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solniwko [45]
3 years ago
7

A compound machine is sued to lift a car. If you apply a force of 60 N to the machine, it lifts the car with a force of 550 N.

Physics
2 answers:
faltersainse [42]3 years ago
7 0

Answer:

a) 9.167

b) 402.5 J

Explanation:

We have a machine, which when applying a force of 60N, it returns 550N.

a) The mechanical advantage of a machine is the number of times that the machine increases an input force.

Mathematically, is the relationship between the output force and the input force.

The mechanical advantage is dimensionless.

In this case, the input force (the force we apply) is 60 N; and the output force (the force that the compound machine exerts) is 550 N.

We solve it this way:

M.A.=\frac{550N}{60N} =9.167

This is the mechanical advantage of the machine in this exercise.

b)In a machine, we define the efficiency as the relationship between the work we receive from the machine, and the work we give to it.

The machine efficiency is a real number between 0 and 1 and is dimensionless.

In this exercise, the efficiency is 35% = 0.35

We solve it this way :

\frac{WorkWeReceive}{WorkWeGive}=0.35

\frac{WorkWeReceive}{1150J}=0.35

WorkWeReceive=(0.35).(1150J)=402.5J

We obtain 402.5 J useful work from the machine.

mrs_skeptik [129]3 years ago
3 0

a) MA=550/60=9.17

b) Wu=1150*0.35=402.5 J

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Q2. You push a crate up a ramp with a force of 10 N. Despite your pushing, the crate slides down the ramp 4 m. How much work did
Ksivusya [100]

Answer:

40 J

Explanation:

From the question given above, the following data were obtained:

Force (F) = 10 N

Distance (s) = 4 m

Workdone (Wd) =?

Work done is simply defined as the product of force and distance moved in the direction of the force. Mathematically, we can express the Workdone as:

Workdone = force × distance

Wd = F × s

With the above formula, we can obtain the workdone as follow:

Force (F) = 10 N

Distance (s) = 4 m

Workdone (Wd) =?

Wd = F × s

Wd = 10 × 4

Wd = 40 J

Thus, 40 J of work was done.

5 0
3 years ago
Can you help quick in science please
nikdorinn [45]
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4 0
3 years ago
The sound intensity at a distance 2.00 m from a sound source is 5.00 Find the total sound energy emitted by the source in each s
notka56 [123]

Answer:

     P = 251, 3 W

Explanation:

The intensity is defined as the power emitted per unit area

           I = P / A

Since sound is distributed in all directions spherical shape, the area of ​​a sphere is

           A = 4π r²

let's clear the power and replace

         P = I A

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6 0
3 years ago
A farmer hitches her tractor to a sled loaded with firewood and pulls it a distance
Delvig [45]

(a) The work done by the force applied by the tractor is 79,968.47 J.

(b) The work done by the frictional force on the tractor is 55,977.93 J.

(c) The total work done by  all the forces is 23,990.54 J.

<h3>Work done by the applied force</h3>

The work done by the force applied by the tractor is calculated as follows;

W = Fd cosθ

W = (5000 x 20) x cos(36.9)

W = 79,968.47 J

<h3>Work done by frictional force</h3>

W = Ffd cosθ

W = (3500 x 20) x cos(36.9)

W = 55,977.93 J

<h3>Net work done by all the forces on the tractor</h3>

W(net) = work done by applied force  -  work done by friction force

W(net) = 79,968.47 J -  55,977.93 J

W(net) = 23,990.54 J

Learn more about work done here: brainly.com/question/25573309

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4 0
2 years ago
Why do some nucleus release electrons?
fredd [130]

Answer:

Electrons are not little balls that can fall into the nucleus under electrostatic attraction

Explanation:

5 0
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