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A train car has a mass of 10,000 kg and is moving at +3.0 m/s. It strikes an identical train car that is at rest. The train cars

Sergeeva-Olga [200]

Answer:

ok what is the question

3 0

3 years ago

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When the switch is closed, what happens to the current in the circuit? (The intrinsic resistance of the inductor is zero.)

Hitman42 [59]

The best and most correct answer among the choices provided by your question is the second choice or letter B.

<span>When the switch is closed, the current instantaneously becomes 1.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

6 0

2 years ago

⚠️ PLEASE HELP ⚠️

Svetllana [295]

Answer:

-79.6

-80

-81.2

Explanation:

5 0

3 years ago

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The strength of a magnetic field follows _____.

ycow [4]

I think its Coulomb's law<span>
</span>

8 0

3 years ago

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1.)Two objects, one of m=20,000 kg, and another of 12,500 kg, are placed at a distance of 5 meters apart. What is the force of g

Delvig [45]

1) $6.67\cdot 10^{-4} N$

The force of gravitation between the two objects is given by:

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} kg^{-1} m^{3} s^{-2}$ is the gravitational constant

m1 = 20,000 kg is the mass of the first object

m2 = 12,500 kg is the mass of the second object

r = 5 m is the distance between the two objects

Substituting the numbers inside the equation, we find

$F=(6.67\cdot 10^{-11})\frac{(20,000 kg)(12,500 kg)}{(5 m)^2}=6.67\cdot 10^{-4} N$

2) $2.7\cdot 10^{-3} N$

From the formula in exercise 1), we see that the force is inversely proportional to the square of the distance:

$F \sim \frac{1}{r^2}$

this means that if we cut in a half the distance without changing the masses, the magnitude of the forces changes by a factor

$F'\sim \frac{1}{(r/2)^2}=4 \frac{1}{r^2}=4F$

So, the gravitational force increases by a factor 4. Therefore, the new force will be

$F' = 4 F=4(6.67\cdot 10^{-4} N)=2.7\cdot 10^{-3} N$

3) 12.5 Nm

The torque is equal to the product between the magnitude of the perpendicular force and the distance between the point of application of the force and the centre of rotation:

$\tau=Fd$