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madam [21]
3 years ago
15

A heat engine does 9200 J of work per cycle while absorbing 22.0 kcal of heat from a hightemperature reservoir. What is the effi

ciency of this engine?
Physics
1 answer:
Morgarella [4.7K]3 years ago
4 0

Answer: 9.9%

Explanation: efficiency = (work output /work input) × 100

Note that, 1 kilocalorie = 4184 joules, hence 22kcal = 22× 4184 = 92048 joules.

Work output = 9200 j and work input = 92048 j

Efficiency = (9200/92048) × 100 = 0.099 × 100 = 9.9%

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Elements of Music<br>Timbre<br>Dynamics<br>Rhythm<br>Form<br>Texture<br>Harmony<br>Style​
rodikova [14]

Explanation:

Rhythm, Harmony, Timbre, Dynamics, Texture, and Form

7 0
3 years ago
During heavy rain, a section of a mountainside measuring 2.5 km horizontally, 0.80 km up along the slope, and 2.0 m deep slips i
Crank

Answer:

The mass of the mud is 3040000 kg.

Explanation:

Given that,

length = 2.5 km

Width = 0.80 km

Height = 2.0 m

Length of valley = 0.40 km

Width of valley = 0.40 km

Density = 1900 Kg/m³

Area = 4.0 m²

We need to calculate the mass of the mud

Using formula of density

\rho=\dfrac{m}{V}

m=\rho\times V

Where, V = volume of mud

\rho = density of mud

Put the value into the formula

m=1900\times4.0\times0.40\times10^{3}

m =3040000\ kg

Hence, The mass of the mud is 3040000 kg.

4 0
3 years ago
A 0.14-MIN baseball is dropped from rest. It has a momentum of 0.90 kg⋅m/skg⋅m/s just before it lands on the ground.
nikitadnepr [17]

The time spent in the air by the ball at the given momentum is 6.43 s.

The given parameters;

  • <em>momentum of the ball, P = 0.9 kgm/s</em>
  • <em>weight of the ball, W = 0.14 N</em>

The impulse experienced by the ball is calculated as follows;

Ft = \Delta P

where;

Ft is impulse

\Delta P is change in momentum

The time of motion of the ball is calculated as follows;

t = \frac{\Delta P}{F} \\\\t = \frac{0.9 - 0}{0.14} \\\\t = 6.43 \ s

Thus, the time spent in the air by the ball at the given momentum is 6.43 s.

Learn more here:brainly.com/question/13468390

7 0
2 years ago
a uniform metre rule is pivoted at its centre and weights of 5 N and 12 N are hung at the 3 cm and 5 cm marks respectively. how
schepotkina [342]

In order to balance the stick on the pivot, the total "moments" must be equal on both sides.  A "moment" is (a weight) x (its distance from the center).

for the 5N weight: Moment = (5N) x (3 cm) = 15 N-cm

for the 12N weight: Moment = (12N) x (5 cm) = 60 N-cm

Sum of the moments trying to pull the stick down on that side = 75 N-cm

Whatever we hang on the other side has to provide a moment of 75 N-cm in the other direction.  We have a 25N weight. Where should we hang it ?  

(25N) x (distance from the pivot) = 75 N-cm

Distance from the pivot = (75 N-cm) / (25 N)

<em>Distance from the pivot =  3 cm </em>

8 0
3 years ago
What will be the restoring force if a spring with a spring constant of 45 newtons per meter is pulled 0.03 meters in the downwar
allsm [11]
To calculate for the force in a spring, we use Hooke's Law which relates force and the displacement of the spring. It is said that the force is directly proportional to the displacement. So, it will have the equation F = kx where k is a constant and it is the spring constant.

F = kx
F = 45 N/m (0.03)
F = 1.35 N
8 0
3 years ago
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