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Lynna [10]
4 years ago
11

A snowball is rolling down a hill at 4.5 m/s and accumulating snow as it goes. Its diameter begins at 0.50 m and ends at the bot

tom of the hill at 2.5 m. What is the snowball's change in centripetal acceleration from the top of the hill and the bottom of the hill?
Physics
1 answer:
Reil [10]4 years ago
6 0
To find the change in centripetal acceleration, you should first look for the centripetal acceleration at the top of the hill and at the bottom of the hill.

The formula for centripetal acceleration is:
Centripetal Acceleration = v squared divided by r

where:
v = velocity, m/s
r= radium, m

assuming the velocity does not change:

at the top of the hill:
centripetal acceleration = (4.5 m/s^2) divided by 0.25 m
                                      = 81 m/s^2

at the bottom of the hill:
centripetal acceleration = (4.5 m/s^2) divided by 1.25 m
                                      = 16.2 m/s^2

to find the change in centripetal acceleration, take the difference of the two.
change in centripetal acceleration = centripetal acceleration at the top of the hill - centripetal acceleration at the bottom of the hill

= 81 m/s^2 - 16.2 m/s^2
= 64.8 m/s^2 or 65 m/s^2
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4 years ago
How are Earth and Venus similar? How is Venus different from Earth? (Provide two similarities and two differences.)
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3 years ago
A merry-go-round a.k.a "the spinny thing" is rotating at 15 RPM, and has a radius of 1.75 m
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Answer:

A.) 4 revolution

B.) 0.2 revolution

C.) 4 seconds

D.) 2.75 m/s

Explanation:

Given that a merry-go-round a.k.a "the spinny thing" is rotating at 15 RPM, and has a radius of 1.75 m

Solution

1 revolution = 2πr

Where r = 1.75m

A. How many revolutions will it make in 3 minutes?

(2π × 1.75) / 3

10.9955 / 3

3.665 RPM

Number of revolution = 15 / 3.665

Number of revolution = 4 revolution

B. How many revolutions will it make in 10.0 seconds?

First convert 10 seconds to minutes

10/60 = 0.167 minute

(2π × 1.75) / 0.167

10.9955 / 0.167

65.973

Number of revolution = 15 / 65.973

Number of revolution = 0.2 revolution

C. How long does it take for a person to make 1 complete revolution?

15 = 1 / t

Make t the subject of formula

t = 1/15

t = 0.0667 minute

t = 4 seconds

D. What is the velocity in m/s of person standing on its edge?

Velocity in m/ s will be:

Velocity = (15 × 2pi × r) / 60

Velocity = 164.9334 / 60

Velocity = 2.75 m/s

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3 years ago
When does sound become damaging to Human hearing?
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Answer:

D) 250 decibels

Explanation:

4 0
3 years ago
A solid sphere, solid cylinder, and a hollow pipe all have equal masses and radii. If the three of them are released simultaneou
Roman55 [17]

Answer:

The solid sphere will reach the bottom first.

Explanation:

In order to develop this problem and give it a correct solution, it is necessary to collect the concepts related to energy conservation. To apply this concept, we first highlight the importance of conserving energy so we will match the final and initial energies. Once this value has been obtained, we will concentrate on finding the speed, and solving what is related to the Inertia.

In this way we know that,

\Delta KE = - \Delta PE

KE_t + KE_r = mgh

We know as well that the lineal and angular energy are given by,

KE_r = \frac{1}{2}I\omega^2

And the tangential kinetic energy as

KE_t = \frac{1}{2} mv^2

Where\omega = \frac{v}{R}

Replacing

\frac{1}{2}mv^2 + \frac{1}{2}I\frac{v}{R} = mgh

Re-arrange for v,

v=\sqrt{\frac{2mgh}{m+I/R^2}}

We have here three different objects: solid cylinder, hollow pipe and solid sphere. We need the moment inertia of this objects and replace in the previous equation found, then,

For hollow pipe:

I_{hp}=mR^2

v_{hp}=\sqrt{\frac{2mgh}{m+(mR^2)/R^2}}

v_{hp}=\sqrt{\frac{2mgh}{m+m)}

v_{hp}=\sqrt{gh}

For solid cylinder:

I_{sc}=\frac{1}{2}mR^2

v_{sc}=\sqrt{\frac{2mgh}{m+(1/2mR^2)/R^2}}

v_{sc}=\sqrt{\frac{2mgh}{m+1/2m}}

v_{sc}=\sqrt{\frac{3}{4}gh}

For solid sphere,

I_{ss}=\frac{2}{5}mR^2

v_{ss}=\sqrt{\frac{2mgh}{m+(2/5mR^2)/R^2}}

v_{ss}=\sqrt{\frac{2mgh}{m+2/5m}}

v_{ss}=\sqrt{\frac{10}{7}gh}

Then comparing the speed of the three objects we have:

v_{hp}

\sqrt{gh}

3 0
4 years ago
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