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earnstyle [38]
2 years ago
10

A machine has a mechanical advantage of 5. What force should be applied to the machine to make it apply 3000 N to an object? 0.

002 N 600 N 3000 N 15,000 N.
Physics
1 answer:
yuradex [85]2 years ago
5 0

The magnitude of force needed to be applied on the machine is of 15,000 N.

How mechanical advantage is related to output force?

The ratio of force output to the applied force gives the value of the mechanical advantage. In other words, the mechanical advantage has a direct relation with the force output.

Given data:

The value of the mechanical advantage is, MA = 5.

The magnitude of applied force is, F = 3000 N.

Since the output force is the force exerted on the machine. Then the relation is given as,

MA = F'/F

here,

F' is the magnitude of the force exerted on the machine.

Solving as,

5 = F'/3000

F' = 5 × 3000

F' = 15,000 N

Thus, we can conclude that the magnitude of force needed to be applied on the machine is of 15,000 N.

Learn more about the mechanical advantage here:

brainly.com/question/200179

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Answer:

150000000000 m

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Explanation:

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Distance=3\times 10^8\times \frac{1000}{2}=150000000000\ m

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Time = Distance / Speed

Time=\frac{75}{3\times 10^8}=0.00000025\ seconds

Echo time will be twice the time

Echo\ time=2\times 0.00000025=0.0000005\ seconds

The echo time will be 0.0000005 seconds

Difference in time = Difference in distance / Speed

\Delta t=\frac{\Delta d}{v}\\\Rightarrow \Delta t=\frac{10}{3\times 10^8}\\\Rightarrow \Delta=33.33\ ns

The accuracy by which I will be able to measure the echo time is 33.33 ns

5 0
4 years ago
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A proton travels with a speed of 5.02×10⁶ m/s in a direction that makes an angle of 60.0° with the direction of a magnetic ficld
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The magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.

Speed of the proton = 5.02 × 10 ⁶ m /a

Angel of between the velocity and the magnetic force = 60 °

The magnitude of magnetic field B = 0.180 T

The magnitude of the magnetic force on the proton is,

F = q(v \times B)

F = qvB \: sin \:  θ

F = 1.6 \times 10 ^{ - 19}  \times 5.02 \times 10 ^{6}  \times 0.180 \times  \: sin \: 60°

= 1.25 \times 10 ^{ - 13}  \: N

Therefore, the magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.

To know more about magnetic force, refer to the below link:

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