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3 years ago

How much kinetic energy does an 80 kg man have while running at 3 m/s?

Physics

1 answer:

Soloha48 [4]3 years ago

7 0

Hello!

$\large\boxed{KE = 360 J}$

Use the equation KE = 1/2mv² to solve for the kinetic energy of the man.

We are given the mass and velocity, so plug these values into the equation:

KE = 1/2(80)(3²)

KE = 1/2(720)

KE = 360 J

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A projectile is launched at an angle of 29 degrees above the horizontal with an initial velocity of 36.6 at an unknown height.

fgiga [73]

The vertical velocity of the projectile upon returning to its original is 17. 74 m/s

<h3>How to determine the vertical velocity</h3>

Using the formula:

Vertical velocity component , Vy = V * sin(α)

Where

V = initial velocity = 36. 6 m/s

α = angle of projectile = 29°

Substitute into the formula

Vy = 36. 6 * sin ( 29°)

Vy = 36. 6 * 0. 4848

Vy = 17. 74 m/s

Thus, the vertical velocity of the projectile upon returning to its original is 17. 74 m/s

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2 years ago

When an object is falling and reaches a constant velocity, the net force on the object is ____ and the weight of the object is e

Maslowich

When an object is falling and reaches a constant velocity, the net force on the object is <em>zero</em> (it's not accelerating), and the weight of the object is equal to <em>the force of air resistance against the object</em>. (choice-D)

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3 years ago

If an object accelerates from rest, with a constant of 8 m/s2, what will its velocity be after 35s?

AlladinOne [14]

Answer:

<em>Its speed will be 280 m/s</em>

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the speed of an object changes by an equal amount in every equal period of time.

If a is the constant acceleration, vo the initial speed, vf the final speed, and t the time, vf can be calculated as:

$v_f=v_o+at$

The object accelerates from rest (vo=0) at a constant acceleration of $a=8\ m/s^2$ . The final speed at t=35 seconds is:

$v_f=0+8*35$

$v_f=280\ m/s$

Its speed will be 280 m/s

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3 years ago

Our milky way galaxy is 100000 lyly in diameter. a spaceship crossing the galaxy measures the galaxy's diameter to be a mere 1.

Sidana [21]

The speed of the spaceship relative to the galaxy is 0.99999995c.

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The Milky way galaxy is 100,000 light years in diameter.

The galaxy's diameter is a mere 1. 0 ly.

We know that ;

$L = L_0 \sqrt{1-\frac{v^2}{c^2} }$

L = 1 light year

L₀ = 100,000 light year

$1 = 100,000 \sqrt{1-\frac{v^2}{c^2} }$

$1 = 100,000 \sqrt{1-\frac{v^2}{(3*10^8)^2} }$

$\frac{1}{100,000} = \sqrt{1-\frac{v^2}{c^2} }$

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Therefore, the speed of the spaceship relative to the galaxy is 0.99999995c.

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2 years ago

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ziro4ka [17]

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So, do not get confused.

Ohm's law is not a universal law, please remember that as well. Some materials do not follow it and we call them non-ohmic conductors. I hope I helped! ^-^

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3 years ago