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3 years ago

How far can a car traveling at a rate of 40 kilometer per hour travelbin 2 1/2 hours

2 answers:

5 0

Supposing it stays traveling at 40 kph you'll travel exactly 40 kilometers in just one hour. So in 2 hours you'll have travel 80 Kilometers. and in half an hour you travel 20 which is the Half of 40. And 20+80 = 100 Kilometers.

CaHeK987 [17]3 years ago

4 0

90 kilometers because you need to multiply 40 by 2 and then you get 80 and finally you add 10 and get 90 kilometers

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How to find instantaneous velocity physics?

evablogger [386]

You should have the velocity as a function of time either given explicitly or implicitly (a graph)

v = ds/dt (differentiating the position vector)

integrating the acceleration.

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NOT SURE IF THAT WHAT YOU WANT.

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3 years ago

The loudness of a sound is determined by the __________, or height, of the sound wave.

balandron [24]

C. amplitude....................

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3 years ago

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What are the different types of topology?

dolphi86 [110]

Answer:

common types of topologies, and we're going to break each of them down in the guide below.

Bus topology. As the simplest design, a bus topology requires nodes to be in a linear order. ...

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Explanation:

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2 years ago

Read 2 more answers

Suppose the entire population of the world gathers in ONE spot and everyone jumps at the sound of a prearranged signal. While ev

puteri [66]

Answer:

(b) Yes, the earth gains momentum but the change in momentum of the earth is much lesser compared to that of everyone in the air. The resistance to motion (inertia of the earth), which is a function of its mass is so great that the earth's acceleration is small in the given time frame.

Explanation:

From Newton's second law which can be stated mathematically as

F = m(v-u)/t = ma.

By Newton's law of gravitation, there is a force between the earth and everyone in the air. This force is responsible for the change in momentum of everyone in the air and this force gives them an acceleration equal to g = 9.80m/s². By Newton's law of gravitation and Newton's third law of motion, this force is also equal to the force exerted by everyone on the earth.

For this to be true,

F = M (everyone) ×a (everyone) = M(earth) × a (earth).

And

a (earth) = {M (everyone) ×a (everyone) }/M (earth)

Then

a (earth) must be lesser than a (everyone) since M(earth) >> M(everyone).

a = change in momentum/ time

Therefore the earth will have a much lesser change in momentum which is the reason we won't notice the earth's movement.

Thank you for reading.

6 0

3 years ago

A rotating flywheel has moment of inertia 18.0 kg⋅m^2 for an axis along the axle about which the wheel is rotating. Initially th

timama [110]

Answer:

The rotational kinetic energy takes 0.430 seconds to become half its initial value.

Explanation:

By the Principle of Energy Conservation and the Work-Energy Theorem we know that flywheel slow down due to the action of non-conservative forces (i.e. friction), the energy losses are equal to the change in the rotational kinetic energy. That is:

$\Delta E = K_{1}-K_{2}$ (1)

Where:

$\Delta E$ - Energy losses, measured in joules.

$K_{1}$ , $K_{2}$ - Initial and final rotational kinetic energies, measured in joules.

By definition of rotational kinetic energy, we expand the equation above:

$\Delta E = \frac{1}{2}\cdot I\cdot (\omega_{1}^{2}-\omega_{2}^{2})$ (2)

Where:

$I$ - Moment of inertia of the flywheel, measured in kilograms per square meter.

$\omega_{1}$ , $\omega_{2}$ - Initial and final angular speed, measured in radians per second.

If we know that $K_{1} = 30\,J$ , $K_{2} = 15\,J$ and $I = 18\,kg\cdot m^{2}$ , then the initial angular speed is:

$K_{1} = \frac{1}{2}\cdot I \cdot \omega_{1}^{2}$ (3)

$\omega_{1}=\sqrt{\frac{2\cdot K_{1}}{I} }$

$\omega_{1} = \sqrt{\frac{2\cdot (30\,J)}{18\,kg\cdot m^{2}} }$

$\omega_{1} \approx 1.825\,\frac{rad}{s}$

$\omega_{1}\approx 0.291\,\frac{rev}{s}$

$K_{2} = \frac{1}{2}\cdot I \cdot \omega_{2}^{2}$ (4)

$\omega_{2}=\sqrt{\frac{2\cdot K_{2}}{I} }$

$\omega_{2} = \sqrt{\frac{2\cdot (15\,J)}{18\,kg\cdot m^{2}} }$

$\omega_{2} \approx 1.291\,\frac{rad}{s}$

$\omega_{2} \approx 0.205\,\frac{rev}{s}$

Under the assumption that flywheel is decelerating uniformly, we get that the time taken for the flywheel to slowdown is:

$t = \frac{\omega_{2}-\omega_{1}}{\alpha}$ (5)

If we know that $\omega_{1}\approx 0.291\,\frac{rev}{s}$ , $\omega_{2} \approx 0.205\,\frac{rev}{s}$ and $\alpha = -0.200\,\frac{rev}{s^{2}}$ , then the time needed is:

$t = \frac{0.205\,\frac{rev}{s}-0.291\,\frac{rev}{s}}{-0.200\,\frac{rev}{s^{2}} }$

$t = 0.43\,s$

The rotational kinetic energy takes 0.430 seconds to become half its initial value.

6 0

3 years ago