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iren2701 [21]
3 years ago
5

In order to increase the kinetic energy of a speeding train by 44%, 42MJ of work must be performed (1 MJ = 106 J). If the final

velocity of the train (after the 42 MJ of work) is 9 m/s, what is the mass of the train in metric tons (1 metric ton = 1000 kg)?
Physics
1 answer:
victus00 [196]3 years ago
7 0

Answer:

Explanation:

100 % increase makes an amount 2 times .

44 % increase = 42 MJ

100 % increase = 95.45 MJ

final kinetic energy = 2 x 95.45 MJ

= 190.9 x 10⁶ J

1/2 m v² = 190.9 x 10⁶

.5 x m x 9² = 190.9 x 10⁶

m = 4.71 x 10⁶ kg

= 4710 metric ton .

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I hope it's help you

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n outer space, a constant net force with a magnitude of 140 N is exerted on a 32.5 kg probe initially at rest. A) What accelerat
musickatia [10]

Answer:

a) 4.31 m/s²

b) 215.5 m

Explanation:

a) According to Newton's first law of motion

The net force applied to particular mass produced acceleration, a, according to

F = ma

F = 140 N

m = 32.5 kg

a = ?

140 = 32.5 × a

a = 140/32.5 = 4.31 m/s²

b) Using the equations of motion, we can obtain the distance travelled by the object in t = 10 s

u = initial velocity of the probe = 0 m/s (since it was initially at rest)

a = 4.31 m/s²

t = 10 s

s = distance travelled = ?

s = ut + at²/2

s = 0 + (4.31×10²)/2 = 215.5 m

7 0
3 years ago
a 100g ice cube at 0 degrees celsius is placed in 650 grams of water at 25 degrees celsius. When the mixture reaches equillibriu
Artyom0805 [142]

Answer:

The latent heat of fusion of water is 334.88 Joules per gram of water.

Explanation:

Let the latent heat of ice be 'x' J/g

1) Thus heat absorbed by 100 gram of ice to get converted into water equals

Q_1=100\times x

2) heat energy required to raise the temperature of water from 0 to 25 degree Celsius equals

Q_2=100\times 4.186\times 11=4604.6Joules

Thus total energy needed equals Q_1+Q_2=100x+4604.6

3) Heat energy released by the decrease in the temperature of water from 25 to 11 degree Celsius is

Q_3=650\times 4.186\times (25-11)\\\\Q_{3}=38092.6Joules

Now by conservation of energy we have

Q_1+Q_2=Q_3\\\\100x+4604.6=38092.6\\\\\therefore x=\frac{38092.6-4604.6}{100}=334.88J/g

6 0
3 years ago
A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Teflon frying p
Semenov [28]

Answer:

f_n=3.75N

Explanation:

From the question we are told that:

Frictional force F=0.150N

Coefficient of kinetic friction \mu=0.04

Generally the equation for Normal for is mathematically given by

 f_n=\frac{F}{\mu}

Therefore

 f_n=\frac{0.150}{0.04}

 f_n=3.75N

5 0
2 years ago
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