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Katen [24]
2 years ago
5

How much tension must a cable withstand if it is used to accelerate a 1200-kg car vertically upwards at 0.7 m/s^2

Physics
1 answer:
Naddik [55]2 years ago
7 0

Answer:

M a = T - M g          acceleration of car where T is tension in cable

T = M (g + a) = M (9.80 + .7) = M * 10.5

T = 1200 * 10.5 = 12,600 N

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A 35.9 g mass is attached to a horizontal spring with a spring constant of 18.4 N/m and released from rest with an amplitude of
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Answer:

7.74m/s

Explanation:

Mass = 35.9g = 0.0359kg

A = 39.5cm = 0.395m

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At equilibrium position, there's total conservation of energy.

Total energy = kinetic energy + potential energy

Total Energy = K.E + P.E

½KA² = ½mv² + ½kx²

½KA² = ½(mv² + kx²)

KA² = mv² + kx²

Collect like terms

KA² - Kx² = mv²

K(A² - x²) = mv²

V² = k/m (A² - x²)

V = √(K/m (A² - x²) )

note x = ½A

V = √(k/m (A² - (½A)²)

V = √(k/m (A² - A²/4))

Resolve the fraction between A.

V = √(¾. K/m. A² )

V = √(¾ * (18.4/0.0359)*(0.395)²)

V = √(0.75 * 512.53 * 0.156)

V = √(59.966)

V = 7.74m/s

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3 years ago
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