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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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a student drops an object from the top of a building which is 19.6 m high. How long does it take the object to fall to the groun
zubka84 [21]

Here's a formula that's simple and useful, and if you're really in
high school physics, I'd be surprised if you haven't see it before. 
This one is so simple and useful that I'd suggest memorizing it,
so it's always in your toolbox.

This formula tells how far an object travels in how much time,
when it's accelerating:

               Distance = (1/2 acceleration) x (Time²).

                           D = 1/2 A T²

For your student who dropped an object out of the window,

     Distance = 19.6 m
     Acceleration = gravity = 9.8 m/s²

                                              D = 1/2 G T²

                                          19.6 =   4.9   T²

Divide each side by 4.9 :       4  =           T²

Square root each side:           2  =          T

When an object is dropped in Earth gravity,
it takes  2  seconds to fall the first 19.6 meters.

8 0
4 years ago
Read 2 more answers
Saturn moves in an orbit around the Sun with radius 10 AU. How many degrees does it move on the Celestial in one year? (Hint: Ca
Lana71 [14]

Answer:

B. About 12 degrees

Explanation:

The orbital period is calculated using the following expression:

T = 2π*(\sqrt{\frac{r^3}{Gm}})

Where r is the distance of the planet to the sun, G is the gravitational constant and m is the mass of the sun.

Now, we don't actually need to solve the values of the constants, since we now that the distance from the sun to Saturn is 10 times the distance from the sun to the earth. We now this because 1 AU is the distance from the earth to the sun.

Now, we divide the expression used to calculate the orbital period of Saturn by the expression used to calculate the orbital period of the earth. Notice that the constants will cancel and we will get the rate of orbital periods in terms of the distances to the sun:

\frac{Tsaturn}{Tearth} = \sqrt{\frac{rSaturn^3}{rEarth^3} } = \sqrt{10^3}}

Knowing that the orbital period of the earth is 1 year, the orbital period of Saturn will be \sqrt{10^3}} years, or 31.62 years.

We find the amount of degrees it moves in 1 year:

1year * \frac{360degrees}{31.62years} = 11.38 degrees

or about 12 degrees.

6 0
3 years ago
How are ocean ridges formed
Basile [38]
The answer is divergent boundaries.
I hope this helps you!
5 0
3 years ago
Read 2 more answers
At a meeting of physics teacher in Montana, the teachers were asked to calculate where a flour sack would land if dropped from a
Harlamova29_29 [7]

At a distance of 469.2 m from the original point below the airplane.

Explanation:

First of all, we have to calculate the time it takes for the sack to reach the ground.

To do so, we just analyze its vertical motion, which is a free-fall motion, so we can use the suvat equation:

s=ut+\frac{1}{2}at^2

where, taking downward as positive direction:

s = 300 m is the vertical displacement

u = 0 is the initial vertical velocity

t is the time

a=g=9.8 m/s^2 is the acceleration of gravity

Solving for t, we find the it takes for the sack to reach the ground:

t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(300)}{9.8}}=7.82 s

Now we analyze the horizontal motion. The horizontal velocity of the pack is constant (since there are no forces along the horizontal direction) and equal to the initial speed of the airplane, so:

v_x = 60 m/s

We also know the total time of flight,

t = 7.82 s

Therefore, we can find the horizontal distance travelled by the sack:

d=v_x t = (60)(7.82)=469.2 m

So, the sack will land 469.2 m from the original point below the airplane.

Learn more about free fall and projectile motion:

brainly.com/question/1748290

brainly.com/question/11042118

brainly.com/question/2455974

brainly.com/question/2607086

brainly.com/question/8751410

#LearnwithBrainly

7 0
3 years ago
What is the kinetic energy of a 120-cm thin uniform rod with a mass of 450 g that is rotating about its center at 3.60 rad/s?
goldfiish [28.3K]

Answer:

1.05 J.

Explanation:

Kinetic Energy: This is the energy possessed by a body due to its motion. The S.I unit of kinetic energy is Joules (J). The formula of kinetic energy is given as

Ek = 1/2mv²................. Equation 1

Where Ek = kinetic energy, m = mass of the uniform rod, v = liner velocity of the rod.

But,

v = αr .......................... Equation 2

Where α = angular velocity of the rod, r = radius of the circle.

Given: α = 3.6 red/s, r = 120/2 = 60 cm = 0.6 m.

Substitute into equation 2

v = 3.6(0.6)

v = 2.16 m/s.

Also given: m = 450 g = 0.45 kg.

Substitute into equation 1

Ek = 1/2(0.45)(2.16²)

Ek = 1.05 J.

4 0
3 years ago
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