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9 months ago

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• An object of mass 2.0 kg moves in uniform circular motion on a horizontal friction less table. The radius of the circle is 0.7

5 m and the centripetal force is 10.0 N. The work done by this force when the object moves through one-half of a complete revolution is

1 answer:

Gekata [30.6K]9 months ago

4 0

Answer:

23.562

Explanation:

First it is good to work out the circumference:

2 pi r

2 pi .75

circumference = 4.7124

Next we need half of this because it is half a resolution. This will be our distance travelled.

half = 2.3562

word done = force x distance

work done = 10 x 2.3562

work done = 23.562

(I didn't see the need for mass...)

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In my trigonometry class, we were assigned a problem on Angular and Linear Velocity.

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1) 0.0011 rad/s

2) 7667 m/s

Explanation:

1)

The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

$\omega=\frac{\theta}{t}$

where

$\theta$ is the angular displacement of the object

t is the time elapsed

$\omega$ is the angular velocity

In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is

$\theta=2\pi$ rad

And the time taken is

$t=95 min \cdot 60 =5700 s$

Therefore, the angular velocity of the telescope is

$\omega=\frac{2\pi}{5700}=0.0011 rad/s$

2)

For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

$v=\omega r$

where

v is the linear velocity

$\omega$ is the angular velocity

r is the radius of the circular orbit

In this problem:

$\omega=0.0011 rad/s$ is the angular velocity of the Hubble telescope

The telescope is at an altitude of

h = 600 km

over the Earth's surface, which has a radius of

R = 6370 km

So the actual radius of the Hubble's orbit is

$r=R+h=6370+600=6970 km = 6.97\cdot 10^6 m$

Therefore, the linear velocity of the telescope is:

$v=\omega r=(0.0011)(6.97\cdot 10^6)=7667 m/s$

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3 years ago

A 1.50-mm-diameter glass sphere has a charge of + 1.60 nC. What speed does an electron need to orbit the sphere 1.60 mm above th

saveliy_v [14]

Answer :

Velocity will be $3.28\times 10^{-11}m/sec$

Explanation:

We have given glass surface has a diameter of 1.5 mm

And charge q = 1.60 nC

Radius of electrons orbit r = height of electron above surface + radius of sphere = $=1.6+\frac{1.5}{2}=2.35mm = 0.00235m$

Force on electron is given by $F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}$ , here q is charge on sphere and e is charge on electron

$F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}=\frac{kqe}{r^2}=\frac{9\times 10^9\times 1.6\times 10^{-9}\times 1.6\times 10^{-19}}{0.00235^2}=4.172\times 10^{-13}N$

This force work as centripetal force

So $F=\frac{mv^2}{r}$

$4.172\times 10^{-13}=\frac{9.11\times 10^{-31}v^2}{0.00235}$

v = $=0.0328\times 10^{-9}=3.28\times 10^{-11}m/sec$

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STALIN [3.7K]

Answer:

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Pls helppp. Is this right?

andreyandreev [35.5K]

Answer:

yes you are totally right

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The kinetic energy of a rocket is increased by a factor of eight after its engines are fired, whereas its total mass is reduced

Rudik [331]

The momentum increases by a factor of 2

Explanation:

We can solve this problem by rewriting the momentum of the rocket in terms of the kinetic energy and the mass.

The kinetic energy of the rocket is:

$K=\frac{1}{2}mv^2$ (1)

where

m is the mass

v is the velocity

The momentum of the rocket is

$p=mv$ (2)

From eq.(1) we get

$v=\sqrt{\frac{2K}{m}}$

and substituting into (2),

$p=\sqrt{2mK}$

Now in this problem we have:

- The kinetic energy of the rocket is increased by a factor 8:

$K' = 8K$

- The mass is reduced by half:

$m'=\frac{m}{2}$

Substituting, we find the new momentum:

$p'=\sqrt{2(\frac{m}{2}(8K)}=\sqrt{4(2mK)}=2\sqrt{2mK}=2p$

So, the momentum increases by a factor of 2.

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