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Anuta_ua [19.1K]
2 years ago
13

How much kinetic energy does a train with a mass of 30,000 kg have moving 2 m/s?

Physics
1 answer:
Fynjy0 [20]2 years ago
6 0

KE = 60,000 J

Explanation:

KE = (1/2)mv^2

= (1/2)(30000 kg)(2 m/s)^2

= 60,000 J

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Car A hits car B (initially at rest and of equal mass) frombehind while going 35 m/s. Immediately after the collision, car Bmove
kolezko [41]

Answer:

The fraction of kinetic energy lost in the collision in term of the initial energy is 0.49.

Explanation:

As the final and initial velocities are known it is possible then the kinetic energy is possible to calculate for each instant.

By definition, the kinetic energy is:

k = 0.5*mV^2

Expressing the initial and final kinetic energy for cars A and B:

ki=0.5*maVa_{i}^2+0.5*mbVb_{i}^2

kf=0.5*maVa_{f}^2+0.5*mbVb_{f}^2

Since the masses are equals:

m=ma=mb

For the known velocities, the kinetics energies result:

ki=0.5*mVa_{i}^2

ki=0.5*m(35 m/s)^2=612.5m^2/s^2*m

kf=0.5*mbVb_{f}^2

kf=0.5*m(25 m/s)^2=312.5m^2/s^2*m

The lost energy in the collision is the difference between the initial and final kinectic energies:

kl=ki-kf

kl = 612.5m^2/s^2*m-312.5 m^2/s^2*m=300 m^2/s^2*m

Finally the relation between the lost and the initial kinetic energy:

kl/ki = 300 m^2/s^2 * m / 612.5 m^2/s^2 * m

kl/ki = 24/49=0.49

7 0
3 years ago
Suppose that the electric field in the Earth's atmosphere is E = 8.60 101 N/C, pointing downward. Determine the electric charge
asambeis [7]

Answer:

q = 3.87 x 10⁵ C

Explanation:

given,

Electric field, E = 8.60 x 10¹ = 86 N/C

radius of earth, R = 6371 Km = 6.371 x 10⁶ m

Coulomb constant, K = 9 x 10⁹ N · m²/C²

Charge on the earth = ?

the electric field at the point

E =\dfrac{kq}{r^2}

q =\dfrac{Er^2}{k}

inserting all the values

q =\dfrac{86\times (6.371\times 10^6)^2}{9\times 10^{9}}

      q = 3.87 x 10⁵ C

The electric charge on the earth is equal to 3.87 x 10⁵ C

4 0
3 years ago
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