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3 years ago

A car sets off from the traffic lights. It reaches a speed of 27 m/s in 18 seconds. What is its acceleration?

Physics

1 answer:

Katarina [22]3 years ago

6 0

Answer:

1.6875m/s²

Explanation:

$acceleration = \frac{final \: velocity \: - \: initial \: velocity}{time \: taken}$

a = unknown.

v (final velocity) = 27m/s

u (initial velocity) = 0m/s (as the car was stationary in the traffic lights).

t (time taken) = 18s

$a = \frac{27 - 0}{16}$

acceleration = 1.6875m/s²

hope it helps. :)

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Because the disturbances are in opposite directions for this superposition, the resulting amplitude is zero for pure destructive interference

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3 years ago

An uncharged metal sphere, A, is on an insulating base. A second sphere, B, of the same size, shape, and material carrying charg

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Answer:

Explanation:

If we bring the charged sphere B close to, but not touching it , to the uncharged sphere A, as charges can move freely on the conductor, a charge -Q will be built on the outer surface of the sphere A, facing to sphere B.
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Q +(-Q) =0.

If we now remove sphere B, and place it far away, there will be a charge redistribution within sphere A, making to disappear the separation between Q and -Q.
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3 years ago

When you lift an object by moving only your forearm, the main lifting muscle in your arm is the biceps. Suppose the mass of a fo

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Answer:

The answers to the questions are;

The force the biceps must exert to hold a 850 g ball at the end of the forearm at distance dball = 34.0 cm from the elbow, with the forearm parallel to the floor is 166.77 N Upwards.

The force the elbow must exert is 145.6785 N downwards.

Explanation:

We are required to analyze the system using the principle of moments as follows.

Mass of forearm = 1.30 kg

Weight of forearm = mass × acceleration due to gravity = 1.30 kg × 9.81 m/s²= 12.753 N

Location of point of action of weight of forearm = midpoint 17 cm

Forearm to biceps distance = 3.00 cm from the elbow.

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Weight of ball = mass × acceleration due to gravity = 0.850 kg × 9.81 = 8.3385 N

Position of the ball = end of the forearm

Therefore, taking moment about the elbow, we have

∑Mₓ = 0, we take clockwise moment to be positive, therefore

8.3385 N× 34.0 cm + 12.753 N× 17 cm - $F_{biceps}$ ×3.00 cm = 0

Therefore $F_{biceps}$ ×3.00 cm = 500.31 N cm

Therefore $F_{biceps}$ = (500.31 N·cm)/(3.00 cm) = 166.77 N Upwards

The force exerted by the elbow is given by,

Taking moment about the point of contact between the biceps and the forearm, we get

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$F_{elbow}$ = 145.6785 N downwards.

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3 years ago