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mixer [17]
3 years ago
9

A baseball on a T-ball stand has no momentum until it is hit with a bat. When Tyler swings the bat, it has a momentum of 12 kg m

/s. After the bat hits the ball, the ball has a momentum of 8 kg m/s. What is the momentum of the bat AFTER it hits the ball?
A) 0 kg m/s
B) 4 kg m/s
C) 5 kg m/s
D) 8 kg m/s
Physics
1 answer:
NeTakaya3 years ago
4 0

Answer:

4 kg m/s (option B)

Explanation:

We use conservation of momentum to solve this problem.

In the initial state, the ball has momentum zero (no momentum), and the bat has a momentum of 12 kg m/s.

Since the momentum of the system that interacts ("bat plus ball"), has to be conserved (that means the final total momentum must equal the initial total momentum), we have:

Total initial momentum = 0 kg m/s + 12 kg m/s = 12 kg m/s

Total final momentum = 8 kg m/s + X

where X is our unknown: the final momentum of the bat, and 8 kg m/s is the final momentum of the ball.

We make these two quantities equal since the total momentum has to e conserved, and solve for the unknown X in the equation:

Total initial momentum = Total final momentum

12 kg m/s = 8 kg m/s + X

12 kg m/s - 8 kg m/s = X

X = 4 kg m/s

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We have 6 levels and we will verify which of them is true.

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According to the conservation of charge, an electric charge can neither be created nor be destroyed.

Therefore this statement is TRUE.

<em>B ) A positive charge and negative charge attract each other</em>

According to the properties of charges, the opposite charges attract each other and similar charges repel each other

Therefore this statement is  TRUE

<em>C) Two positive charges attract each other.</em>

Using the previous law, we can observe that this statement is false, because equal charges repel each other.

Therefore this statement is FALSE.

<em>D ) Two negative charges repel each other.</em>

Using the same principle as the two previous statements, two equal charges repel each other.

Therefore this statement is TRUE

<em>E) Electric charge is quantized.</em>

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Ganezh [65]

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A 221 g cart starts from rest and rolls in the right direction (positive) down an incline. The incline is at a height of 5 cm. A
Julli [10]

Answer:

1) p₀ = 0.219 kg m / s, p = 0, 2)  Δp = -0.219 kg m / s, 3) 100%

Explanation:

For the first part, which is speed just before the crash, we can use energy conservation

Initial. Highest point

            Em₀ = U = mg y

Final. Low point just before the crash

           Emf = K = ½ m v²

          Em₀ = Emf

          m g y = ½ m v²

           v = √ 2 g y

Let's calculate

           v = √ (2 9.8 0.05)

           v = 0.99 m / s

1) the moment before the crash is

           p₀ = m v

           p₀ = 0.221 0.99

           p₀ = 0.219 kg m / s

After the collision, the car's speed is zero, so its moment is zero.

           p = 0

2) change of momentum

           Δp = p - p₀

            Δp = 0- 0.219

            Δp = -0.219 kg m / s

3) the reason is

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In percentage form it is 100%

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3 years ago
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