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Goshia [24]
3 years ago
7

How can a parked car move?

Physics
1 answer:
MariettaO [177]3 years ago
6 0
Add force to unbalance the forces

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How is work calculated when the force applied is not parallel to the displacement
Leno4ka [110]
Pretty sure it’s Force*Distance*Cos(theta)
7 0
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Read 2 more answers
Which of these is an example of a physical change? Question options: wood decaying iron melting paper burning steel rusting
marin [14]
Im pretty sure burning paper

its either that or melting iron but im 90 percent sure its paper burning
4 0
3 years ago
A clay ball with a mass of 0.48kg has an initial speed of 4.08m/s it strikes a 3.04kg clay ball at rest and the two balls stick
Rom4ik [11]
Final velocity = 0, thus final kinetic energy is 0
Initial kinetic energy:
0.5mv²
= 0.5 x 0.48 x 4.08²
= 4.0 J
Decrease in kinetic energy = 4 - 0 = 4 Joules
5 0
3 years ago
A type of friction that occurs when air pushes against a moving object causing it to negatively accelerate
Romashka [77]

Answer:

Air resistance

Answer B is correct

Explanation:

The friction that occurs when air pushes against a moving object causing it to negatively accelerate is called as air resistance.

hope this helps

brainliest appreciated

good luck! have a nice day!

6 0
3 years ago
On a hot summer day, 3.50 ✕ 106 J of heat transfer into a parked car takes place, increasing its temperature from 36.5°C to 44.4
anygoal [31]

Answer:

a) \Delta s=443037.9747\ J.K^{-1}

b) \Delta s=31868131.8681\ J.K^{-1}

Explanation:

a)

Given:

amount of heat transfer occurred,dQ=3.5\times 10^6\ J

initial temperature of car, T_i=36.5+273=309.5\ K

final temperature of car, T_f=44.4+273=317.4\ K

We know that the change in entropy is given by:

\Delta s=\frac{dQ}{T_f-T_i}

\Delta s=\frac{3.5\times 10^6}{(44.4-36.5)} (heat is transferred into the system of car)

\Delta s=443037.9747\ J.K^{-1}

b)

amount of heat transfer form the system of house, dQ=5.8\times 10^8\ J

initial temperature of house, T_i=23.5+273=296.5\ K

final temperature of house, T_f=5.3+273=278.3\ K

\Delta s=\frac{dQ}{T_f-T_i}

\Delta s=\frac{5.8\times 10^8}{278.3-296.5}

\Delta s=31868131.8681\ J.K^{-1}

6 0
3 years ago
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