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pashok25 [27]
2 years ago
12

A pulley system used to lift car tires has a mechanical advantage of 11.2. If you pull on the pulley with a force of 150 N, how

much force does the pulley exert on the tire?
Physics
2 answers:
Novay_Z [31]2 years ago
7 0
1680 is the answer i got because i did this problem in apex
IgorC [24]2 years ago
5 0

Answer:

Load = 1680 N

Explanation:

Mechanical advantage of the pulley system, MA = 11.2

If a person pull on a pulley with a force of 150 N we have to find the amount of force the pulley exert on the tire. Mechanical advantage of a pulley system is given by the ratio of load to the effort force i.e.

IMA=\dfrac{load}{effort}

11.2=\dfrac{load}{150\ N}

Load = 1680 N

Hence, the force the pulley exert on the tire is 1680 Newtons.

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m = mass = 5 kg

v_{i} = initial velocity = 100 m/s

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I = impulse = 30 Ns

Using the impulse-change in momentum equation

I = m(v_{f} - v_{i})

30 = 5 (v_{f} - 100)

v_{f} = 106 m/s

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In the late 19th century, great interest was directed toward the study of electrical discharges in gases and the nature of so-ca
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A) He finds the same value of q / m for different materials , B)      y = ½ (q / m) E L² / v₀ₓ² , C) v = E / B , D)   B = 2.13 10⁻⁶ T, E) For the first part I have two off-center points., For the second part I can center one point but the other is off center

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Explanation:

Part A

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When examining his statements the correct one is: He finds the same value of q / m for different materials

Part B

For this part let's use Newton's second law

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        q E = m a

        a = q / m E

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           y = ½ a (x / v₀ₓ)²

We replace

           y = ½ (q / m) E L² / v₀ₓ²

Part C

If there is no deflection, the electric and magnetic forces are the same and in the opposite direction

         Fm = Fe

         q v B = q E

          v = E / B

Part D

       

        We replace

        y = ½ (q / m) E L² / (E / B)²

         y = ½ (q / m) L² B² / E

If we do not want any deflection the magnetic field has to return the electrons the amount that they lower y = -4.12 cm

      -4.12 10⁻² = ½ q / m 0.12² B² / 1.1 10³

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      q / m = -1.758 10¹¹ C/ kg

      B = √ 0.259 1.758 10¹¹ = √ 4.55 10⁻¹²

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Part E

As the charge that the two particles is different

For the first part I have two off-center points.

For the second part I can center one point but the other is off center

Therefore the third statement is correct

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A historical society is testing an old cannon. They
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