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anyanavicka [17]
2 years ago
14

Cart 1 has a mass of 1 kg. Cart 2 has a mass of 2 kg. They are pushed apart by a spring. The spring exerts 2N of force into cart

1. How much force is exerted into cart 2? Explain.
Physics
1 answer:
almond37 [142]2 years ago
5 0

The force exerted on the cart 2 during the collision is 2 N.

The given parameters:

  • <em>Mass of cart 1 = m1 = 1 kg</em>
  • <em>Mass of cart 2 = m2 = 2kg</em>
  • <em>Force applied on cart 1 = 2 N</em>

According to Newton's third law of motion, action and reaction are equal and opposite. The force exerted on cart 1 is equal in magnitude to the force exerted on cart 2 but in opposite direction.

F_1 = -F_2

F_2 = -F_1\\\\F_2 = -2 \ N\\\\|F_2| = 2 \ N

Thus, the force exerted on the cart 2 during the collision is 2 N.

Learn more about Newton's third law of motion here: brainly.com/question/13874955

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Explanation:

The total moment of inertia of the system is the sum of the moment of inertia of the rod + the moment of inertia of the two balls.

The moment of inertia of the rod about its centre is given by

I_r = \frac{1}{12}ML^2

where

M = 24 kg is the mass of the rod

L = 0.96 m is the length of the rod

Substituting,

I_r = \frac{1}{12}(24)(0.96)^2=1.84 kg m^2

The moment of inertia of one ball is given by

I_b = mr^2

where

m = 50 kg is the mass of the ball

r=\frac{L}{2}=\frac{0.96}{2}=0.48 m is the distance of each ball from the axis of rotation

So we have

I_b = (50)(0.48)^2=11.5 kg m^2

Therefore, the total moment of inertia of the system is

I=I_r + 2I_b = 1.84+ 2(11.5)=24.8 kg m^2

Learn more about inertia:

brainly.com/question/2286502

brainly.com/question/691705

#LearnwithBrainly

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What cause earthquakes?
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A 150 g sample of brass at 100 °C is placed in a Styrofoam cup of water containing 120 mL of water at 10 °C. No heat is lost to
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Answer:

≈19.144°C.

Explanation:

all the details are in the attachment.

Note, that c₁, m₁, t₁ are the parameters of the sample of brass; c₂, m₂ and t₂ are  the parameters of the sample of water.

P.S. change the provided design according Your requirements.

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A 6.00-μf parallel-plate capacitor has charges of 40.0 μc on its plates. how much potential energy is stored in this capacitor?
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Thew energy stored in a capacitor of capacitance C and voltage between the plates V is

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Substituting numerical value

E=\frac{1}{2*6*10^{-6}} (40*10^{-6})^2\\ E=133.33\; \mu J

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