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erik [133]
3 years ago
6

The air in a car tire la compressed when the car rolls over a rock. If the air

Physics
1 answer:
stealth61 [152]3 years ago
8 0

Answer:

the signs of heat and work are; -Q and -W

Explanation:

The first law of thermodynamics is given by; ΔU = Q − W

where;

ΔU is the change in internal energy of a system,

Q is the net heat transfer (the sum of all heat transfer into and out of the system)

W is the net work done (the sum of all work done on or by the system).

Now, The system in this case is the tire and since the air gets warmer, heat must have left the system. Therefore Q is negative (-Q).

Since work is done by the system, W remains negative.

Thus, the signs of heat and work are; -Q and - W

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The mass of a car is 2,400 kg and its momentum is 22,240 kg<br> m/s. What is its velocity?
Crazy boy [7]

Answer:

9.27

Explanation:

6 0
3 years ago
What is electrical resistance and types of electrical resistance
bazaltina [42]

Answer:

Electrical Resistance is a measure of the opposition to current flow in an electrical circuit

Types: variable resistance and set resistance

Explanation:

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3 years ago
An ore car of mass 39000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 19 m lower vert
cupoosta [38]

Answer:

The compression in the spring is 5.88 meters.                

Explanation:

Given that,

Mass of the car, m = 39000 kg

Height of the car, h = 19 m

Spring constant of the spring, k=4.2\times 10^5\ N/m

We need to find the compression in the spring in stopping the ore car. It can be done by balancing loss in gravitational potential energy and the increase in elastic energy. So,

mgh=\dfrac{1}{2}kx^2

x is the compression in spring

x=\sqrt{\dfrac{2mgh}{k}} \\\\x=\sqrt{\dfrac{2\times 39000\times 19\times 9.8}{4.2\times 10^5}} \\\\x=5.88\ m

So, the compression in the spring is 5.88 meters.                                                                                                                  

6 0
3 years ago
Rank the six combinations of electric charges on the basis of the electric force acting on q1. Define forces pointing to the rig
riadik2000 [5.3K]

Answer:

Plss see attached file

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3 0
3 years ago
Two identical loudspeakers 2.00 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standi
Ganezh [65]

Answer:

242.85 Hz

Explanation:

For maximum intensity of sound, the path difference,ΔL = (n + 1/2)λ/2 where n = 0,1,2...

Since Abby is standing perpendicular to one speaker, the path length for the sound from the other speaker to him is L₁ = √(2.00² + 5.50²) = √(4.00 + 30.25) = √34.25 = 5.85 m.

The path difference to him is thus ΔL = 5.85 m - 5.50 m = 0.35 m.

Since ΔL = (n + 1/2)λ/2 and for lowest frequency n = 0,

ΔL = (n + 1/2)λ/2 = (0 + 1/2)λ/2 = λ/4

ΔL = λ = v/f and f = v/4ΔL where f = frequency of wave and v = velocity of sound wave = 340 m/s.

f = 340/(4 × 0.35) = 242.85 Hz

5 0
3 years ago
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