Answer:
1 N
Explanation:
From coulomb's law,
The force of attraction between two charges is inversely proportional to the square of the distance between the charges.
From the question,
Assuming the charges are the same in both case,
F ∝ /r²....................... Equation 1
Fr² = k
F'r'² = Fr²........................... Equation 2
Where F' = First Force, r'² = First distance, F = second force, r² = second distance.
make F the subject of the equation,
F = F'r'²/r².................... Equation 3
Given: F' = 4 N, r' = 3 m, r = 6 m
Substitute into equation 3
F = 4(3²)/6²
F = 36/36
F = 1 N
Answer:
Gamma rays are the most energetic form of rays. A gamma ray burst can emit a large amount of energy in a very short interval of time. In fact, a short gamma-ray burst occurs in less than 2 seconds. A leading hypothesis states that when two neutron stars merge or when a neutron star merges with a black hole, short gamma-ray burst originates. Such merger are expected to produce kilonovoe.
Answer:
50J
Explanation:
At the top you have(A)
KE_a = O
PE_a = 100J
KE + PE = 100J
At the bottom you have (C)
KE_c= 100J
PE_c=0J
KE+PE = 100J
At point C:
You are at half the height.
We know that at H, PE =100J
PE_c = mgH
At C,
PE_c= mg (H/2) *at half the height
*m and g stay the same
Intuitively, the higher you are, the more potential energy you have.
If you decrease the height by a half, your PE will also decrease
At A:
PE_a / (mg) = H
At B:
PE_b / (mg) = H/2
to also get H on the right hand side, multiply by 2
2 (PE_b/ (mg))= H
2PE_b / (mg) = H
Ok, now that we have set up 2 equations (where H is isolated), find PE at B
AT A = AT B *This way you are saying that H = H (you compare both equations)
PE_a / (mg) = 2x PE_b / (mg)
*mg are the same for both cancel them (you can do that because of the = sign)
PE_a = 2PE_b
We know that PE_a = 100J
100J/2 = PE_b
PE at b = 50J
**FIND KE at b
We know that
KE_b + PE_b is always 100J
100J = 50J + KE_b
KE_b = 50J
Answer:
E = 440816.32 N/C
Explanation:
Given data:
Three point charge of charge equal to +3.0 micro coulomb
fourth point charge = - 3.0 micro coulomb
side of square = 0.50 m
N.m^2/c^2
Due to having equal charge on center of square, 2 charge produce equal electric field at center and other two also produce electric field at center of same value
So we have



[
[
]
plugging all value



E = 440816.32 N/C
The power required to force the current of 4.13 A to flow through the conductor is 1927.43 watts
<h3>What is power? </h3>
This is defined as the rate in which energy is consumed. Electrical power is expressed mathematically as:
Power (P) = square current (I²)× resistancet (R)
P = I²R
<h3>How to determine the power</h3>
- Current (I) = 4.13 A
- Resistance (R) = 113 ohms
- Power (P) =?
P = I²R
P = 4.13² × 113
P = 1927.43 watts
Thus, the power required is 1927.43 watts
Learn more about electrical power:
brainly.com/question/64224
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