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fredd [130]
3 years ago
9

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?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:
Hunter-Best [27]3 years ago
6 0

Answer:

b) True. potencial diferencie does not depend on orientation

Explanation:

In this exercise we are asked to show which statements are true.

The expression the potential with respect to earth or the electric field with respect to earth refers to the potential or electric charge of the planet that is assumed to be very large and does not change in value during work.

It does not refer to the height of the system.

We can now review the claims

a) False. Potential not to be refers to height

b) True. Does not depend on orientation

c) False The potential does not refer to the altitude but to the Earth's charge

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The force of gravity is twice as great on a 2-kg rock as on a 1-kg rock. Why then, dose the 2-kg rock not fall with twice the ac
Phoenix [80]

Answer:

because all objects fall at a rate of 9.8m/s²

8 0
2 years ago
Which statement is correct?
NISA [10]

Answer:

1st statement is true

Explanation:

Here statement 1 is correct

Let think about it, if you push down the bar then you are lifting your weight off the pedals.

Obviously, this question does not take into account of racing bikes with straps on pedals, where you would push on one side and pull on the other to match your legs are doing, with straps your other leg can pull pedals upward.

5 0
3 years ago
The electric field between the plates of the cathode ray tube of an older television set can be as high as 2.5x104 N/C. Determin
pentagon [3]

Answer:

(a) F = -4.01 * 10^{-15} N

(b) a = 4.40 * 10^{15} m/s^2

Explanation:

Parameter given:

Electric field, E = 2.5 * 10^4 N/C

(a) Electric force is given (in terms of electric field) as a product of electric charge and electric field.

Mathematically:

F = qE

Electric charge, q, of an electron = - 1.602 * 10^{-19} C

F = -1.602 * 10^{-19} * 2.5 * 10^4\\\\\\F = -4.01 * 10^{-15} N

(b) This electrostatic force causes the electron to accelerate with an equivalent force:

F = -ma

where m = mass of an electron

a = acceleration of electron

(Note: the force is negative cos the direction of the force is opposite the direction of the electron)

Therefore:

-ma =  -4.01 * 10^{-15} N\\\\\\a = \frac{-4.01 * 10^{-15}}{-m}

Mass, m, of an electron = 9.11 * 10^{-31} kg

=> a = \frac{-4.01 * 10^{-15}}{-9.11 * 10^{-31}}\\\\\\a = 4.40 * 10^{15} m/s^2

The acceleration of the electron is 4.40 * 10^{15} m/s^2

5 0
3 years ago
A flagpole is a simple device on which a flag can be raised or lowered. Which simple machine is usually part of a flag pole?
kozerog [31]
The anwser is C it can also be A but C would make the most sense
4 0
3 years ago
Read 2 more answers
How to I isolate t? (time value)
Vlad1618 [11]
Heya!!!

Answer to your question:

All you need to do is to compare both the equations...
On comparing, you'll get
d=6m
v=6.5m/s
a=-9.81m/s^2
we know,

6= 6.5t -0.5 (9.81)t^2
6=6.5t-4.905 t^2
4.905 t^2-6.5t +6=0
now you can use quadratic eq. to solve this.

Hope it helps ^_^
4 0
3 years ago
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