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Maksim231197 [3]
3 years ago
10

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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An object of mass 2.0 kg is attached to the top of a vertical spring that is anchored to the floor. The unstressed length of the
poizon [28]

Answer:

The value is A  =  0.014 \  m

Explanation:

From the question we are told that

    The mass of the object is  m  =  2.0 \  kg

    The unstressed length of the string is  l  =  0.08 \  m

    The length of the spring when it is  at equilibrium is  l_e = 5.9 \  cm  =  0.059 \  m

      The initial speed (maximum speed)of the spring when given a downward blow v  =  0.30 \  m/s

Generally the maximum speed  of the spring  is mathematically represented as

           u =  A *  w

Here A is maximum height above the floor (i.e the maximum amplitude)

            and w is the angular frequency which is mathematically represented as

       w = \sqrt{\frac{k}{m} }

So

        u =  A *   \sqrt{\frac{k}{m} }

=>      A  =  u *   \sqrt{\frac{m}{k} }

Gnerally the length of the compression(Here an assumption that the spring was compressed to the ground by the hammer is made) by the hammer is mathematically represented as

           b  =  l -l_e

=>         b  = 0.08 - 0.05 9

=>         b  = 0.021 \  m

Generally at equilibrium position the net force acting on the spring is  

            k *  b  -  mg  =  0

=>         k *  0.021   -   2 * 9.8  =  0

=>        k =  933 \  N/m

So

            A  =  0.30  *   \sqrt{\frac{2}{933} }

=>          A  =  0.014 \  m

8 0
3 years ago
What is the magnitude of the gravitational force exerted by earth on a 9.0-kg brick when the brick is in free fall?
GarryVolchara [31]
The magnitude is 9.000kg
4 0
3 years ago
You are pulling a sled using a horizontal rèpe, as shown in the diagram. The rope pulls the sled. exerting a force of 50 N to th
geniusboy [140]

Answer:

Part 1

20 N

Part 2

0.4 m/s²

Part 3

4 m/s

Explanation:

The force which pulls the sled right = 50 N

The friction force exterted towards left by the snow = -30 N

The mass of the sled = 50 kg

Part 1

The sum of the forces on the sled, F = 50 N + (-30) N = 20 N

Part 2

The acceleration of the sled is given as follows;

F = m·a

Where;

m = The mass of the sled

a = The accelertion

a = F/m

∴ a = (20 N)/(50 kg) = 0.4 m/s²

The acceleration of the sled, a = 0.4 m/s²

Part 3

The initial velocity of the sled, u = 2 m/s

The kinematic equation of motion to determine the speed of the sled is v = u + a·t

The speed, <em>v</em>, of the sled after t = 5 seconds is therefore;

v = 2 m/s + 0.4 m/s² × 5 s = 4 m/s.

7 0
3 years ago
Is boiling a pot of water an example of Thermodynamics?
garri49 [273]

Answer: I think I have the answer to your problem!!

A pot of boiling water is a nice everyday example to explain internal energy, heat, and work. However, it's not a good choice for applying the first law of thermodynamics, because the boiling pot isn't a closed system.

Explanation:

Have a great day and don't forget to rate!!!! and give brainlist!

7 0
3 years ago
What is meant by an acceleration of negative 2metre per second square​
soldi70 [24.7K]

means that a body is in motion, and its velocity is measured in meters per second. And, that velocity is increasing by two meters per second, every second.

8 0
3 years ago
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