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nadezda [96]
3 years ago
5

The power lines are at a high potential relative to the ground, so there is an electric field between the power lines and the gr

ound. To maximize the potential difference between one end of the fluorescent tube and the other, how should the tube be held? Select the best answer from the choices provided. A. The tube should be held horizontally, parallel to the ground. B.The potential difference between the ends of the tube does not depend on the tube's orientation. C.The tube should be held vertically, perpendicular to the ground.
Physics
1 answer:
s344n2d4d5 [400]3 years ago
7 0

Answer:C.The tube should be held vertically, perpendicular to the ground.

Explanation: Power lines are mainly overhead power installation that transfer electric energy from one place to another. Electric power lines contains both Magnetic field and Electric field.

Potential difference is the change in the amount of energy carried by an electric circuit from one point to another. TO MAXIMIZE THE POTENTIAL DIFFERENCE BETWEEN ONE END OF THE TUBE AND ANOTHER? THE TUBE SHOULD BE HELD VERTICALLY,PERPENDICULAR TO THE GROUND.

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A muon is a short-lived particle. It is found experimentally that muonsmoving at speed 0.9ccan travel about 1500 meters between
Monica [59]

Answer:

a)t=5.5\mu s

b)\Delta t'=12.5\mu s

Explanation:

a) Let's use the constant velocity equation:

v=\frac{\Delta x}{\Delta t}

  • v is the speed of the muon. 0.9*c
  • c is the speed of light 3*10⁸ m/s

t=\frac{\Delta x}{v}=\frac{1500}{0.9*(3*10^{8})}

t=5.5\mu s

b) Here we need to use Lorentz factor because the speed of the muon is relativistic. Hence the time in the rest frame is the product of the Lorentz factor times the time in the inertial frame.

\Delta t'=\gamma\Delta t

\gamma =\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

v is the speed of muon (0.9c)

Therefore the time in the rest frame will be:

\Delta t'=\frac{1}{\sqrt{1-\frac{(0.9c)^{2}}{c^{2}}}}\Delta t

\Delta t'=\frac{1}{\sqrt{1-0.9^{2}}}\Delta t

\Delta t'=\frac{1}{0.44}\Delta t

No we use the value of Δt calculated in a)

\Delta t'=2.27*5.5=12.5\mu s

I hope it helps you!

     

 

8 0
3 years ago
A 1500kg car traveling at 25m/s skids to a stop. The force of friction between the tires and the road is 10500N. How far does th
ale4655 [162]

44.64m

Explanation:

Given parameters:

Mass of the car = 1500kg

Initial velocity = 25m/s

Frictional force = 10500N

Unknown:

Distance moved by the car after brake is applied = ?

Solution:

The frictional force is a force that  opposes motion of a body.

To solve this problem, we need to find the acceleration of the car. After this, we apply the appropriate motion equation to solve the problem.

     -Frictional force = m x a

the negative sign is because the frictional force is in the opposite direction

m is the mass of the car

 a is the acceleration of the car

    a = \frac{frictional force}{mass} = \frac{10500}{1500} = -7m/s²

Now using;

   V² = U² + 2as

   V is the final velocity

   U is the initial velocity

   a is the acceleration

   s is the distance moved

  0² = 25² + 2 x 7 x s

  0 = 625 - 14s

  -625 = -14s

      s = 44.64m

   

learn more:

Velocity problems brainly.com/question/10932946

#learnwithBrainly

7 0
3 years ago
Why is the answer C for this problem?
shusha [124]

Answer:

Explanation:

If you look closely, force 1 does not reach 0.2 until 0.4 force 2 reaches 0.2 at about 0.2 - hope that made sense :P

4 0
3 years ago
Define wind ward side
Rasek [7]

Answer:

The windward side is that side which faces the prevailing wind (upwind), whereas the leeward, or "lee" side, is the side sheltered from the wind by the mountain's very elevation (downwind)

7 0
3 years ago
I NEEDHELP ASAP !!! BEIN TIMED AND I HAVE 5 MINS LEFT AND 8 LEFT TO ANSWER !
inn [45]

i think its a, good luck on your test

3 0
3 years ago
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