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2 years ago

15

Derive, with full working, the symbol equation for the power output P watts of the energy transfer that occurs when a mass of m

kg falls through a height of h metres in t seconds.

1 answer:

Alexxandr [17]2 years ago

3 0

I agree that the height is kg or an maybe

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A cyclotron with a magnetic field of 0.47 t is designed to accelerate protons in a circle with a radius 0.68 m. What is the angu

Ghella [55]

The angular frequency of the cyclotron is 0.07 x $10^{8}$ Hz.

<h3>What is angular frequency?</h3>

Angular frequency, abbreviated "ω" is a scalar measure of rotation rate in physics.
It describes the rate of change of the argument of the sine function, the rate of change of the phase of a sinusoidal waveform, or the angular displacement per unit of time.

<h3>What is cyclotron?</h3>

The cyclotron device is made to accelerate charge particles to extremely high speeds by applying crossed electric and magnetic fields.

<h3>Calculation of angular frequency:</h3>

Given,

B = 0.47 T

r = 0.68

mass of proton = 1.6x $10^{-27}$

q = 1.6 x $10^{-19}$

so, the frequency is:

f = qB/2 $\pi$ m

f = 1.6 x $10^{-19}$ x 0.47/2x3.14x1.6x $10^{-27}$

f = 0.07 x $10^{8}$

Hence, the angular frequency of the cyclotron is 0.07 x $10^{8}$ Hz.

Learn more about angular frequency here:

brainly.com/question/14244057

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7 0

1 year ago

Calculate the location xcm of the center of mass of the Earth-Moon system. Use a coordinate system in which the center of the Ea

Jet001 [13]

Answer:

The center of mass of the Earth-Moon system is 4.673 kilometers away from center of Earth.

Explanation:

Let suppose that planet and satellite can be treated as particles. The masses of Earth and Moon ( $m_{E}$ , $m_{M}$ ) are $5.972\times 10^{24}\,kg$ and $7.349\times 10^{22}\,kg$ , respectively. The distance between centers is 384,403 kilometers. The location of the center of mass can be found by using weighted averages:

$\bar x = \frac{x_{E}\cdot m_{E}+x_{M}\cdot m_{M}}{m_{E}+m_{M}}$

If $x_{E} = 0\,km$ and $x_{M} = 384,403\,km$ , then:

$\bar x = \frac{(0\,km)\cdot (5.972\times 10^{24}\,kg)+(384,403\,km)\cdot (7.349\times 10^{22}\,kg)}{5.972\times 10^{24}\,kg+7.349\times 10^{22}\,kg}$

$\bar x = 4.673\,km$

The center of mass of the Earth-Moon system is 4.673 kilometers away from center of Earth.

8 0

3 years ago

A pet-store supply truck moves at 25.0 m/s north along a highway. inside, a dog moves at 1.75 m/s at an angle of 35.0Â° east of

koban [17]

<u>Answer:</u>

Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.

<u>Explanation:</u>

Let the east point towards positive X-axis and north point towards positive Y-axis.

Speed of truck = 25 m/s north = 25 j m/s

Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s

= (1.43 i + 1.00 j) m/s

Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j

Magnitude of velocity = 26.04 m/s

Angle from positive horizontal axis = 86.85⁰

So Velocity of the dog relative to the road = 26.04 m/s 86.85⁰ east of north = 26.04 m/s 3.15⁰ north of east.

4 0

3 years ago

You push your couch a distance of 3.9 m across the living room floor with a horizontal force of 220.0 n. the force of friction i

Mrac [35]

The general formula to calculate the work is:

$W=Fd \cos \theta$

where F is the force, d is the displacement of the couch, and $\theta$ is the angle between the direction of the force and the displacement. Let's apply this formula to the different parts of the problem.

(a) Work done by you: in this case, the force applied is parallel to the displacement of the couch, so $\theta=0^{\circ}$ and $\cos \theta=1$ , therefore the work is just equal to the product between the horizontal force you apply to push the couch and the distance the couch has been moved:

$W=Fd=(220.0 N)(3.9 m)=858 J$

(b) work done by the frictional force: the frictional force has opposite direction to the displacement, therefore $\theta=180^{\circ}$ and $\cos \theta=-1$ . Therefore, we must include a negative sign when we calculate the work done by the frictional force:

$W=-Fd=-(144.0 N)(3.9 m)=-561.6 J$

(c) The work done by gravity is zero. In fact, gravity (which points downwards) is perpendicular to the displacement of the couch (which is horizontal), therefore $\theta=90^{\circ}$ and $\cos \theta=0$ : this means

$W=0$ .

(d) Work done by the net force:

The net force is the difference between the horizontal force applied by you and the frictional force:

$F=220 N-144 N=76 N$

And the net force is in the same direction of the displacement, so $\theta=0^{\circ}$ and $\cos \theta=1$ and the work done is

$W=Fd=(76 N)(3.9 m)=296.4 J$

4 0

3 years ago

Question 15 plz picture above

Pie

15) C. The amount of each element that begins....

8 0

3 years ago