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Anton [14]
2 years ago
6

a steam engine works on its vicinity and 285 k heat is released with the help of 225 degree centigrade energy absorbed to the sy

stem. what is the efficiency of steam engine
Physics
1 answer:
Studentka2010 [4]2 years ago
7 0

The efficiency of the steam engine is 78.9%.

<h3>What is the efficiency of a machine?</h3>

The efficiency of a machine is a measure of the useful work done by the machine as against the work input into the machine.

  • Efficiency of a machine = work output/work input × 100%

Work output of the steam engine = 225 K

Work input of the steam engine = 285 K

The efficiency of the machine = 225/275 × 100% = 78.9%

Therefore, the efficiency of the steam engine is 78.9%.

Leran more about efficiency of machines at: brainly.com/question/7536036

#SPJ1

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A ray of light traveling in air strikes the surface of a liquid. if the angle of incidence is 29.7◦ and the angle of refraction
lana66690 [7]
When light moves from a medium with higher refractive index to a medium with lower refractive index, the critical angle is the angle above which there is no refracted ray, and it is given by:
\theta_c = \arcsin ( \frac{n_r}{n_i} ) (2)
where n_r is the refractive index of the second medium and n_i is the refractive index of the first medium.

We can find the ratio n_r / n_i by using Snell's law:
n_i \sin \theta_i = n_r \sin \theta_r (1)
where
\theta_i is the angle of incidence
\theta_r is the angle of refraction

By using the data of the problem and re-arranging (1), we find
\frac{n_r}{n_i} =  \frac{\sin \theta_i}{\sin \theta_r} = \frac{\sin 16.3^{\circ}}{\sin 29.7^{\circ}} =0.566

and if we use eq.(2) we can now find the value of the critical angle:
\theta_c = \arcsin ( \frac{n_r}{n_i} ) = \arcsin (0.566) = 34.5^{\circ}
3 0
3 years ago
A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A cons
Virty [35]

Complete Question

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?

Answer:

H_m=1.65m

H_E=1.16307m

Explanation:

From the question we are told that

Mass of ball M=2kg

Length of string L= 2m

Wind force F=13.2N

Generally the equation for \angle \theta is mathematically given as

tan\theta=\frac{F}{mg}

\theta=tan^-^1\frac{F}{mg}

\theta=tan^-^1\frac{13.2}{2*2}

\theta=73.14\textdegree

Max angle =2*\theta= 2*73.14=>146.28\textdegree

Generally the equation for max Height H_m is mathematically given as

H_m=L(1-cos146.28)

H_m=0.9(1+0.8318)

H_m=1.65m

Generally the equation for Equilibrium Height H_E is mathematically given as

H_E=L(1-cos73.14)

H_E=0.9(1+0.2923)

H_E=1.16307m

8 0
2 years ago
Why does the earth bulge at the equator?
sattari [20]

centrifugal force is a fictitious force. What is happening is that since the earth itself is not a rigid body it will deform when under motion. Although gravity attempts to make the earth spherical, as it is rotating the earth deforms, in such away that it flattens to become an oblique spheroid. This happens as the material at the equator must have a net resultant centripetal force (not centrifugal) which causes its position of equilibrium from the center of the earth to be further away than at the poles as they do not have this force as they are not rotating around the center of mass.

4 0
3 years ago
What is a Non-Example of Newton's 1st law?
guajiro [1.7K]

Answer:

Say a 14 year old girl was at a construction site and she was asked to move something like a 10,000 pound brick( one brick). She would be acting on it as the unbalanced force but they would still not change their position.

so to say the girl would be doing everything she could to move that brick but the brick would still be in that same spot so the unbalanced force (the girl) would be acting on the thing that was at rest but it wouldn't move.

so the unbalanced force would not really be acting on the thing at rest; even though the unbalanced force was doing something to the brick.

( just think about it and you will eventually get it...just imagine in your head...)

Explanation:

5 0
2 years ago
A car enters a level, unbanked semi-circular hairpin turn of 100 m radius at a speed of 28 m/s. The coefficient of friction betw
meriva

Answer:

As  28m/s = 28m/s

Explanation:

r = the radius of the curve

m =  the mass of the car

μ = the coefficient of kinetic friction

N = normal reaction

When rounding the curve, the centripetal acceleration is

a = \frac{v^{2}}{r}

therefore

\mu mg = m \frac{v^{2}}{r} \\\\ \mu =  \frac{v^{2}}{rg}

v = \sqrt{\mu rg}

\mu = \sqrt{0.8 \times 100\times9.8} \\\\= 28m/s

As  28m/s = 28m/s

8 0
3 years ago
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