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hammer [34]
3 years ago
7

When discussing engines, the ratio of output work to input work expressed as a percentage is called .

Physics
2 answers:
trapecia [35]3 years ago
4 0

The Answer Is Efficiency

Nikolay [14]3 years ago
3 0
Hello Friend!

Efficiency η is the ratio of input work over the output work. It is defined as:
η=work done/heat absorbed=(Q₂-Q₁)/Q₁ where Q₁ is the heat absorbed and Q₂-Q₁ is the work done.

This ratio is never greater or equal to 1, or 100% because the second law of thermodynamics doesn't allow it.

I Hope my answer has come to your Help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead! :)

(Ps. Mark As Brainliest IF Helped!)

-TheOneAboveAll :D
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RUDIKE [14]

Answer:

k

Explanation:

umm really

5 0
3 years ago
1 What conclusion can you draw about semicircular canals from their name ? A Their shape B Their size Their location in your bod
oee [108]

Answer:

A. Their shape

Explanation:

The name clearly shows that the shape of these canals are semi-circular. The semi-circular canals consist of three tubes filled with fluid and located in the inner ear. They help to maintain balance and transmit impulses through the movement of the fluids. The impulses sent through these fluids are sent to the brain for interpretation.

The function of the canals are not indicated by the name, rather the shape is hinted through the name.

7 0
3 years ago
fuel was consumed at a certain rate of 0.05Kg\s in a rocket engine and ejected as a gas with a speed of4000m\s . Determine the t
ivann1987 [24]

Answer:

Thrust = 200 N

Explanation:

The engine thrust can be found by using the following formula:

Thrust = mv

where,

m = mass flow rate of the fuel = 0.05 kg/s

v = velocity of ejected gases = 4000 m/s

Therefore, using the given values in the equation, we get:

Thrust = (0.05\ kg/s)(4000\ m/s)

<u>Thrust = 200 N</u>

5 0
3 years ago
A generator connected to an RLC circuit has an rms voltage of 140 V and an rms current of 31 mA. Part A
vredina [299]

Answer:

53.06 ohm

Explanation:

We have given the rms voltage V =140 V

And the rms current i=31 mA

So the impedance Z=\frac{V}{i}=\frac{140}{3\times 10^{-3}}=46.666kohm

Resistance is given as R = 3 kohm

And capacitive reactance X_C=6.5kohm

We know that Z=\sqrt{R^2+(X_L-X_C)^2}

46.666=\sqrt{3^2+(X_L-6.5)^2}

Squaring both side 2177.1556=9+(X_L-6.5)^2

(X_L-6.5)^2=2168.155

(X_L-6.5)=46.5634

X_L=53.063ohm

8 0
3 years ago
A 0.453 kg pendulum bob passes through the lowest part of its path at a speed of 2.58 m/s. What is the tension in the pendulum c
harkovskaia [24]

Answer with Explanation:

Mass of pendulum bob, m=0.453 kg

Speed, v_1=2.58 m/s

a.r=75.1 cm=75.1\times 10^{-2}m=0.751 m

1cm=10^{-2} m

Tension in the pendulum cable is given  by

Tension=Centripetal force+force due to gravity

T=\frac{mv^2}{r}+mg

Where g=9.8 m/s^2

Substitute the values

T=\frac{0.453(2.58)^2}{75.1\times 10^{-2}}+0.453\times 9.8

T=8.45 N

b.When the pendulum reaches its highest point,then

Final velocity, v_2=0

According to law of conservation of energy

mgh_1+\frac{1}{2}mv^2_1=mgh_2+\frac{1}{2}mv^2_2

gh_1+\frac{1}{2}v^2_1=gh_2+\frac{1}{2}v^2_2

h_1=0

Substitute the values

9.8\times 0+\frac{1}{2}(2.58)^2=9.8\times h_2+\frac{1}{2}(0)^2

3.3282=9.8h_2

h_2=\frac{3.3282}{9.8}=0.34 m

The angle mad  by cable with the vertical=cos\theta=\frac{0.751-0.34}{0.751}=0.55

\theta=cos^{-1}(0.55)=56.6^{\circ}

c.When the pendulum reaches at highest point then

Acceleration, a=0

Therefore, the tension  in the pendulum cable

T=mgcos\theta

Substitute the values

T=0.453\times 9.8cos56.6

T=2.4 N

8 0
4 years ago
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