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3 years ago

What is the kinetic energy of a 150gram object moving at a velocity of 100m/s?

Physics

1 answer:

allochka39001 [22]3 years ago

8 0

Answer:

kinetic energy = 1 ÷ 2 × mass ×velocity ^2

kinetic energy = 1 ÷ 2 × 0.15 × 100^2

Kinetic energy = 750 Joules

Explanation:

NOTE ; S.I unit of mass is kilograms

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Shadow and eclipses result from

Basile [38]

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${\green{\dashrightarrow}}$ When an opaque obstacle is placed between a source of light and a screen, a shadow of the obstacle is formed on the screen. The kind of shadow depends on the size of the source of light. In other words, the earth casts its shadow on the moon. The solar eclipse occurs when the moon comes between the sun and the earth.

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2 years ago

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A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacemen

PilotLPTM [1.2K]

Answer:

The average angular acceleration is 0.05 radians per square second.

Explanation:

Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})$

Where:

$\omega_{o}$ , $\omega$ - Initial and final angular velocities, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

Then,

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}$

Given that $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\theta-\theta_{o} = 4.9\,rad$ , the angular acceleration is:

$\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}$

$\alpha = 0.05\,\frac{rad}{s^{2}}$

Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:

$\omega = \omega_{o} + \alpha \cdot t$

Where $t$ is the time measured in seconds.

The time is cleared and obtain after replacing every value:

$t = \frac{\omega-\omega_{o}}{\alpha}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\alpha = 0.05\,\frac{rad}{s^{2}}$ , the required time is:

$t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }$

$t = 14\,s$

Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:

$\bar \alpha = \frac{\omega-\omega_{o}}{t}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $t = 14\,s$ , the average angular acceleration is:

$\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}$

$\bar \alpha = 0.05\,\frac{rad}{s^{2}}$

The average angular acceleration is 0.05 radians per square second.

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3 years ago

How much conventional current must you run in a solenoid with radius = 0.05 m and length = 0.39 m to produce a magnetic field in

Ugo [173]

Answer:

Explanation:

radius of the solenoid, r = 0.05 m

length of the solenoid, l = 0.39 m

Magnetic field of the solenoid, B = 2 x 10^-5 T

Number of turns, N = 200

The magnetic field of the solenoid is given by

$B=\mu _{0}ni$

where, i be the current and n be the number of turns per unit length

n = N / l = 200 / 0.39 = 512.8

$2\times 10^{-5}=4 \times 3.14\times 10^{-7}\times 512.8\times i$

i = 0.031 A

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gavmur [86]

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Therefore, statement A is correct.

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You have given a power supply, copper wire, and an iron nail. What should you do to decrease the strength of the electro magnet?

Shkiper50 [21]

If you wrap some of the wire around the nail in one direction and some of the wire in the other direction, the magnetic fields from the different sections fight each other and cancel out, reducing the strength of your magnet.

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