ Power = 30 W

⇒ Time = 1 minute = 60 seconds

=============================================================

Solving :

⇒ Energy used = 30 W × 60 s

⇒ Energy used = 1,800 J

⇒ Energy used = <u>1.8kJ</u>

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2 years ago

Read 2 more answers

Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i

s2008m [1.1K]

Answer:

The final speed for the mass 2m is $v_{2y}=-1,51\ v_{i}$ and the final speed for the mass 9m is $v_{1f} =0,85\ v_{i}$ .

The angle at which the particle 9m is scattered is $\theta = -66,68^{o}$ with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

<u>Conservation of linear momentum:</u>

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

$\vec{p}=m\vec{v}$

$p_{xi} =p_{xf}$

In the x axis before the collision we have

$p_{xi}=9m\ v_{i} - 2m\ v_{i}$

and after the collision we have that

$p_{xf} =9m\ v_{1x}$

In the y axis before the collision $p_{yi} =0$

after the collision we have that

$p_{yf} =9m\ v_{1y} - 2m\ v_{2y}$

so

$p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}$

then

$p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}$

<u>Conservation of kinetic energy:</u>

$\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}$

so

$\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}$

Putting in one side of the equation each speed we get

$\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}$

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the speed of the 9m particle:

$v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}$

the magnitude of the final speed of the particle 9m is

$v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }$

$v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}$

The tangent that the speed of the particle 9m makes with the -y axis is

$tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}$

As a vector the speed of the particle 9m is:

$\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}$

As a vector the speed of the particle 2m is:

$\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}$

8 0

3 years ago