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gregori [183]
4 years ago
14

a machine with a mechanical advantage of 2.5 requires an input force of 120 newtons. What output force is applied by this machin

e
Physics
1 answer:
creativ13 [48]4 years ago
4 0
If your machine has a mechanical advantage of 2.5, then WHATEVER force you apply to the input, the force at the output will be 2.5 times as great.

If you apply 1 newton to the machine's input, the output force is

                 (2.5 x 1 newton)  =  2.5 newtons.

If you apply 120 newtons to the machine's input, the output force is

                 (2.5 x 120 newtons)  =  300 newtons.

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A particle that is smaller than an atom or a cluster of particles.
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A block with mass m =6.4 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.28 m.
Zanzabum

Answer

given,

mass of block (m)= 6.4 Kg

spring is stretched to distance, x = 0.28 m

initial velocity = 5.1 m/s

a) computing weight of spring

    k x = m g

k = \dfrac{mg}{x}

k = \dfrac{6.4 \times 9.8}{0.28}

      k = 224 N/m

b) f = \dfrac{\omega}{2\pi}

    \omega = \sqrt{\dfrac{k}{m}}= \sqrt{\dfrac{224}{6.4}} = 5.92 \ rad/s

   f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

   f = \dfrac{1}{2\pi}\sqrt{\dfrac{224}{6.4}}

  f =0.94\ Hz

c)  v_b = -v cos \omega t

    v_b = -5.1 \times cos (5.92 \times 0.42)

    v_b = 4.04\ m/s

d)  a_{max} = v \omega

    a_{max} = 4.04 \times 5.92

    a_{max} =23.94\ m/s^2

e)  Y =- A sin (\omega t)

    A = \dfrac{v}{\omega}

    A = \dfrac{4.04}{5.92}

        A = 0.682 m

    Y =- 0.682 \times sin (5.92 \times 0.42)

    Y =- 0.42

Force =m \omega^2 |Y|

          =6.4 \times 5.92^2\times 0.42

F = 94.20 N

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4 years ago
When you displace an object from its equilibrium position and the force pushing it back toward equilibrium is _____
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Explanation:

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The velocity of a car changes from 20 m/s east to 5 m/s east in 5 seconds. What is the acceleration of the car?
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acceleration =  \frac{ v_{2}- v_{1}  }{t} = \frac{5-20}{5} =-3m/s^{2}
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3 years ago
An automobile with a mass of 1180 kg is traveling at a speed v = [06] ____________________ m/s. a. What is its kinetic energy in
iogann1982 [59]

Complete Question:

An automobile with a mass of 1180 kg is traveling at a speed v =2.51 m/s. What is its kinetic energy in SI units? What speed (m/s) must an 82.7-kg person move to have the same kinetic energy? At what speed (m/s) must is 12.1-g bullet move to have the same kinetic energy? What would be the speed (m/s) of the automobile if its kinetic energy were doubled?

Answer:

a) 3717.1 J b) 9.48 m/s c) 783.8 m/s d) 3.55 m/s

Explanation:

a)

  • By definition, the kinetic energy of a mass m with a speed v, is as follows:

      K = \frac{1}{2} * m *v^{2}

  • if m= 1180 Kg, and v= 2.51 m/s, the kinetic energy can be calculated as follows:

       K = \frac{1}{2} * m *v^{2} =  \frac{1}{2} * 1180 kg*(2.51 m/s)^{2} = 3717.1 J

b)

  • If the kinetic energy must be the same, and m= 82,7 Kg, we can write the following expression:

       K = \frac{1}{2} * m *v^{2} =  \frac{1}{2} * 82.7 kg*((v)(m/s))^{2} = 3717.1 J

  • We can solve the above equation as follows:

        v =\sqrt{\frac{2*K}{m} } = \sqrt{\frac{2*3717.1J}{82.7kg} } = 9.48 m/s

c)

  • If K remains the same, and m = 12.1 g = 0.0121 kg (in SI units). we can solve for v  as follows:

       v =\sqrt{\frac{2*K}{m} } = \sqrt{\frac{2*3717.1J}{0.0121kg} } = 783.8 m/s

d)

  • Now, if the kinetic energy were doubled, we would have the following equation:

       K = \frac{1}{2} * m *v^{2} =  \frac{1}{2} * 1180 kg*((v) m/s)^{2} = 3717.1 J * 2 = 7434.2 J

  • We can solve for the new speed v as follows:

        v =\sqrt{\frac{2*K}{m} } = \sqrt{\frac{2*7434.2J}{1180kg} } = 3.55 m/s

8 0
3 years ago
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