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3 years ago

A 70.0-kilogram man is walking at a speed if 2.0 m/s. What is the kinetic energy?

Physics

1 answer:

Paraphin [41]3 years ago

3 0

Answer:

140 J

Explanation:

From the the question, the mass of the man =70.0 kg and the speed at which the man is walking =2.0 m/s.

$K.E = \frac{1}{2}m {v}^{2}$

where K.E = the kinetic energy, m=mass and v= speed.

By substitution,

$K.E = \frac{1}{2} \times 70 \times {2}^{2}$

$\implies \: K.E = \frac{280}{2}$

$\implies K.E =140$

Hence the kinetic energy of the man is 140J

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A 1500 kg car traveling east at 40 km/hr turns a corner and speeds up to a velocity of 50 km/hr due north. What is the change in

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Dejamos caer un objeto desde lo alto de una torre y medimos el tiempo que tarda en llegar al suelo que resulta ser de 0,02 minut

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Explanation:

The described situation is as follows:

An object is dropped from the top of a tower and when measuring the time it takes to reach the ground that turns out to be 0.02 minutes.

This situation is related to free fall, this also means we have constant acceleration, hence the equations we will use are:

$V_{f}=V_{o}+at$ (1)

${V_{f}}^{2}={V_{o}}^{2}+2ad$ (2)

Where:

$V_{f}$ Is the final velocity of the object

$V_{o}=0$ Is the initial velocity of the object (it was dropped)

$a=9.8 m/s^{2}$ is the acceleration due gravity

$d$ is the height of the tower

$t=0.02min=1.2 s$ is the time it takes to the object to reach the ground

b) Begining with (1):

$V_{f}=0+at$ (3)

$V_{f}=at=(9.8 m/s^{2})(1.2 s)$ (4)

$V_{f}=11.76 m/s$ (5) This is the final velocity of the object

a) Substituting (5) in (2):

$(11.76 m/s)^{2}=0+2(9.8 m/s^{2})d$ (6)

Clearing $d$ :

$d=\frac{(11.76 m/s)^{2}}{2(9.8 m/s^{2})}$ (7)

$d=7.056 m$ (8) This is the height of the tower

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4 years ago