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lyudmila [28]
2 years ago
8

Two charged objects attract each other with a force of F. What happens to the force between them if one charge is doubled, the o

ther charge is tripled, and the separation distance between their centers is reduced to ¼ its original value?
Physics
1 answer:
Ray Of Light [21]2 years ago
3 0

Answer:

The force is increased by a factor of 96. (F1 = 96F)

Explanation:

Let the first charge be Q1

Let the second charge be Q2

Let the distance between their centers be r

The electrostatic force, F, between them is:

F = (k*Q1*Q2) / r²

If the first charge is doubled = 2Q1

If the second charge is tripled = 3Q2

The separation between their centers is reduced to ¼ = ¼r

The force between them becomes:

F1 = (k* 2Q1 * 3Q2) / (¼r)²

F1 = (6 * k * Q1 * Q2) / (r²/16)

F1 = 96(k * Q1 * Q2) /r²

F1 = 96F

The force is increased by a factor of 96.

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An American traveler in China carries a transformer to convert China's standard 220 V to 120 V so that she can use some small ap
Arisa [49]

Answer:

(a) The ratio of turns in the primary and secondary coils of her transformer is 1.833

(b) The ratio of input to output current is 0.55

(c) To increase the output voltage, you can either increase the number of turns in the secondary coil (step-up) or increase the input current. Therefore, the Chinese person has to increase the input current of the transformer to achieve an increased output voltage that can power her 220 V appliances.

Explanation:

Given;

input voltage, V_p = 220 V

output voltage, V_s = 120 V

General transformer equation is given as;

\frac{V_p}{V_s} = \frac{N_p}{N_s} = \frac{I_s}{I_p}

where;

Np is number of turns in the primary coil

Ns is number of turns in the secondary coil

Is - is the secondary current or output current

Ip - is the primary current or input current

(a) The ratio of turns in the primary and secondary coils of her transformer;

\frac{N_p}{N_s} = \frac{V_p}{V_s} \\\\\frac{N_p}{N_s} = \frac{220}{120} = 1.833

(b) The ratio of input to output current;

\frac{I_p}{I_s} = \frac{V_s}{V_p} \\\\\frac{I_p}{I_s} = \frac{120}{220} \\\\\frac{I_p}{I_s} = 0.55

(c) To increase the output voltage, you can either increase the number of turns in the secondary coil (step-up) or increase the input current. Therefore, the Chinese person has to increase the input current of the transformer to achieve an increased output voltage that can power her 220 V appliances.

8 0
2 years ago
Which two of the following involve the same energy transfer. Assume that the same substance and the same mass is involved in all
Elanso [62]
B. evaporation
c. condensation

They are opposite processes that involve the same transfer of energy
3 0
2 years ago
A solid sphere has a radius of 0.200 m and a mass of 150.0 kg. how much work is required to get the sphere rolling with an angul
Allisa [31]

Here in this case we can use work energy theorem

As per work energy theorem

Work done by all forces = Change in kinetic Energy of the object

Total kinetic energy of the solid sphere is ZERO initially as it is given at rest.

Final total kinetic energy is sum of rotational kinetic energy and translational kinetic energy

KE = \frac{1}{2}Iw^2 +\frac{1}{2} mv^2

also we know that

I = \frac{2}{5}mR^2

w= \frac{v}{R}

Now kinetic energy is given by

KE = \frac{1}{2}(\frac{2}{5}mR^2)w^2 +\frac{1}{2} m(Rw)^2

KE = \frac{1}{5}mR^2w^2 +\frac{1}{2} mR^2w^2

KE = \frac{7}{10}mR^2w^2

KE = \frac{7}{10}*150*(0.200)^2(50)^2

KE = 10500 J

Now by work energy theorem

Work done = 10500 - 0 = 10500 J

So in the above case work done on sphere is 10500 J

7 0
2 years ago
Can someone help me its really hard to do d is stuff​
Tasya [4]

Answer:

direction, speed

means the object is staying still, 0

newtons, N

the sum of all the forces acting on an object

Explanation:

3 0
3 years ago
A 44.0 kg uniform rod 4.90 m long is attached to a wall with a hinge at one end. The rod is held in a horizontal position by a w
pychu [463]

Answer:

x ≤ 3.6913 m

Explanation:

Given

Mrod = 44.0 kg

L = 4.90 m

Tmax = 1450 N

Mman = 69 kg

A: left end of the rod

B: right end of the rod

x = distance from the left end to the man

If we take torques around the left end as follows

∑τ = 0   ⇒   - Wrod*(L/2) - Wman*x + T*Sin 30º*L = 0

⇒   - (Mrod*g)*(L/2) - (Mman*g)*x + Tmax*Sin 30º*L = 0

⇒  -  (44*9.8)*(4.9/2) - (69*9.8)*x + (1450)*(0.5)*(4.9) = 0

⇒ x ≤ 3.6913 m

4 0
3 years ago
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