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3 years ago

A 71 kg man is walking at 2 m/s. Calculate the Kinetic Energy for the man

Physics

1 answer:

Westkost [7]3 years ago

7 0

Answer:

Explanation:

The kinetic energy of an object can be found by using the formula

$k = \frac{1}{2} m {v}^{2} \\$

m is the mass

v is the velocity

From the question we have

$k = \frac{1}{2} \times 71 \times {2}^{2} \\ = \frac{1}{2} \times 71 \times 4 \\ = 71 \times 2$

We have the final answer as

Hope this helps you

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Answer:

Explanation:

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3 years ago

With respect to the earth, object 1 is moving at speed 0.80 c to the right. Object 2 is moving in the same direction at speed 0.

viktelen [127]

Answer:

0.976 c

Explanation:

$v_{1e}$ = velocity of object 1 relative to earth = 0.80 c

$v_{21}$ = velocity of object 2 relative to object 1 = 0.80 c

$v_{2e}$ = velocity of object 2 relative to earth

Velocity of object 2 relative to earth is given as

$v_{2e}= \frac{v_{1e} + v_{21}}{1 + \frac{v_{1e}v_{21}}{c^{2}}}$

$v_{2e}= \frac{0.80 c + 0.80 c}{1 + \frac{(0.80c)(0.80c)}{c^{2}}}$

$v_{2e}$ = 0.976 c

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3 years ago

4. Name three things that all games, no matter where they are played around the world, have in common.

lilavasa [31]

Answer:

Key components of games are goals, rules, challenge, and interaction.

Explanation:

Hope I helped.

3 0

3 years ago

Read 2 more answers

A diffraction grating produces a first-order bright fringe that is 0.18 m away from the central bright fringe on a flat screen.

Georgia [21]

Answer:

The wavelength of the light is 562.5 nm

Solution:

As per the question:

Order, n = 1

Slit separation, d = $2.5\times 10^{- 6}\ m$

Distance from the bright fringe, y = 0.18 m

Distance between the screen and the grating, D = 0.8 m

Now,

We know from the eqn for diffraction:

$n\lambda = dsin\theta$

n = 1

$\lambda = dsin\theta$ (1)

Also,

For very small angle, $\theta$ :

$sin\theta$ ≈ $tan\theta = \frac{y}{D} = \frac{0.18}{0.8} = 0.225$

Using the above value in eqn (1):

$\lambda = 2.5\times 10^{- 6}\times 0.225 = 5.625\times 10^{- 7}\ m = 562.5\ nm$

3 0

3 years ago

A 15.5 kg mass vibrates in simple harmonic motion with a frequency of 9.73 Hz. It has a maximum displacement from equilibrium of

algol [13]

Answer:

12.14 cm

Explanation:

mass, m = 15.5 kg

frequency, f = 9.73 Hz

maximum amplitude, A = 14.6 cm

t = 1.25 s

The equation of the simple harmonic motion

y = A Sin ωt

y = A Sin (2 x π x f x t)

put, t = 1.25 s, A = 14.6 cm, f = 9.73 Hz

y = 14.6 Sin ( 2 x 3.14 x 9.73 x 1.25)

y = 14.6 Sin 76.38

y = 12.14 cm

Thus, the displacement of the particle from the equilibrium position is 12.14 cm.

6 0

4 years ago