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zheka24 [161]
3 years ago
14

Class II levers like ankles and wheelbarrows are useful because they provide mechanical advantage, by amplifying the input force

to provide a greater output force. In other words, we can lift a load without having to lift the full weight of the load. Prove using a mathematical relation that Class II levers provide mechanical advantage. You will need to use the rotational equilibrium condition. It will be helpful to draw a picture indicating the movement of the load. Also, to simplify we can assume the force is still perpendicular to by considering the case when the load just starts to move.

Physics
1 answer:
marusya05 [52]3 years ago
8 0

Answer:

The solution and the explanation are in the Explanation section.

Explanation:

According to the diagram that is in the attached image, the EFFORT force at point A and the load is at O point. The torque due to weight is:

TA = W * (a * cosθ)

The torque due to effort at C point is:

TC = E * (b * cosθ)

The net torque is equal to 0, we have:

Tnet = 0

W * (a * cosθ) - E * (b * cosθ) = 0

E=W\frac{a}{b}

From the figure, you can observe that a/b < 1, thus E < W

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A 2.0 g particle moving at 5.2 m/s makes a perfectly elastic head-on collision with a resting 1.0 g object.
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Answer:

(a) The speed of the first particle is 1.75 m/s. The speed of the second particle is 6.9 m/s after the collision.

(b) The speed of the first particle is 3.45 m/s in the negative direction. The speed of the second particle is 1.73 m/s.

(c) The final kinetic energy of the incident particle in part (a) and part(b) is 0.0031 J and 0.011 J, respectively.

Explanation:

(a)

In an elastic collision, both momentum and energy is conserved.

\vec{P}_{initial} = \vec{P}_{final}\\m_1v_1 = m_1v_1' + m_2v_2'\\K_{initial} = K_{final}\\\frac{1}{2}m_1v_1^2 = \frac{1}{2}m_1v_1'^2 + \frac{1}{2}m_2v_2'^2

Combining these equations will give the speed of the second particle.

v_2' = \frac{2m_1}{m_1 + m_2}v_1 = \frac{2*2}{2+1}(5.2) = 6.9~m/s

We can use this to find the speed of the first particle.

m_1v_1 = m_1v_1' + m_2v_2'\\2(5.2) = 2v_1' + (1)(6.9)\\v_1' = 1.75~m/s

(b)

If m_2 = 10g.

v_2' = \frac{2m_1}{m_1 + m_2}v_1 = \frac{2*2}{2+10}(5.2) = 1.73~m/s

m_1v_1 = m_1v_1' + m_2v_2'\\2(5.2) = 2v_1' + (10)(1.73)\\v_1' = -3.45~m/s

The minus sign indicates that the first particle turns back after the collision.

(c)

The final kinetic energy of the particle in part (a) and part (b) is

K_a = \frac{1}{2}m_1v_1'^2 = \frac{1}{2}(2\times10^{-3})(1.75)^2 = 0.0031 ~J\\K_b = \frac{1}{2}m_2v_1'^2 = \frac{1}{2}(2\times10^{-3})(3.45)^2 = 0.011~J

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A parallel-plate capacitor, with air dielectric, is charged by a battery, after which the battery is disconnected. A slab of gla
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Complete Question

A parallel-plate capacitor, with air dielectric, is charged by a battery, after which the battery is disconnected. A slab of glass dielectric is then slowly inserted between the plates. As it is being inserted,  

A :

a force repels the glass out of the capacitor.  

B :

a force attracts the glass into the capacitor.    

C :

no force acts on the glass.      

D :

a net charge appears on the glass.      

E :

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Answer:

The correct option is B

Explanation:

Generally when the glass dielectric is slowly inserted between the plated,

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