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2 years ago

Which two options would INCREASE the electric force between two charged particles?

1 answer:

7 0

The electric force between two charged particles can be increased by decreasing the distance between the two particles.

<h3>How to increase electric force between two charged particles.</h3>

The technique of decreasing the separation distance between objects increases the force of attraction or repulsion between the objects. while

increasing the separation distance between objects decreases the force of attraction or repulsion between the objects.

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Two identical charged pith balls are brought together to touch each other. They are then

sergejj [24]

Answer:

-17.5 nC

Explanation:

charge A = -30 nC

charge B = -5 nC

After adding them it would be the average of the two charges because of the getting same voltage difference. so

c = (-30+(-5)) / 2 nC

c= -17.5 nC

answer is -17.5 nC

4 0

3 years ago

A defibrillator is used during a heart attack to restore the heart to its normal beating pattern. A defibrillator passes 18 A of

miskamm [114]

Answer:

q = 0.036 C

Explanation:

Given that,

Current passes through a defibrillator, I = 18 A

Time, t = 2 ms

We need to find the charge moved during this time. We know that,

Electric current = charge/time

$q=It$

Put all the values,

$q=18\times 0.002\\\\q=0.036\ C$

So, 0.036 C of charge moves during this time.

8 0

3 years ago

Please help with 4 and 5, thank you :)

allochka39001 [22]

Answer: #4

Sally is faster.

Explanation:

If you multiply Sallies it is going to be less than Jessica's.

6 0

3 years ago

Ask Your Teacher Two long, straight wires are parallel and 11 cm apart. One carries a current of 2.9 A, the other a current of 5

dsp73

Answer:

The magnitude of force per unit length of one wire on the other is $2.7945\times 10^{-5}\ N$ and the direction is away from one another

The magnitude of force per unit length of one wire on the other is $2.7945\times 10^{-5}\ N$ and the direction is towards each other.

Explanation:

$\mu_0$ = Vacuum permeability = $4\pi\times 10^{-7}\ N/A^2$

$i_1$ = Current in first wire = 2.9 A

$i_2$ = Current in second wire = 5.3 A

r = Gap between the wires = 11 cm

Force per unit length

$F_{12}=F_{21}=\frac{\mu_0 i_1i_2}{2\pi r}\\ =\frac{4\pi\times 10^{-7}\times 2.9\times 5.3}{2\pi 0.11}\\ =2.7945\times 10^{-5}\ N$

The magnitude of force per unit length of one wire on the other is $2.7945\times 10^{-5}\ N$ and the direction is away from one another

$F_{12}=F_{21}=\frac{\mu_0 i_1i_2}{2\pi r}\\ =\frac{4\pi\times 10^{-7}\times 2.9\times 5.3}{2\pi 0.11}\\ =2.7945\times 10^{-5}\ N$

The magnitude of force per unit length of one wire on the other is $2.7945\times 10^{-5}\ N$ and the direction is towards each other.

7 0

3 years ago

A 2000-kg railway freight car coasts at 4.4 m/s underneath a grain terminal, which dumps grain directly down into the freight ca

ruslelena [56]

Answer:

933. 3kg

Explanation:

We are given that

Mass,m=2000 kg

Initial speed,u=4.4 m/s

We have to find the maximum mass of grain if the speed of loaded freight car must not go below 3.0 m/s.

Final speed,v=3 m/s

By conservation of momentum

Initial momentum=Final momentum

$mu=m'v$

Substitute the values

$2000\times 4.4=m'(3)$

$m'=\frac{2000\times 4.4}{3}$

$m'=2.93\times 10^3 kg$

Mass of freight loaded car=2933.3 kg

Mass of grains=2933.3-2000=933.3 kg

Hence, the maximum mass of grains that it can accept=933.3 kg

7 0

3 years ago