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Svetllana [295]
2 years ago
15

What are the magnitude and direction of electric field at the center of square?

Physics
1 answer:
Kobotan [32]2 years ago
6 0

Answer:

hello the diagram related to this question is missing attached below is the missing diagram

Answer :

The magnitude of the electric field = 4KQ / L^2

direction = 45° east to south

Explanation:

The magnitude of the electric field = 4KQ / L^2

direction = 45° east to south

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Answer:

Internal resistance refers to the opposition to the flow of current offered by the cells and batteries themselves resulting in the generation of heat. Internal resistance is measured in Ohms. ... Example: 1 The potential difference across the cell when no current flows through the circuit is 3 V.

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8 0
3 years ago
Why are Watson’s experiments considered controversial?
iris [78.8K]
D, it is considered unethical today
4 0
3 years ago
Read 2 more answers
A juggler throws a bowling pin straight up with an initial speed of 8.20 m/s. How much time elapses until the bowling pin return
tensa zangetsu [6.8K]

Answer:

1.68 s

Explanation:

From newton's equation of motion,

a = (v-u)/t.................................. Equation 1

Making t the subject of the equation

t =(v-u)g............................. Equation 2

Where t = time taken for the bowling pin to reach the maximum height, v = final velocity bowling pin, u = initial velocity of the bowling pin, g = acceleration due to gravity.

Note: Taking upward to be negative and down ward to be positive,

Given: v = 0 m/s ( at the maximum height), u = 8.20 m/s, g = -9.8 m/s²

t = (0-8.20)/-9.8

t = -8.20/-9.8

t = 0.84 s.

But,

T = 2t

Where T = time taken for the bowling pin to return to the juggler's hand.

T = 2(0.84)

T = 1.68 s.

T = 1.68 s

7 0
3 years ago
At what height h above the ground does the projectile have a speed of 0.5v?
maw [93]

Answer:

h=\dfrac{3v^2}{8g}

Explanation:

It is given that,

Speed of the projectile is 0.5 v. Let h is the height above the ground. Using the first equation of motion to find it.

v=u+at

v=u-gt

Initial speed of the projectile is v and final speed is 0.5 v.

0.5v=v-gt

t=\dfrac{v}{2g}

g is the acceleration due to gravity

Let h is the height above the ground. Using the second equation of motion as :

h=vt-\dfrac{1}{2}gt^2

h=v\dfrac{v}{2g}-\dfrac{1}{2}g(\dfrac{v}{2g})^2

h=\dfrac{3v^2}{8g}

So, the height of the projectile above the ground is \dfrac{3v^2}{8g}. Hence, this is the required solution.

6 0
3 years ago
The acceleration due to gravity on the moon is about 1/6 of the acceleration due to gravity on the earth. A net force F acts hor
uysha [10]

Answer:

c.a_m

Explanation:

We are given that

Acceleration due to gravity on the moon=a_m

Acceleration due to gravity on the earth=a_e

g_m=\frac{1}{6}g_e

Net force due to am on an object on moon=F_{net}=ma_m

There is no friction and no drag force and there is no gravity involved

Then, the force acting on an object on earth=F=ma_e

F=F_{net}(given)

ma_m=ma_e

a_e=a_m

Hence, option c is true.

3 0
2 years ago
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