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user100 [1]
4 years ago
5

An engine has an energy input of 125 J, and 35 J of that energy is transformed into useful energy. What is the efficiency of the

engine?
Physics
2 answers:
Korolek [52]4 years ago
8 0
Efficiency = (useful output) / (input)

Efficiency = (35 J) / (125 J) = 0.28  =  28%
Anna007 [38]4 years ago
8 0

Efficiency is equal to 35 / 125 * 100 which is equal to 28%.


Efficiency is the ratio of the useful work performed by a machine or in a process to the total energy that it consumed.


So basically it is the useful output which in this case was 35 J divided by the total input given to the machine which was 125 J in this case multiplying by 100 to get the percentage.


If we did not need the percentage then we could have simply not multiplied it by 100.

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Answer: barometer.

A cyclone is a storm or system of winds that rotates around a center of low atmospheric pressure. An anticyclone is a system of winds that rotates around a center of high atmospheric pressure.

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3 years ago
3 ways how natural disasters impact environment.
Trava [24]
1. When sewage treatment plants flood or debris reaches reservoirs and streams, the quality of the water is affected.

2. Storm surges cause beaches to shift and alter shape.

3. During flash floods, riverbanks erode.
5 0
2 years ago
A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from
Ulleksa [173]

Answer:

1270.64\ \text{J}

Explanation:

m = Mass of object = \dfrac{mg}{g}

mg = Weight of object = 20 N

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

\theta = Angle of slope = 30^{\circ}

f = Force of friction

fd = Work done against friction

The force balance of the system is

\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}

The work done against friction is 1270.64\ \text{J}.

8 0
3 years ago
A busy chipmunk runs back and forth along a straight line of acorns that has been set out between his burrow and a nearby tree.
saveliy_v [14]

Answer: a = 1.32m/s2

Therefore, the average acceleration is 1.32m/s2

Explanation:

Acceleration is the rate of change in the velocity per time

a = change in velocity/time

a = ∆v/t

average acceleration a = (v2 -v1)/t. ....1

Given;

Final velocity v2 = 1.63m/s

Initial velocity v1 = -1.15ms

time taken t = 2.11s

Substituting into eqn 1

a = [1.63 - (-1.15)]/2.11

a = (1.63+1.15)/2.11

a = 2.78/2.11

a = 1.32m/s2

Therefore, the average acceleration is 1.32m/s2

6 0
4 years ago
Arrange the examples in order, starting with the object that has the least amount of energy. In each case, assume there’s no fri
Artemon [7]
First example: book, m= 0.75 kg, h=1.5 m, g= 9.8 m/s², it has only potential energy Ep,

Ep=m*g*h=0.75*9.8*1.5=11.025 J

Second example: brick, m=2.5 kg, v=10 m/s, h=4 m, it has potential energy Ep and kinetic energy Ek,

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Third example: ball, m=0.25 kg, v= 10 m/s, it has only kinetic energy Ek

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Fourth example: stone, m=0.7 kg, h=7 m, it has only potential energy Ep,

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The order of examples starting with the lowest energy:

1. book, 2. ball, 3. stone, 4. brick 


4 0
3 years ago
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