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Lena [83]
2 years ago
12

d. e. Study the given diagram and calculate the following: i. work done by load ii. work done by effort iii. M.A iv. V.R v. effi

ciency [Friction is neglected] ​

Physics
1 answer:
il63 [147K]2 years ago
4 0

i. The work done by the load is load x distance moved by load.

ii. The work done by effort is effort applied x distance moved by effort.

iii. The mechanical advantage of the simple machine is Load/effort.

iv. The velocity ratio of the simple machine is 2.

v. The efficiency of the machine is M.A/V.R x 100%.

<h3>Work done by the load</h3>

The work done by the load is the product of the load and the distance through which the load is moved. The magnitude is calculated as follows;

Work done by the load = load x distance moved by load

<h3>Work done by effort</h3>

The work done by the effort is the product of the effort and the distance through which the effort is applied. The magnitude is calculated as follows;

Work done by effort = effort applied x distance moved by effort

<h3>Mechanical advantage of the simple machine</h3>

M.A = Load/Effort

<h3>Velocity ratio of the simple machine</h3>

V.R = distance moved by effort / distance moved by load

V.R  = 30 cm/15 cm

V.R = 2

<h3>Efficiency of the machine</h3>

E = (M.A/V.R) x 100%

Thus, the work done by the load is load x distance moved by load.

The work done by effort is effort applied x distance moved by effort.

The mechanical advantage of the simple machine is Load/effort.

The velocity ratio of the simple machine is 2.

The efficiency of the machine is M.A/V.R x 100%.

Learn more about mechanical advantage here: brainly.com/question/25984831

#SPJ1

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So, we can write the following:

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If the track is horizontal, this means that thre is no change in gravitational potential energy, so any loss of energy must be kinetic energy.

Before the collision, the total kinetic energy of the system was the following:

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