1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
gogolik [260]
3 years ago
10

The thermal energy in a heat engine is used to move a piston. Which best describes why this is possible?

Physics
2 answers:
vichka [17]3 years ago
7 0

Answer:

the answer is a

Explanation:

just took the test

fiasKO [112]3 years ago
3 0
Hot air always goes for the top of air and its making pressure on piston which makes piston move up.
You might be interested in
A force of 20 N acts upon a 5 kg block. What is the acceleration of the object.
horsena [70]

Answer:

4m/s/s

Explanation:

a=f/m

m=5kg

f=20N

20/5=4

(N=kg-m/s/s)

7 0
2 years ago
Read 2 more answers
As the pendulum swings through the origin, it has a speed of v0 = 1.1 m/s. If the mass on the end of the pendulum is 15 g, and t
serg [7]

Answer:

1: 6.18 cm

2: 52.5609 degrees

Explanation:

We have the pendulum speed at the origin, and in that moment, all energy is kinetic, so we can calculate the pendulum energy by:

Ec = 0.5*m*v^2 = 0.5*0.015*1.1^2 = 0.0091 J

Now with that energy, we can calculate the height the pendulum will reach, as in that moment, the kinetic energy is totally converted to gravitational potencial energy:

Eg = m*g*h = 0.0091

0.015 * 9.81 * h = 0.0091

h = 0.0091 / (0.015 * 9.81 ) = 0.0618 m = 6.18 cm

Looking at the image attached, we can see that the pendulum will form a triangle, and one of the cathetus will be the length of the pendulum minus the height it went up, and the hypotenusa will be the pendulum length.

So, we know that the sine of the angle will be the division between the opposite cathetus and the hypotenusa:

sin(angle) = (30-6.18)/30 = 23.82/30 = 0.794 -> angle = 52.5609 degrees

4 0
3 years ago
Four kilograms of carbon monoxide (co) is contained in a rigid tank with a volume of 1 m3. the tank is fitted with a paddle whee
Alenkasestr [34]
Part (a): Specific volume
Specific volume, v = V/m, V = Volume of the tank, m = mass of CO in the tank
Therefore,
v = 1/4 = 0.25 m^3/kg

Part (b): Energy transferred in kJ
Work done, W = PΔt, where P = power = 14 W = 14 J/s, t = time = 1 hour = 60*60 = 3600 seconds
Therefore,
W = 14*3600 = 50400 J = 50.4 kJ

Part (c): Energy transferred by heat
ΔU = Q + W
Then,
Heat transferred by heat, Q = ΔU - W
But, ΔU = mΔu = 4 kg*10 kJ/kg = 40 kJ
Therefore,
Q = 40 - 50.4 = -10.4 kJ (negative sign indicates that heat is removed from the CO).
6 0
3 years ago
It has been proposed that we could explore Mars using inflated balloons to hover just above the surface. The buoyancy of the atm
klio [65]

Answer:

a) mb = 0.0596 kg ; r = 0.974 m

b) a = 754 m/s^2 .. (Upward)

c) mL = 5.96 kg

Explanation:

Given:-

- The density of Mars atmosphere , ρ = 0.0154 kg/m^3

- The surface density of ballon, σ = 5.0g/m^2

Solution:-

(a) What should be the radius and mass of these balloons so they just hover above the surface of Mars?

- We will first isolate a balloon in the Mar's atmosphere and consider the forces acting on the balloon. We have two forces acting on the balloon.

- The weight of the balloon - "W" - i.e ( Tough plastic weight + Gas inside balloon). Since, the balloon is filled with a very light gas we will assume the weight due to gas inside to be negligible. So we have:

                            W = mb*g

Where,  mb: Mass of balloon

             g: Gravitational constant for Mars

- The mass of the balloon can be determined by using the surface density of the tough plastic given as "σ" and assuming the balloon takes a spherical shape when inflated with surface area "As".

