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Ulleksa [173]
3 years ago
7

Calculate the mechanical advantage of a lever that has an input force of 15 N and an output force of 60 N. *

Physics
2 answers:
svp [43]3 years ago
5 0

Answer:

Explanation:

Given that, .

Input effort is 15N

Fi = 15N

Output effort is 60N

Fo = 60N

Then,

Mechanical advantage(M•A) is the ratio of two forces, and it is also referred to has force ratio.

M•A = Output Force / Input Force

M•A = Fo / Fi

M•A = 60 / 15

M•A = 4

The mechanical advantage is 4

andreev551 [17]3 years ago
4 0

Answer:

4

Explanation:

Mechanical Advantage is given as a ratio of the force a machines gives out to the force that it receives.

Simply put, it is the ratio of output force to input force:

MA = \frac{OF}{IF}

We are given that:

OF = 60 N

IF = 15 N

Hence, the Mechanical Advantage of the lever is:

MA = \frac{60}{15} \\\\\\MA = 4

It is a ratio so it has no unit.

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2,4

Explanation:

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7 0
3 years ago
While eating lunch high up in a skyscraper, two construction workers calculate their gravitational potential
Maurinko [17]

Answer:

The mass of the other worker is 45 kg

Explanation:

The given parameters are;

The gravitational potential energy of one construction worker = The gravitational potential energy of the other construction worker

The mass of the lighter construction worker, m₁ = 90 kg

The height level of the lighter construction worker's location = h₁

The height level of the other construction worker's location = h₂ = 2·h₁

The gravitational potential energy, P.E.,  is given as follows;

P.E. = m·g·h

Where;

m = The mass of the object at height

g = The acceleration due to gravity

h = The height at which is located

Let P.E.₁ represent the gravitational potential energy of one construction worker and let P.E.₂ represent the gravitational potential energy of the other construction worker

We have;

P.E.₁ = P.E.₂

Therefore;

m₁·g·h₁ = m₂·g·h₂

h₂ = 2·h₁

We have;

m₁·g·h₁ = m₂·g·2·h₁

m₁ = 2·m₂

90 kg = 2 × m₂

m₂ = (90 kg)/2 = 45 kg

The mass of the other construction worker is 45 kg.

8 0
3 years ago
It's not D
DanielleElmas [232]
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5 0
2 years ago
You attach a meter stick to an oak tree, such that the top of the meter stick is 2.27 meters above the ground. later, an acorn f
Alexandra [31]

The acorn was at a height of <u>4.15 m</u> from the ground before it drops.

The acorn takes a time t to fall through a distance h₁, which is the length of the scale. When the acorn reaches the top of the scale, its velocity is u.

Calculate the speed of the acorn at the top of the scale, using the equation of motion,

s=ut+ \frac{1}{2} at^2

Since the acorn falls freely under gravity, its acceleration is equal to the acceleration due to gravity g.

Substitute 2.27 m for s (=h₁), 0.301 s for t and 9.8 m/s² for a (=g).

s=ut+ \frac{1}{2} at^2\\ (2.27 m)=u(0.301s)+\frac{1}{2}(9.8m/s^2)(0.301s)^2\\ u=\frac{1.8261m}{0.301s} =6.067m/s

If the acorn starts from rest and reaches a speed of 6.067 m/s at the top of the scale, it would have fallen a distance h₂ to achieve this speed.

Use the equation of motion,

v^2=u^2+2as

Substitute 6.067 m/s for v, 0 m/s for u, 9.8 m/s² for a (=g) and h₂ for s.

v^2=u^2+2as\\ (6.067m/s)^2=(0m/s)^2+2(9.8m/s^2)h_2\\ h_2=\frac{(6.067m/s)^2}{2(9.8m/s^2)} =1.878 m

The height h above the ground at which the acorn was is given by,

h=h_1+h_2=(2.27 m)+(1.878 m)=4.148 m

The acorn was at a height <u>4.15m</u> from the ground before dropping down.

3 0
3 years ago
An object travels with velocity v = 4.0 meters/second and it makes an angle of 60.0° with the positive direction of the y-axis. 
Darya [45]
Goooooooaaaaalllll , C 2.5 meters
7 0
3 years ago
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