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Llana [10]
2 years ago
13

Which of the following examples describes a situation where a car is experiencing a net force?

Physics
1 answer:
Lapatulllka [165]2 years ago
8 0
Where are the options?

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50200 J of heat are removed from
Dmitry_Shevchenko [17]

Correct Answer:

3.1375

Explanation:

Use equation Q=mcΔT to find m

Plug in all variables -50200=x\cdot 2000\cdot -8

Answer: 3.1375

4 0
3 years ago
I'm trying to find the potential energy of a planet using formula G M(sun) x m(planet) / r. I have given G - newton's graviation
ludmilkaskok [199]

Answer:

-5.39\times10^{33} J

Explanation:

Potential energy =

-\frac{GMm}{r}\\=-\frac{6.67\times10^{-11}\times1.99\times10^{30}\times5.97\times10^{24}}{147/1\times10^{9}}\\=-5.39\times10^{33} J

7 0
2 years ago
Select all that apply.
Jobisdone [24]
\text{Acceleration}=\frac{d^2}{dt^2}\text{displacement}
So the dimensions of acceleration is LT^{-2}
Any answer that comes under that definition is correct.
6 0
3 years ago
Read 2 more answers
Which statement describes the kinetic energy of a particle?
worty [1.4K]
Kinetic energy has nothing to do with anything other than motion of the particle.
When a particle with velocity v collides another particle(suppose it is at rest for simplication), assuming that there is perfectly elastic collision between them, the velocity of particle which was at rest becomes mv/M ( assuming mass of particle in motion to be m and at rest to be M) from convervation of linear momentum. And all this transfer of energy happens in a fraction of seconds which is not visible to naked eyes.

Hence 1st option is correct!
5 0
3 years ago
A locomotive accelerates a 25-car train along a level track. Every car has a mass of 7.7 ✕ 104 kg and is subject to a friction f
MaRussiya [10]

To solve the problem it is necessary to apply the concepts related to Force of Friction and Tension between the two bodies.

In this way,

The total mass of the cars would be,

m_T = 25(7.7*10^4)Kg

m_T = 1.925*10^6Kg

Therefore the friction force at 29Km / h would be,

f=250v

f= 250*29Km/h

f = 250*29*(\frac{1000m}{1km})(\frac{1h}{3600s})

f = 2013.889N

In this way the tension exerts between first car and locomotive is,

T=m_Ta+f

T=(1.925*10^6)(0.2)+2013.889

T= 3.8701*10^5N

Therefore the tension in the coupling between the car and the locomotive is 3.87*10^5N

6 0
3 years ago
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