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Veseljchak [2.6K]
3 years ago
5

A gasoline engine absorbs 2 500 J of heat and performs 1 000 J of mechanical work in each cycle. The efficiency of the engine is

:
Physics
1 answer:
larisa86 [58]3 years ago
5 0

Answer:

The efficiency of the engine is 40 %

Explanation:

Given;

input energy, = 2500 J

output energy, = 1000 J

The efficiency of the engine is given as;

Efficiency = output energy / input energy

Efficiency = 1000 / 2500

Efficiency = 0.4

Efficiency(%) = 0.4 x 100%

Efficiency = 40 %

Therefore, the efficiency of the engine is 40 %

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A flea jumps straight up to a maximum height of 0.490 m . What is its initial velocity v0 as it leaves the ground?
zzz [600]

Answer:

\huge\boxed{\sf v_o=3.1\ m/s}

Explanation:

<u>Given Data:</u>

Acceleration due to gravity = g = -9.8 m/s²

Maximum Height = h = 0.490 m

At h, v_f = 0

<u>Required:</u>

v_o=?

<u>Formula:</u>

2gh=v_f^2-v_o^2

<u>Solution:</u>

Put the givens

2 (-9.8) (0.490) = (0)\² - v_o^2\\\\-9.604=-v_o^2\\\\9.604=v_o^2\\\\Take \ sqrt\ on \ both \ sides\\\\\sqrt{9.604}=v_o^2\\\\3.1 \ m/s=v_o\\\\v_o=3.1\ m/s\\\\\rule[225]{225}{2}

3 0
2 years ago
Suppose a metal sphere is launched up a ramp with Vi=1.5 m/s. The end of the ramp is 1.20 m above the floor. The ramp is at an a
Mekhanik [1.2K]
The sphere will go up until all the initial kinetic energy be transformed into potential energy.

Intital kinetic energy = m*(vi)^2 / 2


Final potential energy = mgh

mgh = m(vi)^2 / 2 => h = (vi)^2 / (2g)

g = 9.81 m/s^2
vi = (1.5m/s)^2

h = (1.5m/s)^2 / (2*9.81m/s)^2 = 0.115 m

The range is the distance run over the ramp

Using trigonometry, sin(20°) = h /run => run = h / sin(20) = 0.115m / sin(20) = 0.336 m

Answer: 0.336 m
3 0
4 years ago
Heat engines convert which type of energy into mechanical work
KonstantinChe [14]

Answer: Heat engines convert chemical energy to mechanical energy

Explanation:

7 0
3 years ago
If a car used 260,000 W of power to complete a race in 15 s, how much work did the car do?
suter [353]

Answer: 3.9 MW

Explanation:

1 W = 1 J/s

260000 J/s (15 s) = 3,900,000 = 3.9 MW

3 0
3 years ago
Explain how low atmospheric oxygen slowed alligator growth in terms of cellular respiration.
sveta [45]

Answer:

Explanation:

The quantity of oxygen in the atmosphere influences the development trajectory, cardiovascular allometry, and metabolic rate of the American alligator. The growth of hypoxic alligators is hampered by a lack of oxygen in the atmosphere, which may impair their food use capabilities.

5 0
3 years ago
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