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Novay_Z [31]
4 years ago
14

What sentence(s) is/are true when we talk about equipotential lines?

Physics
1 answer:
rodikova [14]4 years ago
6 0

Answer:

a. True

Explanation:

Equipotential lines are the imaginary lines in the space where actually the electric potential is same at each and every point.

Work is not required to move along such points of the equipotential line because the movement is always perpendicular to the electric field lines because these lines are always perpendicular to the electric field lines.

The electric potential for a point charge is given mathematically as:

V=\frac{1}{4\pi.\epsilon_0}\times \frac{Q}{r}

where:

Q= magnitude of the point charge

r= radial distance form the charge

\epsilon_0= permittivity of free space

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A 12-V battery maintains an electric potential difference between two parallel metal plates separated by 10 cm. What is the elec
Sauron [17]

Answer:

The electric field between the plates is 120 V/m.

(c) is correct option.

Explanation:

Given that,

Potential difference = 12 volt

Distance = 10 cm = 0.1 m

We need to calculate the electric field between the plates

Using formula of electric field

E = \dfrac{V}{d}

Where, V = potential difference

d = distance between the plates

Put the formula

E =\dfrac{12}{0.1}

E=120\ V/m

Hence, The electric field between the plates is 120 V/m.

4 0
4 years ago
In an "atom smasher," two particles collide head on at relativistic speeds. If the velocity of the first particle is 0.741c to t
USPshnik [31]

Answer:

W_x = 0.9156\ c

Explanation:

given,

velocity of particle 1 = 0.741 c to left

velocity of second particle = 0.543 c to right

relative velocity between the particle = ?

for the relative velocity calculation we have formula

W_x = \dfrac{|u_x - v_x|}{1-\dfrac{u_xv_x}{c^2}}

u_x = 0.543 c

v_x = - 0.741 c

W_x = \dfrac{0.543 c - (-0.741 c)}{1-\dfrac{(0.543 c)(-0.741 c)}{c^2}}

W_x = \dfrac{0.543 c +0.741 c)}{1+\dfrac{(0.543)(0.741)c^2}{c^2}}

W_x = \dfrac{1.284c}{1+0.402363}

W_x = 0.9156\ c

Relative velocity of the particle is W_x = 0.9156\ c

5 0
3 years ago
A player throws a football 50.0 m at 61.0° north of west. what is the westward component of the displacement of the football?
-BARSIC- [3]

Answer: 24.24 m

Explanation:

A player throws football 50.0 m at 61° North of west. we will write this in terms of horizontal and vertical components.

Horizontal component: 50 cos 61° = 24.24 m which is westwards

Vertical component: 50 sin 61° = 43.73 m which towards North.

Refer to diagram below.

Thus, the westward component of displacement of the football is the horizontal component of the displacement = 24.24 m.

7 0
3 years ago
Please help me .....explain an experiment of phenomenon of rainfall​
barxatty [35]

Answer:

Hindi ko alam pasensya ka ha godbless

8 0
3 years ago
A hammer is used to hit a nail into a board. Which statement is correct about the forces at play between the nail and the hammer
emmasim [6.3K]

Answer:

Hewo My Lovelys!!

Answer is down below!!

Explanation:

The answer is C) The nail exerts an equal force on the hammer in the opposite direction.

Reason: The Newtons third law states that there is an equal an opposite reaction for every action. When hammer pushes the nail, the nail will push the hammer back in opposite direction. When the hammer hits a nail then nail will exert the equal and opposite force to the hammer. These both objects will exert force on each other in opposite directions.

Hope this helps!! =3

Have a great day, evening, of night!! <3

~ XxGhostMosskitxX

5 0
3 years ago
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