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avanturin [10]
3 years ago
14

Two identical freight cars roll without friction (one at 1 m/s, the other at 2 m/s) toward each other on a level track. They col

lide, couple together, and roll away in the direction that _________.
Physics
1 answer:
Kazeer [188]3 years ago
7 0

Answer:

They collide, couple together, and roll away in the direction that <u>the 2m/s car was rolling in.</u>

Explanation:

We should start off with stating that the conservation of momentum is used here.

Momentum = mass * speed

Since, mass of both freight cars is the same, the speed determines which has more momentum.

Thus, the momentum of the 2 m/s freight car is twice that of the 1 m/s freight car.

The final speed is calculated as below:

mass * (velocity of first freight car) + mass * (velocity of second freight car) = (mass of both freight cars) * final velocity

(m * V1) + (m * V2) = (2m * V)

Let's substitute the velocities 1m/s for the first car, and - 2m/s for the second. (since the second is opposite in direction)

We get:

m*1 + m*(-2) = 2m*V

solving this we get:

V = - 0.5 m/s

Thus we can see that both cars will roll away in the direction that the 2 m/s car was going in. (because of the negative sign in the answer)

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Answer:

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Explanation:

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5 0
2 years ago
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What is the formula to determine the mass of the Earth?​
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Answer:

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Explanation:

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2 years ago
Spaceship 1 and Spaceship 2 have equal masses of 300kg. They collide. Spaceship 1's final speed is 3 m/s, and Spaceship 2's fina
fiasKO [112]

Answer:

B. 1500 kg*m/s

Explanation:

Momentum p = m* v

In any type of collision, the total momentum is preserved!

The total momentum before and the total momentum after the collision is the same. We know the mass and speed after the collision so we can calculate the total momentum.

p1 + p2 =

m1*v1 + m2*v2

m1 = me = 300 kg

v1 = 3 m/s

v2 = 2 m/s

Substitute the given numbers:

300*3 + 300+2

900 + 600

1500 kg*m/s, which is answer B.

3 0
2 years ago
A circular parallel plate capacitor is constructed with a radius of 0.52 mm, a plate separation of 0.013 mm, and filled with an
olya-2409 [2.1K]

Answer:

289282

Explanation:

r = Radius of plate = 0.52 mm

d = Plate separation = 0.013 mm

A = Area = \pi r^2

V = Potential applied = 2 mV

k = Dielectric constant = 40

\epsilon_0 = Electric constant = 8.854\times 10^{-12}\ \text{F/m}

Capacitance is given by

C=\dfrac{k\epsilon_0A}{d}

Charge is given by

Q=CV\\\Rightarrow Q=\dfrac{k\epsilon_0AV}{d}\\\Rightarrow Q=\dfrac{40\times 8.854\times 10^{-12}\times\pi \times (0.52\times 10^{-3})^2\times 2\times 10^{-3}}{0.013\times 10^{-3}}\\\Rightarrow Q=4.6285\times 10^{-14}\ \text{C}

Number of electron is given by

n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{4.6285\times 10^{-14}}{1.6\times10^{-19}}\\\Rightarrow n=289281.25\ \text{electrons}

The number of charge carriers that will accumulate on this capacitor is approximately 289282.

6 0
2 years ago
When landing after a spectacular somersault, a 35.0 kg gymnast decelerates by pushing straight down on the mat. calculate the fo
Sliva [168]

The deceleration experienced by the gymnast is the 9 times of the acceleration due to gravity.

Now from Newton`s  first law, the net force on gymnast,

F_{net} =F-W=ma

Here, W is the weight of the gymnast and a is the acceleration experienced by the gymnast (9\times g acceleration due to gravity)  

Therefore,

F= ma+W OR F=ma+mg=m(g+a)

Given m = 30 kg anda=9\times g=9\times 9.8 m/s^{2} =88.2 m/s^{2}

Substituting these values in above formula and calculate the force exerted by the gymnast,  

F=(40 kg) (88.2 m/s^{2} +9.8 m/s^{2} )

F=3.537\times10^{3}N

6 0
2 years ago
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