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3 years ago

Difference between centrifugal and semi-centrifugal clutches

Physics

2 answers:

Bezzdna [24]3 years ago

6 0

Answer:

Semi-centrifugal clutches are used in high powered race cars, to reduce the driver effort. Working is just like semi-centrifugal clutch.

The clutch action is purely under centrifugal force.

At low engine rpm the centrifugal force is low, so there is slip, as engine rpm increases, so does the centrifugal force.

grandymaker [24]3 years ago

4 0

Centrifugal clutch: These are automatic clutches under the action of centrifugal forces. ... Semi – centrifugal clutch: The manual type of clutches which have a pedal assist to it and the clutches becomes free as the system gains rpm time to time. These clutches are used to remove driver effort in racing cars.

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A ball is thrown vertically upward from the top of a building 80 feet tall with an initial velocity of 64 feet per second. The d

-BARSIC- [3]

Answer:

a) $t=6.37s$

b) $t=3.3333s$

Explanation:

The knowable variables are the initial hight and initial velocity

$s_{o}=80ft$

$v_{os}=64ft/s$

The equation that describes the motion of the ball is:

$s=80+64t-16t^{2}$

If we want to know the time that takes the ball to hit the ground, we need to calculate it by doing s=0 that is the final hight.

$0=80+64t-12t^{2}$

a) Solving for t, we are going to have two answers

$t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}$

a=-16

b=64

c=80

$t=-1.045 s$ or $t=6.378s$

<em><u>Since time can not be negative the answer is t=6.378s </u></em>

b) To find the time that takes the ball to pass the top of the building on its way down, we must find how much does it move too

First of all, we need to find the maximum hight and how much time does it take to reach it:

$v_{y}=v_{o}+gt$

at maximum point the velocity is 0

$0=64-32.2t$

Solving for t

$t=1.9875 s$

Now, we must know how much distance does it take to reach maximum point

$s=0+64t-16t^{2} =64(1.9875)-12(1.9875)^{2} =80ft$

So, the ball pass the top of the building on its way down at 160 ft

$160=80+64t-16t^{2}$

Solving for t

$t=2s$ or $t=3.333s$

Since the time that the ball reaches maximum point is almost t=2s that answer can not be possible, so the answer is t=3.333s for the ball to go up and down, passing the top of the building

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3 years ago

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Harlamova29_29 [7]

Answer:

That is true

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3 years ago

The change in momentum of an object is equal to the Question 4 options: Force acting on it times its velocity. impulse acting on

lesya [120]

Answer:

impulse acting on it

Explanation:

The impulse is defined as the product between the force applied to an object (F) and the time interval during which the force is applied ( $\Delta t$ ):

$I=F\Delta t$

We can prove that this is equal to the change in momentum of the object. In fact, change in momentum is given by:

$\Delta p = m \Delta v$

where m is the mass and $\Delta v$ is the change in velocity. Multiplying and dividing by $\Delta t$ , we get

$\Delta p = m \frac{\Delta v}{\Delta t} \Delta t$

and since $\frac{\Delta v}{\Delta t}$ is equal to the acceleration, a, we have

$\Delta p = ma \Delta t$

And since the product (ma) is equal to the force, we have

$\Delta p = F \Delta t$

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3 years ago

7. A force of 100 N acting on a body gives it a speed of 200 m/s in 2

alekssr [168]

Answer:

Choice a. 1 kg, assuming that all other forces on the object (if any) are balanced.

Explanation:

By Newton's Second Law,

$\displaystyle a = \frac{\Sigma F}{m}$ ,

where

$a$ is the acceleration of the object in $\text{m}\cdot\text{s}^{-2}$ ,
$\Sigma F$ is the net force on the object in Newtons, and
$m$ is the mass of the object in kilograms.

As a result,

$\displaystyle m = \frac{\Sigma F}{a}$ .

Assume that all other forces on this object are balanced. The net force on the object will be $100\;\text{N}$ . The net force is constant. Acceleration should also be constant and the same as the average acceleration in the two seconds.

<h3>What is the average acceleration of this object?</h3>

$\displaystyle \begin{aligned}\text{Acceleration} &= \text{Average Acceleration}=\frac{\text{Change in Velocity}}{\text{Time Taken}}\end{aligned}$ .

$\displaystyle {a} = \frac{200\;\text{m}\cdot\text{s}^{-1}}{2\;\text{s}}=100\;\text{m}\cdot\text{s}^{-2}$ .

<h3>Apply Newton's Second Law to find the mass of the object.</h3>

$\displaystyle m = \frac{\Sigma F}{a} = \frac{100\;\text{N}}{100\;\text{m}\cdot\text{s}^{-2}} = 1\;\text{kg}$ .

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Difference between centrifugal and semi-centrifugal clutches​

Difference between centrifugal and semi-centrifugal clutches