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fenix001 [56]
3 years ago
14

a 1.2x10^3 kilogram car is accelerated uniformly from 10. meters per second to 20 meters per second in 5.0 seconds. what is the

magnitude of the net force acting on the car during this 5.0 second interval
Physics
1 answer:
irinina [24]3 years ago
7 0
Force , F = ma

F =  m(v - u)/t               

Where m = mass in kg, v= final velocity in m/s, u = initial velocity in m/s
t = time, Force is in Newton.

m= 1.2*10³ kg,  u = 10 m/s,  v = 20 m/s, t = 5s

F =  1.2*10³(20 - 10)/5

F = 2.4*10³ N = 2400 N


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Which line represents a stationary object?
sineoko [7]

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I think its B but I may be wrong

3 0
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Read 2 more answers
1 poir
djverab [1.8K]

Answer:

D) 735 J(oules)

Explanation:

Work is defined as force * distance

Force is defined as mass * acceleration

Given a mass of 15 kg and a gravitational acceleration of 9.8 m/s² since the box is being lifted up, the force being applied to the box is 15 kg * 9.8 m/s² = 147 N

Since the distance is 5 meters, the work done is 147 N * 5 m = 735 N/m = 735 J, making D the correct answer.

4 0
3 years ago
What mass of water must evaporate from the skin of a 70.0 kg man to cool his body 1.00 ∘C? The heat of vaporization of water at
romanna [79]

Answer:

100 cc

Explanation:

Heat released in cooling human body by t degree

= mass of the body x specific heat of the body x t

Substituting the data given

Heat released by the body

= 70 x 3480 x 1

= 243600 J

Mass of water to be evaporated

= 243600 / latent heat of vaporization of water

= 243600 / 2420000

= .1 kg

= 100 g

volume of water

= mass / density

= 100 / 1

100 cc

1 / 10 litres.

6 0
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On the kelvin scale what is the freezing point of water
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Hope this helps!
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