Answer:
Explanation:
From the question;
We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.
We are to calculate the following task, i.e. to determine the electric field at the distances:
a) at 4.75 cm
b) at 20.5 cm
c) at 125.0 cm
Given that:
the charge (q) = 33.3 nC/m
= 33.3 × 10⁻⁹ c/m
radius of rod = 5.75 cm
a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.
Then, the electric field will be zero.
b) The electric field formula 
E = 1461.95 N/C
c) The electric field E is calculated as:
E = 239.76 N/C
Answer:
Power of the string wave will be equal to 5.464 watt
Explanation:
We have given mass per unit length is 0.050 kg/m
Tension in the string T = 60 N
Amplitude of the wave A = 5 cm = 0.05 m
Frequency f = 8 Hz
So angular frequency 
Velocity of the string wave is equal to 
Power of wave propagation is equal to 
So power of the wave will be equal to 5.464 watt
The mass would be the same
47kg on the moon as well
Answer:
5.619×10⁶ N
Explanation:
Applying,
F = kqq'/r²................... Equation 1
Where F = electrostatic force between the charges, k = coulomb's constant, q = first charge, q' = second charge, r = distance btween the charges
From the questiion,
Given: q = 2.5 C, q' = 2.5 C, r = 100 m
Constant: 8.99×10⁹ Nm²/C²
Substitute these values into equation 1
F = (2.5×2.5×8.99×10⁹)/100²
F = 56.19×10⁵
F = 5.619×10⁶ N
Answer:
Explanation:
= Initial diameter = 4 cm
= Initial velocity = 1 m/s
= Final diameter = 7.8 m
= Final velocity
= Area = 
From the continuity equation we get
The speed of water at the second floor is .
