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3 years ago

What is the acceleration of a softball if it has a mass of 0.5 kg and hits the cathers glove with a force of 25n

1 answer:

ioda3 years ago

3 0

Acceleration is found if we have the force and mass.

With the following equation: F = ma, we can find the missing values.

F = 25n
M = 0.5 kg
a = ?

a = f/m
a = 25/0.5
a = 50

a = 50 m/s

So, the acceleration is 50 m/s^2

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Akhtar, Kiran and Rahul were riding in a motorocar that was moving with a high velocity on

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Answer:

According to the law of conservation of momentum:

Momentum of the car and insect system before collision = Momentum of the car and insect

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Hence, the change in momentum of the car and insect system is zero.

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reversed. As a result, the velocity of the insect changes to a great amount. On the other hand,

the car continues moving with a constant velocity. Hence, Kiran’s suggestion that the insect

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Therefore, the momentum gained by the insect is equal to the momentum lost by the car.

Akhtar made a correct conclusion because the mass of the car is very large as compared to

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Rahul gave a correct explanation as both the car and the insect experienced equal forces

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3 years ago

Why is the light bulb not a closed system?

Nana76 [90]

Answer:An incandescent light bulb gives only energy of the system in the form of heat. ... The roof of a house is stable but it receives energy from the surroundings and transfers energy through it towards lower temperature side which includes no mass transfer. So, it is best considered as closed system.

Explanation:

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3 years ago

In scientific notation, 213,000 is equal to:

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The answer is 2.13 × 10⁵

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4 years ago

Read 2 more answers

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2 years ago

At the very end of Wagner's series of operas The Ring of Nibelung, Brunnhilde takes the golden ring form the finger of the dead

Blababa [14]

Answer:

a) 404 m² b) apparent height = 7.5 m

Explanation:

This question is about refraction and total internal refraction.

Here I will take refractive index of air and water

$n_{air}=1\\ n_{water}=1.33=4/3$

Now let's look at the diagram I have attached here

At some angle A, the light from the ring (yellow point) under water will be totally internally refracted (B = 90°), which means that rays of light (yellow arrow) that make large enough angle A will not be able to escape from the water. Since we assumed that the ring is a point, there will be a critical cone of angle A with the ring at its apex which traces a circle of radius R on the surface of water, which, beyond this radius, no light could escape.

According to snell's law

$\frac{sin(B)}{sin(A)} = \frac{n_{water}}{n_{air}} = 4/3$

At critical angle B = 90°

$\frac{3)}{4}sin(B) = [tex]\frac{3}{4} sin(90^\circ ) = 0.75 = sin(A)$

Therefore

$A = 48.6^\circ$

With this, we can find the radius of the circle (refer to my diagram)

$h* tan (A) = R\\R =11.3 m$

And with that we can find the area

$A = \pi R^2=404\ m^2$

Additional Problem

For apparent depth from above, we can think that, since we are accustomed to seeing light at the speed of c in air, our brain interpret light from <em>any</em> source to be traveling at c. This causes light that originated under water, which has the speed of

$v_{water} = \frac{c}{n_{water}} = 0.75c$

to appear as if it has traveled with the same duration as light with speed c

In order for this to happen our brain perceive shortened length which is the apparent depth.

To put it in mathematical term

$t_{travel}=\frac{h_{apparent}}{v_{water}} =\frac{h}{c}$

So we get apparent depth

$h_{apparent}=0.75h = 7.5\ m$

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3 years ago