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Mkey [24]
3 years ago
10

For how long a time must a tow truck pull with a force of 550N on a stalled 1200kg car to give it a forward velocity of 2.0m/s?

Physics
1 answer:
horsena [70]3 years ago
5 0

Answer:

4.36 seconds

Explanation:

According to the question;

  • Force is 550 N
  • Mass of the car is 1200 kg
  • Velocity of the car is 2.0 m/s

We are needed to find the time the car must the tow track pull the car.

  • From Newton's second law of motion;
  • Impulsive force, F = Mv÷t , where m is the mass, v is the velocity and t is the time.

Rearranging the formula;

t = mv ÷ F

Thus;

Time = (1200 kg × 2.0 m/s²) ÷ 550 N

        = 4.36 seconds

Thus, the time needed to pull the car is 4.36 seconds

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An ideal monatomic gas at temperature T is held in a container. If the gas is compressed isothermally, that is at constant tempe
OlgaM077 [116]

Answer:

a) 0 J

b) W = nRTln(Vf/Vi)

c) ΔQ = nRTln(Vf/Vi)

d) ΔQ = W

Explanation:

a) To find the change in the internal energy you use the 1st law of thermodynamics:

\Delta U=\Delta Q-W

Q: heat transfer

W: work done by the gas

The gas is compressed isothermally, then, there is no change in the internal energy and you have

ΔU = 0 J

b) The work is done by the gas, not over the gas.

The work is given by the following formula:

\\W=nRTln(\frac{V_f}{V_i})

n: moles

R: ideal gas constant

T: constant temperature

Vf: final volume

Vi: initial volume

Vf < Vi, then W < 0 and the work is done on the gas

c) The gas has been compressed. Thus, its temperature increases and heat has been transferred to the gas.

The amount of heat is equal to the work done W

d)

\Delta U = \Delta Q-W\\\\0=\Delta Q-W\\\\\Delta Q=W=nRTln(\frac{V_f}{V_i})

5 0
3 years ago
Suppose the battery in a clock wears out after moving thousand coulombs of charge through the clock at a rate of 0.5 Ma how long
Ksivusya [100]

Answer:

Hello your question is poorly written below is the complete question

Suppose the battery in a clock wears out after moving Ten thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?

answer :

a) 231.48 days

b) n = 3.125 * 10^15

Explanation:

Battery moved 10,000 coulombs

current rate = 0.5 mA

<u>A) Determine how long the clock run on the battery. use the relation below</u>

q = i * t ----- ( 1 )

q = charge , i = current , t = time

10000 = 0.5 * 10^-3 * t

hence  t = 2 * 10^7 secs

hence the time = 231.48  days

<u>B) Determine how many electrons per second flowed </u>

q = n*e ------ ( 2 )

n = number of electrons

e = 1.6 * 10^-19

q = 0.5 * 10^-3 coulomb ( charge flowing per electron )

back to equation 2

n ( number of electrons ) = q / e = ( 0.5 * 10^-3 ) / ( 1.6 * 10^-19 )

hence : n = 3.125 * 10^15

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p +
I think is that
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