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3 years ago

In the following lever system, what is the counter-clockwise torque?

Physics

2 answers:

djyliett [7]3 years ago

8 0

<h2>Hello!</h2>

The answer is: 300 N.m

Torque is equal to the product between the applied force and the distance from the pivot (where the lever is supported), we are given a lever system with 2 applied forces at different distances from the pivot, and we are asked to find what is the counter-clockwise torque, so:

The counter-clockwise torque is located at the left end of the lever, so:

$Torque=Force*Distance=300N*1m=300N.m$

So, the counter-clockwise torque is equal to 300 N.m.

Also, we can calculate the clockwise torque:

$Torque=Force*Distance=200N*6m=1200N.m$

Have a nice day!

Anna35 [415]3 years ago

5 0

B) 300Nm

I know its 300Nm because I had this on my usatestprep quiz

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You are correct, the answer is C.

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3 years ago

You place a 10 kg block on a ramp with an angle of 20 degrees. You push the block up the ramp giving it an initial velocity of 1

Ainat [17]

Answer:

L = 15.97 m

Explanation:

Given:-

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- The inclination of ramp, θ = 20°

- The initial speed, Vi = 15 m/s

- The coefficient of friction u = 0.4

Find:-

find the total distance the block travels before it turns around and slides back down the ramp.

Solution:-

- The total distance travelled by the block up the ramp is defined when all the kinetic energy is converted into potential energy and work is done against the friction. Final velocity V2 = 0.

- Develop a free body diagram of the block. Resolve the weight "W" of the block normal to the surface of ramp. Then apply equilibrium condition for the block in the direction normal to the surface:

N - W*cos( θ ) = 0

Where, N : The contact force between block and ramp.

N = m*g*cos ( θ )

- The friction force (Ff) is defined as:

Ff = u*N

Ff = u*m*g*cos ( θ )

- Apply the work-energy principle for the block which travels a distance of "L" up the ramp:

K.E i = P.E f + Work done against friction

Where, K.E i = 0.5*m*Vi^2

P.E f = m*g*L*sin( θ )

Work done = Ff*L

- Evaluate "L":

0.5*m*Vi^2 = m*g*L*sin( θ ) + u*m*g*cos ( θ )*L

0.5*Vi^2 = g*L*sin( θ ) + u*g*cos ( θ )*L

0.5*Vi^2 = L [ g*sin( θ ) + u*g*cos ( θ ) ]

L = 0.5*Vi^2 / [ g*sin( θ ) + u*g*cos ( θ ) ]

L = 0.5*15^2 / [ 9.81*sin( 20 ) + 0.4*9.81*cos ( 20 ) ]

L = 15.97 m

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4 years ago

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Nataliya [291]

Answer:

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3 years ago

A rope pulls a 20 kg box up. If the tension in the rope is 275 N, what is the acceleration of the box?

Nana76 [90]

Answer:

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