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3 years ago

A 70 kg man is walking at a speed of 2 m/s. What is his Kinetic Energy?

Physics

2 answers:

Ilya [14]3 years ago

8 0

Kinetic energy = 1/2 (mass) (speed squared)

KE = 1/2 (70kg) (2m/s)^2

KE = (35 kg) (4 m^2/s^2)

KE = (140) kg-m^2/s^2

That's 140 Newton-meter

140 Joules

den301095 [7]3 years ago

5 0

140 Jules. Is the correct answer

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Explanation:

Given parameters:

Force = 30N

Weight Susan = 45kg

Weight of Dad = 100kg

Unknown:

Acceleration of Susan = ?

Acceleration of Dad = ?

Solution:

Force = mass x acceleration

Acceleration = $\frac{force}{mass}$

Acceleration of Susan = $\frac{30}{45}$ = 0.67m/s²

Acceleration of Dad = $\frac{30}{100}$ = 0.3m/s²

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3 years ago

A wave has a wavelength of 5 meters and a frequency of 3 hertz. What is the speed of the wave?

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A sound wave in a steel rail has a frequency of 620 Hz and a wavelength of 10.5 ... Find the speed of a wave with a wavelength of 5 m and a frequency of 68 Hz.

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2 years ago

A tuning fork labeled 392 Hz has the tip of each of its two prongsvibrating with an amplitude of 0.600mma) What is the maximum s

DanielleElmas [232]

a) 1.48 m/s

The tuning fork is moving by simple harmonic motion: so, the maximum speed of the tip of the prong is related to the frequency and the amplitude by

$v_{max}=\omega A$

where

$v_{max}$ is the maximum speed

$\omega$ is the angular frequency

A is the amplitude

For the tuning fork in the problem, we have

$\omega=2\pi f=2 \pi(392 Hz)=2462 rad/s$ , where f is the frequency

$A=0.600 mm=6\cdot 10^{-4} m$ is the amplitude

Therefore, the maximum speed is

$v_{max}=(2462 rad/s)(6\cdot 10^{-4}m)=1.48 m/s$

b) $3.0\cdot 10^{-5} J$

The fly's maximum kinetic energy is given by

$K=\frac{1}{2}mv_{max}^2$

where

$m=0.0270 g=2.7\cdot 10^{-5} kg$ is the mass of the fly

$v_{max}=1.48 m/s$ is the maximum speed

Substituting into the equation, we find

$K=\frac{1}{2}(2.7\cdot 10^{-5}kg)(1.48 m/s)^2=3.0\cdot 10^{-5} J$

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2 years ago

A bicycle racer inflates their tires to 7.1 atm on a warm autumn afternoon when temperatures reached 27 °C. By morning the tempe

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Answer:

The required pressure is 6.4866 atm.

Explanation:

The given data : -

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Initial temperature ( T₁ ) = 27°C = (27 + 273) K = 300 K

In the morning .

Final temperature ( T₂ ) = 5°C = ( 5 + 273 ) K = 278 K

Given that volume remains constant.

To find final pressure ( p₂ ).

Applying the ideal gas equation.

p * v = m * R * T

$\frac{p}{T} = constant$

$\frac{p_{1} }{T_{1} } = \frac{p_{2} }{T_{2} }$

$p_{2} = \frac{T_{2} }{T_{1} } *p_{1}$

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2 years ago