What is the acceleration experiance by a car that takes 10s to reach 27m/s from rest

1 answer:

lina2011 [118]3 years ago

4 0

Magnitude of acceleration = (change in speed) / (time for the change).

Change in speed = (27 - 0) = 27 m/s
Time for the change = 10 s

Magnitude of acceleration = (27 m/s) / (10 s) = 2.7 m/s² .

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The system below has a friction force of 25 N acting on the cart which 8 kg. The mass hanging off the edge has a mass of 6 kg. F

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The cart will be pulled to the right by the hanging mass, so by Newton's second law, the net force on the cart is

T - 25 N = (8 kg) a

where T is the tension in the rope and a is the acceleration.

The hanging mass has a net force of

(6 kg) g - T = (6 kg) a

where g = 9.8 m/s².

Adding these equations together eliminates T, and we can solve for a :

(T - 25 N) + ((6 kg) g - T ) = (14 kg) a

33.8 N = (14 kg) a

a = (33.8 N) / (14 kg) ≈ 2.4 m/s²

Then the tension in the rope is

T - 25 N = (8 kg) (2.4 m/s²)

T ≈ 25 N + 19.31 N ≈ 44 N

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An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. What is the minimum coef

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An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7

In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.

As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.

Using the Newton second law;

$\sum F_x = \sum F_y = 0$