                           As = 4πr^2

Where,  r: The radius of balloon

So,                      mb =  4σπr^2

- Substitute the mass of balloon "mb" in the expression developed for weight of the balloon:

                         W = 4*σ*g*πr^2    ......... Eq1

- The weight of the balloon is combated by the buoyant force - "Fb" produced by the volume of Mars atmosphere displaced by the balloon acting in the upward direction:

                        Fb = ρ*Vs*g

Where,    Vs : Volume of sphere = 4/3 πr^3

So,                    Fb = ρ*g*4/3 πr^3   ....... Eq 2        

- Apply the Newton's equilibrium conditions on the balloon in the vertical direction:

                       Fb - W = 0

                       Fb = W

                       ρ*g*4/3 πr^3 = 4*σ*g*πr^2        

                       r = 3σ / ρ

                       r = 3*0.005 / 0.0154

                       r = 0.974 m           .... Answer            

- Use the value of radius "r" and compute the "mb":

                       mb =  4σπr^2

                       mb =  4*0.005*π (0.974)^2  

                       mb = 0.0596 kg   ... Answer  

(b) If we released one of the balloons from part (a) on earth, where the atmospheric density ρ = 1.20kg/m^3, what would be its initial acceleration assuming it was the same size as on Mars? Would it go up or down?

- The similar analysis is to be applied when the balloon of the same size i.e r = 0.974 m and mass mb = 0.0596 kg is inflated on earth with density  ρ = 1.20kg/m^3.

- Now see that the buoyant force acting on the balloon due to earth's atmosphere is different from that found on Mars. So the new buoyant force Fb using Eq2 is:

                       Fb = ρ*g*4/3 πr^3

Where,   g: Gravitational constant on earth = 9.81 m/s^2

                       Fb = (1.20)*(9.81)*(4/3)* π*(0.974)^3

                       Fb = 45.5 N

- Apply the Newton's second law of motion in the vertical direction on the balloon:

                      Fb - W = mb*a

Where,          a: The acceleration of balloon

                     a = (Fb - W) / mb

                     a = Fb/mb - g

                     a = 45.5/0.0596 - 9.81

                    a = 754 m/s^2  (upward) ..... Answer

c), d) If on Mars these balloons have five times the radius found in part (a), how heavy an instrument package could they carry?

- The new radius of the balloon - "R" -is five times what was calculated in part (a):

- Apply the Newton's equilibrium conditions in the vertical direction on the balloon with the addition of downward weight of load "WL":

                     Fb - W - WL = 0

                     WL = Fb - W

                     mL*g = ρ*g*4/3 πR^3 - 4*σ*g*πR^2      

Where,          mL : The mass of load due to instrument package

                     mL =  ρ*4/3 πR^3 - 4*σ*πR^2

                     mL = 0.0154*4/3*π*(5*0.974)^3 - 4*(0.005)*π*(5*0.974)^2    

                     mL = 7.45 - 1.45

                     mL = 5.96 kg   ..... Answer                      

6 0
3 years ago
Read 2 more answers
What is POWER?
Serjik [45]
D.) POWER is the rate at which work is done
7 0
3 years ago
Other questions:
  • Fiberglass, an insulator, can be found in the wals and roofs of some houses and buildings. Why would an insulator be needed insi
    5·2 answers
  • How fast must a bug swim to keep up with the waves it produces? How fast must it move to produce a bow wave?
    8·1 answer
  • A sound source emits 20.0 W of acoustical power spread equally in all directions. The threshold of hearing is 1.0 × 10-12 W/m2.
    12·1 answer
  • The answer<br> May someone please help me out.
    11·2 answers
  • What part of a thunderstorm kills the most people each year?
    15·2 answers
  • When light is reflected by a mirror, the angle of an incidence is always?
    5·1 answer
  • "If we increase the amplitude of transverse waves on a string, what happens to the speed of a wave along the string?"
    13·1 answer
  • PLEASE HELP ASAP!!!
    9·1 answer
  • Write any three applicataion pressure in our daily life ?​
    15·1 answer
  • When you serve the ball, if the ball does not land in the opposite side rectangle this is called a ______. question 2 options: f
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!