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jeka94
3 years ago
13

When a car travels around a curve in the road , what supplies the centripetal force needed ?

Physics
1 answer:
Maksim231197 [3]3 years ago
6 0
The centrifugal force working against the car would be the entire car's weight, so the centripetal force that is supplied to the car must be the car's tires traction, aka the friction of the car's tires and the surface.
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An airplane during departure has a constant acceleration of 3 m / s².
Rama09 [41]

Constant acceleration of plane = 3m/s²

a) Speed of the plane after 4s

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b) Flight point will be termed as the point the plane got initial speed, u, 20m/s

Find speed after 8s, v

a = 3m/s²

from,

a = <u>v</u><u> </u><u>-</u><u> </u><u>u</u>

t

3 = <u>v</u><u> </u><u>-</u><u> </u><u>2</u><u>0</u>

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6 0
2 years ago
A box is moving along the x-axis and its position varies in time according to the expression:
Colt1911 [192]

Answer:

38.4 m/s

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a) at t = 3.2s. x = 6 * 3.2^2 = 61.44 m

b) at t = 3.2 + Δt. x = 6*(3.2 + \Delta t)^2

c) As Δt approaches 0. We can find the velocity by the ratio of Δx/Δt

v = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{\Delta t}

v = \frac{6*(3.2 + \Delta t)^2 - 61.44}{\Delta t}

v = \frac{6(3.2^2 + 6.4\Delta t + \Delta t^2) - 61.44}{\Delta t}

v = \frac{61.44 + 38.4\Delta t + \Delta t^2 - 61.44}{\Delta t}

v = \frac{\Delta t(38.4 + \Delta t)}{\Delta t}

v = 38.4 + \Delta t

As Δt approaches 0, v = 38.4 + 0 = 38.4 m/s

3 0
3 years ago
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