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2 years ago

How long with it take an object to accelerate from 6 m/s to 13 m/s at a rate of 1.4 m/s2. *

Physics

1 answer:

Blizzard [7]2 years ago

7 0

Explanation:

hello,

a = ( v - u ) / t

where u is the initial velocity.

and v is the final velocity.

t represents time,

and a represents acceleration.

in this case,

a = 1.4 m/s²

u = 6 m/s

v = 13 m/s

hence,

1.4 = (13 - 6)/t

1.4t = 7

t = 7/1.4

t = 5 s

thank you!

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A 86g ball is dropped vertically to the floor from a height of 2.87m and bounces to a height of 1.28. What is the magnitude of t

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Answer:

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Explanation:

The given mass of the ball, m = 86 g = 0.089 kg

The height from which the ball is dropped, H = 2.87 m

The height to which the ball bounces, h = 1.28 m

Mathematically, we have;

Δp = F·Δt

Where;

Δp = The change in momentum = m·Δv

F = The applied force

Δt = The time of contact with the force

The velocity of the ball just before it touches the ground, v₁ = -√(2·g·H)

The velocity with which the ball leaves, v₂ = √(2·g·h)

The change in momentum, Δp = m·(v₂ - v₁)

∴ Δp = m·(√(2·g·h) - (-√(2·g·H))) = m·(√(2·g·h) +√(2·g·H) )

The impulse, Δp, received by the ball from the floor during the bounce is given as follows;

Δp = 0.089 kg × (√(2 × 9.8 m/s² × 1.28 m) + √(2 × 9.8 m/s² × 2.87 m)) ≈ 1.11329438 m·kg/s

The impulse received by the ball from the floor during the bounce, Δp ≈ 1.11329438 m·kg/s

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3 years ago

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Hence, we can clearly see from the foregoing that the kinetic energy of the ejected photoelectron is proportional to the frequency of the incident photon.

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Aconstant current of 3 Afor 4 hours is required to charge an automotive battery. If the terminal voltage is V, where t is in hou

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Answer:

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Part (b):

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