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Rama09 [41]
3 years ago
6

Before the collision, the mass of the car is 1500kg, and the speed is 36km/hr, and the car is going to the north. What is its mo

mentum? Is its momentum a vector? If it is a vector, indicate it's direction and magnitude. (We can define the north as the positive direction in the following calculation)
Physics
2 answers:
blagie [28]3 years ago
4 0

Answer:

Answer is given in the attachment.

Explanation:

Download pdf
Delvig [45]3 years ago
4 0

Answer:

Momentum = 15000 kg·m/s

It is a vector.

Its direction is north and its magnitude is 15000 kg·m/s

Explanation:

Momentum is the product of mass and velocity.

P = mv

We have to convert to SI units.

P = (1500\ \text{kg})\left(\dfrac{36000\ \text{m}}{3600\ \text{s}}\right) = 15000\ \text{kg}\cdot\text{m/s}

Momentum is a vector because it is the product of a scalar (mass) and a vector (velocity). It points in the direction of the velocity, in this case, north. Its magnitude is as found above.

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The impulse given to the ball is equal to the change in its momentum:

J = ∆p = (0.50 kg) (5.6 m/s - 0) = 2.8 kg•m/s

This is also equal to the product of the average force and the time interval ∆t :

J = F(ave) ∆t

so that if F(ave) = 200 N, then

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2 years ago
Suppose we hang a heavy ball with a mass 13 kg (so the weight is ) from a steel wire 3.9 m long that is 3.1 mm in diameter (radi
Katena32 [7]

Answer:

1.635×10^-3m

Explanation:

Young modulus is the ratio of the tensile stress of a material to its tensile strain.

Young modulus = Tensile stress/tensile strain

Tensile stress = Force/Area

Given force = 130N

Area = Πr² = Π×(1.55×10^-3)²

Area = 4.87×10^-6m²

Tensile stress = 130/4.87×10^-6 = 8.39×10^7N/m²

Tensile strain = extension/original length

Tensile strain = e/3.9

Substituting in the young modulus formula given young modulus to be 2×10¹¹N/m²

2×10¹¹N/m² = 8.39×10^7/{e/3.9)}

2×10¹¹ = (8.39×10^7×3.9)/e

2×10¹¹e = 3.27×10^8

e = 3.27×10^8/2×10¹¹

e = 1.635×10^-3m

The stretch of the steel wire will be

1.635×10^-3m

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Answer:

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Answer: Before the jump, the snowboarder would carry potential energy.

During the jump he will carry kinetic energy.

And after the jump, assuming hes at a full stop, he will carry potential energy once again.

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Answer:

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Explanation:

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