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3 years ago

How much kinetic energy does a person of mass 50kg running at 3m/s have? Provide

Physics

1 answer:

ddd [48]3 years ago

4 0

Answer:

0.225 KJ kinetic energy is used for a person of mass 50 Kg running at speed of 3 m/s.

Explanation:

Given:

mass = 50 kg

velocity = v = 3 m/s

To Find :

Kinetic energy = ?

Solution:

Kinetic energy of an object is the energy that it possesses due to its motion.

The formula for kinetic energy is given as

$\textrm{kinetic energy} = \frac{1}{2}\times mass\times velocity^{2}\\= \frac{1}{2}\times m\times v^{2}\\= \frac{1}{2}\times 50\times 3^{2}\\= 25\times 9\\= 225\ Joules.$

But we required kinetic energy in KiloJoules i.e KJ

1 K = 1000 = $10^{3}$

$\textrm{kinetic energy} = 225\ J\\= 0.225\times 1000\\= 0.225 \times 10^{3}\\= 0.225\ KJ$

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Hey can someone send me the answer to this

bagirrra123 [75]

Cars 'A' and 'C' look like they're moving at the same speed. If their tracks are parallel, then they're also moving with the same velocity.

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3 years ago

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

Eduardwww [97]

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 1,000 kg car is travelling at 6.5 m/s to the North. A 3,500 kg truck is travelling South at the same velocity. What is the tot

Molodets [167]

Answer:

16250 kgm/s due south

Explanation:

Applying,

M = mv................. Equation 1

Where M = momentum, m = mass, v = velocity.

From the car,

Given: m = 1000 kg, v = 6.5 m/s

Substitute these values into equation 1

M = 1000(6.5)

M = 6500 kgm/s

For the truck,

Given: m = 3500 kg, v = 6.5 m/s

Substitute these values into equation 1

M' = 3500(6.5)

M' = 22750 kgm/s.

Assuming South to be negative direction,

From the question,

Total momentum of the two vehicles = (6500-22750)

Total momentum of the two vehicles = -16250 kgm/s

Hence the total momentum of the two vehicles is 16250 kgm/s due south

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When two or more identical capacitors are connected in series across a potential difference?

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3 years ago