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borishaifa [10]
3 years ago
10

Fluid clutches are used on equipment that is subjected to?

Physics
1 answer:
mamaluj [8]3 years ago
4 0
Fluid clutches are used on equipment that is subjected to higher torque per unit volume. This type of clutch makes use of an incompressible fluid like oil in order to transfer the input of a movement from a pedal and cylinder to an actuating cylinder. Releasing the this clutch would stop the transfer of power but still allowing the engine to turn continuously. 
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The larger the push, the larger the change in velocity. This is an example of Newton's Second Law of Motion which states that th
Paul [167]
According to Newton, an object will only accelerate if there is a net or unbalanced forceacting upon it. The presence of an unbalanced force will accelerate an object - changing its speed, its direction, or both its speed and direction.
6 0
3 years ago
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On a balanced seesaw, a boy three times as heavy as his partner sits
slega [8]

Answer:

1/3 the distance from the fulcrum

Explanation:

On a balanced seesaw, the torques around the fulcrum calculated on one side and on another side must be equal. This means that:

W_1 d_1 = W_2 d_2

where

W1 is the weight of the boy

d1 is its distance from the fulcrum

W2 is the weight of his partner

d2 is the distance of the partner from the fulcrum

In this problem, we know that the boy is three times as heavy as his partner, so

W_1 = 3 W_2

If we substitute this into the equation, we find:

(3 W_2) d_1 = W_2 d_2

and by simplifying:

3 d_1 = d_2\\d_1 = \frac{1}{3}d_2

which means that the boy sits at 1/3 the distance from the fulcrum.

8 0
3 years ago
A 1,000 kg car is driving on a 15 m high bridge at 5 m/s. What is the kinetic energy of the car?
yanalaym [24]

Answer:

KE=12,500J

Explanation:

The formula for kinetic energy is:

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

We can plug in the given values into the equation:

KE = \frac{1}{2}*1000kg*(5m/s)^2

KE = 500kg*25m^2/s^2

KE=12,500J

7 0
2 years ago
Identify the breed of goat in the picture.
Step2247 [10]

Answer:

look it up im not a sheaperd sorry

Explanation:

4 0
2 years ago
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Objects with masses of 165 kg and a 465 kg are separated by 0.340 m. (a) Find the net gravitational force exerted by these objec
Dmitrij [34]

Answer:

(a) F_{net} = 4.19 x 10^{-5} N, and its direction is towards m_{2}.

(b) It must be placed inside a hollow shell.

Explanation:

Let, m_{1} = 165 kg, m_{2} = 465 kg, m_{3} = 60 kg, and the distance between m_{1} and m_{2} is 0.340 m.

(a) Since m_{3} is placed midway between m_{1} and m_{2}, then its distance to both masses is 0.170 m.

From the Newton's law of universal gravitation,

F = \frac{Gm_{1}m_{2}  }{r^{2} }

Where all variables have their usual meaning.

Then,

a. F_{net} = F_{23} - F_{13}

F_{13} = \frac{6.67*10^{-11}*165*60 }{(0.17)^{2} }

     = 2.25 x 10^{-5} N

F_{23} = \frac{6.67*10^{-11}*465*60 }{(0.17)^{2} }

     = 6.44 x 10^{-5} N

∴ F_{net} =  = 6.44 x 10^{-5} - 2.25 x 10^{-5}

              = 4.19 x 10^{-5} N

The net force exerted by the two masses on the 60 kg object is 4.19 x 10^{-5}  N.

(ii) /F_{net}/ = /F_{23}/ - /F_{13}/

              = 6.44 x 10^{-5} - 2.25 x 10^{-5}

              = 4.19 x 10^{-5} N

(iii) The direction of the net force is to the right i.e towards m_{2}.

(b) For the net force experienced by the 60 kg object to be zero, it must be placed inside a hollow shell.

7 0
3 years ago
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