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Paladinen [302]
3 years ago
15

When the charges in the rod are in equilibrium, what is the magnitude of the electric field within the rod?

Physics
1 answer:
Anarel [89]3 years ago
3 0

Answer: If we have equilibrium, the magnitude must be zero.

Explanation:

If the charges are in equilibrium, this means that the total charge is equal to zero.

And as the charges must be homogeneously distributed in the rod, we can conclude that the electric field within the rod must be zero, so the magnitude of the electric field must be zero

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s344n2d4d5 [400]

Answer:

Is the amount of water evaporated in a day more significant than the amount of rain falling in a day?

Explanation:

Don't overcomplicate it

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3 years ago
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How far can you get away from your little sister with a squirt gun filled with paint if you can travel at 3 m/s and you have 15s
Kay [80]
It would be what 3•15 is so that's 45 m i think
5 0
3 years ago
The uniform 100-kg beam is freely hinged about its upper end A and is initially at rest in the vertical position with theta = 0.
max2010maxim [7]

Answer:

ac = 2.86 m / s²

Explanation:

Image can detail the system to determine the force in the FA to understand the system into the applicated force

m = 100 kg ,  L = 3 m

∑ F = 0 ⇒ Ay - 100 kg + P * cos (45) = 0

Ay = 768.86 N

∑ Mₐ = α * I ₐ

I ₐ = m * L² / 3  ⇒  I ₐ = 100 kg * 4² m / 3

Replacing

P * sin (45) * 3 = α * 100 kg * 4² m / 3  

α = 1.193 rad / s²

ac = α *2    ⇒  ac = 1.193 rad / s²  * 2

ac = 2.86 m / s²

4 0
3 years ago
two charges having the same charge magnitude experiencing an attracting force of 3.60N when the charges are 30cm apart.what is t
Tomtit [17]

The charges have opposite sign and magnitude 6 \mu C

Explanation:

The magnitude of the electrostatic force between two electric charges is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the two charges

r is the separation between the two charges

In this problem, we have:

F = 3.60 N is the force between the two charges

r = 30 cm = 0.30 m is their separation

The two charges have same magnitude, so

q_1 = q_2 = q

So we can rewrite the equation as

F=\frac{kq^2}{r^2}

And solving for q:

q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(3.60)(0.30)^2}{8.99\cdot 10^9}}=6\cdot 10^{-6} C = 6\mu C

Moreover, the force between the charges is attractive: we know that charges of same sign repel each other while charges of opposite sign attract each other, therefore the charges in this problem have opposite sign, so

q_1 = 6 \mu C\\q_2 = -6 \mu C

Learn more about electric force:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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3 years ago
Which of the following statements is TRUE?
Marrrta [24]

Answer:

I think C......................

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