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3 years ago

A racecar accelerates from rest at 6.5 m/s2 for 4.1 s. How fast will it be going at the end of that time?

Physics

1 answer:

Pie3 years ago

6 0

Answer:

The final velocity of the car is 26.65 m/s.

Explanation:

Given;

acceleration of the racecar, a = 6.5 m/s²

initial velocity of the car, u = 0

time of motion, t = 4.1 s

The final velocity of the car is given by;

v = u + at

where;

v is the final velocity of the car

suvstitute the givens

v = 0 + (6.5)(4.1)

v = 26.65 m/s.

Therefore, the final velocity of the car is 26.65 m/s.

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The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by

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Complete Question:

A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.

Answer:

Power = 54.07 W

Explanation:

Mass of the block = 10 kg

Angle made with the horizontal, θ = 60°

Distance covered, d = 5 m

Tension in the rope, T = 40 N

Coefficient of kinetic friction, $\mu = 0.2$

Let the Normal reaction = N

The weight of the block acting downwards = mg

The vertical resolution of the 40 N force, $f_{y} = 40sin \theta$

$\sum f(y) = 0$

$N + 40 sin \theta - mg = 0\\N = -40sin60 + 10*9.81 = 0\\N = 63.46 N$

$\sum f(x) = 0$

$40 cos 60 - f_{r} - ma = 0\\ f_{r} = \mu N\\ f_{r} = 0.2 * 63.46\\ f_{r} = 12.69 N\\40cos 60 - 12.69-10a = 0\\7.31 = 10a\\a = 0.731 m/s^{2}$

$v^{2} = u^{2} + 2as\\u = 0 m/s\\v^{2} = 2 * 0.731 * 5\\v^{2} = 7.31\\v = \sqrt{7.31} \\v = 2.704 m/s$

Power, $P = Fvcos \theta$

$P = 40 *2.704 cos60\\P = 54.074 W$

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3 years ago

A 10 kg rock that has been dropped from a 60 meter high cliff experiences an average force of air resistance of 30 N. Calculate

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Answer:

4086 J

Explanation:

The potential energy is transformed to kinetic energy less the frictional energy. Potential energy= mgh where m represent mass, g is acceleration due to gravity and h is the height of cliff

Since we have force of air resistance, work done due to air resistance will be product of force and distance

$mgh-Fh= 0.5mv^{2}= KE$

Substituting 10 Kg for m, 9.81 for g and 60 m for F then the kinetic energy at the bottom will be

KE= 10*9.81*60- (30*60)=4086 J

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