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german
3 years ago
15

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. View Available Hint(s) Select the best answer from the choices provided. The potential difference between the ends of the tube does not depend on the tube's orientation. The tube should be held horizontally, parallel to the ground. The tube should be held vertically, perpendicular to the ground.
Physics
1 answer:
Amanda [17]3 years ago
6 0

Answer:

The tube should be held vertically and perpendicular to the ground.

Explanation:

Answer: The tube should be held vertically and perpendicular to the ground. The reason is as follows:

Reasoning:

The power lines are parallel to the ground hence, their electric field will be perpendicular to the ground and equipotential surface will be cylindrical.

Hence, if you will put fluorescent tube parallel to the ground then both the ends of the tube will lie on the same equipotential surface and the potential difference will be zero.

So, to maximize the potential the ends of the tube must be on different equipotential surfaces. The surface which is near to the power line has high potential value and the surface which is farther from the line has lower potential value.

hence, to maximize the potential difference, the tube must be placed perpendicular to the ground.

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Which particle would have the slowest rate of deposition? A.round particle B.very large particle C.particle with sharp ends D.pa
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The particle with sharp ends have the slowest rate of deposition  

Answer: Option C  

<u>Explanation:</u>

          As per aerosol physics, deposition is a process where aerosol particles accumulate or settle on solid surfaces. Thereby, it reduces the concentration of particles in the air. Deposition velocity (rate of deposition) defines from F = vc, where v is deposition rate, F denotes flux density and c refers concentration.  

          Deposition velocity is slowest for particles of intermediate-sized particles because the frictional force offers resistance to the flow. Density is directly proportional to the deposition rate so clearly shows that high-density particles settle faster. Due to friction, round and large-sized particles deposit faster than oval/flattened sediments.  

3 0
3 years ago
Read 2 more answers
An air-track glider attached to a spring oscillates between the 10 cm mark and the 60 cm mark on the track. The glider completes
Assoli18 [71]

Answer:

a) Time period is 3.3 seconds

b) The frequency is 0.3030 Hz

c) amplitude is 0.25 m

d) maximum speed is 0.476 m/s

Explanation:

Given the data in the question;

a) Period

Time Period T = Time taken for one oscillation

T = 33s / 10 = 3.3 seconds

Therefore, Time period is 3.3 seconds

b) Frequency

we know that frequency is the inverse of time period

so;

Frequency f = 1/T = 1 / 3.3 s

Frequency f = 0.3030 Hz

Therefore, The frequency is 0.3030 Hz

c) amplitude

amplitude A = \frac{1}{2}( 60 cm - 10 cm )

A = \frac{1}{2} × 50 cm

A = 25 cm

A = 0.25 m

Therefore, amplitude is 0.25 m

d) maximum speed of the glider

maximum speed V_{max} = ωA

and ω = 2π/T

so maximum speed V_{max} = \frac{2\pi }{T}A

so we substitute

so maximum speed V_{max} = \frac{2\pi }{3.3} × 0.25 m

so maximum speed V_{max} = 0.476 m/s

Therefore, maximum speed is 0.476 m/s

5 0
3 years ago
Compute the dot product of the vectors u and v​, and find the angle between the vectors. Bold v equals 7 Bold i minus Bold j and
OLga [1]

Answer:

\theta = 106.3 degree

Explanation:

As we know that

\vec w = -\hat i + 7\hat j

\vec v = 7\hat i - \hat j

also we know that

\vec v. \vec w = -14

it is given as

\vec v. \vec w = (-\hat i + 7\hat j).(7\hat i - \hat j)

\vec v. \vec w = - 7 - 7 = -14

also we can find the magnitude of two vectors as

|v| = \sqrt{(-1)^2 + (7)^2}

|v| = \sqrt{50}

similarly we have

|w| = \sqrt{(7^2) + (-1)^2}

|w| = \sqrt{50}

now we know the formula of dot product as

\vec v. \vec w = |v||w| cos\theta

-14 = (\sqrt{50})^2cos\theta

\theta = cos^{-1}(\frac{-14}{50})

\theta = 106.3 degree

3 0
3 years ago
Write a short description of how the motion of the racers might change from the start of the race to the finish line
Ad libitum [116K]
The motion of the racers might change from the start because the pressure goes up so all the racer wants is to speed up and win, so when the racer first starts he or she is calm because he's not driving yet and when he or she is on his/hers way to he finish line he/she just wants to win and gets under pressure so he speeds up even more and drifts. Your welcome
6 0
3 years ago
Consider massive gliders that slide friction-free along a horizontal air track. Glider A has a mass of 1 kg, a speed of 1 m/s, a
mamaluj [8]

Answer:

0.167m/s

Explanation:

According to law of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision. The bodies move with a common velocity after collision.

Given momentum = Maas × velocity.

Momentum of glider A = 1kg×1m/s

Momentum of glider = 1kgm/s

Momentum of glider B = 5kg × 0m/s

The initial velocity of glider B is zero since it is at rest.

Momentum of glider B = 0kgm/s

Momentum of the bodies after collision = (mA+mB)v where;

mA and mB are the masses of the gliders

v is their common velocity after collision.

Momentum = (1+5)v

Momentum after collision = 6v

According to the law of conservation of momentum;

1kgm/s + 0kgm/s = 6v

1 =6v

V =1/6m/s

Their speed after collision will be 0.167m/s

6 0
3 years ago
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