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Goshia [24]
2 years ago
12

A proton orbits a long charged wire, making 1.80 ×106 revolutions per second. The radius of the orbit is 1.20 cm What is the wir

e's linear charge density?.
Physics
1 answer:
Fantom [35]2 years ago
8 0

Answer:

linear charge density = -9.495 × 10^{-34} C/m

Explanation:

given data

revolutions per second = 1.80 × 10^{6}

radius = 1.20 cm

solution

we know that when proton to revolve around charge wire then centripetal force is require to be in orbit of radius around provide by electric force

so

- q × E = m × w² × r     ..................1

- 9 × 10^{9}  × \frac{2*linear\ charge\ density}r} q =  m × w² × r   ............2

and w = \frac{2*\pi}{T}  

w = \frac{d\theta }{dt}

w = 1.80 × 10^{6} × \frac{2*\pi}{1}

w = 11304000 rad/s

so here from equation 2

- 9 × 10^{9}  × \frac{2*linear\ charge\ density}{0.012} 1.80 × 10^{6} =  1.672 × 10^{-27} × 11304000² × 0.0120  

linear charge density = -9.495 × 10^{-34} C/m

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If a machine increases force it must apply the force over a long distance? True or false
Eduardwww [97]

Answer: The machine must apply the force over a shorter distance. That's because a machine doesn't change the amount of work and work equals force times distance. Therefore, if force increases, distance must decrease

FALSE

HOPE THIS HELPS

4 0
2 years ago
A bug is 12 cm from the center of a turntable that is rotating with a frequency of 45 rev/min . What minimum coefficient frictio
Agata [3.3K]

Answer:

The minimum coefficient of friction is 0.27.

Explanation:

To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.

Centripetal force is written as

F_c = m\frac{v^2}{r}

with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as

v = r\omega

with ω denoting the angular velocity, which we are given. With that, the above becomes:

F_c = m\frac{v^2}{r}=m\omega^2 r

Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?

F_r=\mu mg \geq m\omega^2 r = F_c\implies\\\mu \geq \frac{\omega^2 r}{g}

Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:

\frac{1 rev}{1 min}\cdot\frac{\frac{2\pi rad}{rev}}{\frac{60s}{1 min}}=\frac{2\pi}{60}\frac{rad}{s}

and so 45 rev/min = 4.71 rad/s.

\mu \geq \frac{\omega^2 r}{g}=\frac{4.71^2\frac{1}{s^2}\cdot 0.12m}{9.8\frac{m}{s^2}}=0.27

A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.



3 0
2 years ago
In general, if the temperature of a chemical reaction is increased, the reaction rateA. IncreasesB. decreasesC. remains the same
In-s [12.5K]
A. Increases

I would assume this to be the answer because heat is another form of energy. If there is more energy the molecules will become more active. This makes A the most logical answer.
6 0
3 years ago
The 'little brain' attached to the rear of the brainstem is called the: (2 points)
11111nata11111 [884]

Answer:

The 'little brain' attached to the rear of the brainstem is called the cerebellum. See the explanation below, please.

Explanation:

The cerebellum is in the back of the brain, it is small in size. It is responsible for functions such as integrations of sensory and motor pathways, balance, coordination of movements, posture.

6 0
3 years ago
Read 2 more answers
A-10A twin-jet close-support airplane is approximately rectangular with a wingspan (the length perpendicular to the flow directi
Sidana [21]

Solution :

Given :

Rectangular wingspan

Length,L = 17.5 m

Chord, c = 3 m

Free stream velocity of flow, $V_{\infty}$ = 200 m/s

Given that the flow is laminar.

$Re_L=\frac{\rho V L}{\mu _{\infty}}$

      $=\frac{1.225 \times 200 \times 3}{1.789 \times 10^{-5}}$

    $= 4.10 \times 10^7$

So boundary layer thickness,

$\delta_{L} = \frac{5.2 L}{\sqrt{Re_L}}$

$\delta_{L} = \frac{5.2 \times 3}{\sqrt{4.1 \times 10^7}}$

    = 0.0024 m

The dynamic pressure, $q_{\infty} =\frac{1}{2} \rho V^2_{\infty}$

                                           $ =\frac{1}{2} \times 1.225  \times 200^2$

                                          $=2.45 \times 10^4 \ N/m^2$

The skin friction drag co-efficient is given by

$C_f = \frac{1.328}{\sqrt{Re_L}}$

     $=\frac{1.328}{\sqrt{4.1 \times 10^7}}$

     = 0.00021

$D_{skinfriction} = \frac{1}{2} \rho V^2_{\infty}S C_f$

                  $=\frac{1}{2} \times 1.225 \times 200^2 \times 17.5 \times 3 \times 0.00021$

                  = 270 N

Therefore the net drag = 270 x 2

                                      = 540 N

7 0
2 years ago
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