Because charges are positioned on a square the force acting on one charge is the same as the force acting on all others.
We will use superposition principle. This means that force acting on the charge is the sum of individual forces. I have attached the sketch that you should take a look at.
We will break down forces on their x and y components:
Let's figure out each component:
Total force acting on the charge would be:
We need to calculate forces along x and y axis first( I will assume you meant micro coulombs, because otherwise we get forces that are huge).
Now we can find the total force acting on a single charge:
As said before, intensity of the force acting on charges is the same for all of them.
All you need to know for this question is Ohm’s Law:
V=IR
I=V/R
R=V/I
So, the answer is (3) Resistance, since it is inversely proportional to Current (I=V/R)
Answer:
Power = 180 Watt.
Explanation:
W = Work done = m x g x H.
m = mass of the body.
g = acceleration due to gravity = 9.8 m/s^2.
W = weight of the body = m x g = 720 N.
H = height of the body = 5 m.
t = Time = 20 s.
Plugging the above values in the formula we get:
W = 720 x 5 = 3600 J.
Power = 
= 
= 180 Watt.
Therefore the required value of power = 180 Watt.
The type of waves used by bats are sound waves. Most of the species use their larynx to produce ultrasound waves in the frequency range of 20 to 200 kilohertz.
These sound waves are echoed, reflected, by surroundings, in this case food or prey. These reflections are received by the specialized receptor cells in the ears of bats. The reflections are analyzed by the brain to make an image.
Fun fact: The brain cells of bats are also specialized to better analyze the frequency of ultrasound used by the bat.
Given
Weight of the block A, Wa = 20 lb, weight of block B Wb = 50 lb. Applied
force to block A, P = 6lb, coefficient of static friction µs = 0.4, coefficient
of kinetic friction µk = 0.3. If a force P
is applied to the body, no relative motion will take place until the applied
force is equal to the force of friction Ff, which is acting opposite to the
direction of motion. Magnitude of static force of friction between block A and
block B, Fs = µsN, where N is
reaction force acting on block A. Now, resolve the forces Fx = max. P = (mA +
mB)a,
6 = (20 / 32.2 + 50 / 32.2)a
2.173a = 6
A = 2.76 ft/s^2
To check slipping occurs between block A and block B, consider block A:
P – Ff = mAaA
6 – Ff = 1.71
Ff = 4.29 lb
And also,
N = wA. We know static friction,
Fs = µsN
Fs = 0.4 x 20
Fs = 8lb
Frictional force is less than static friction. Ff < Fs
<span>Therefors, acceleration of block A, aA = 2.76 ft/s^2, acceleration of
block B aB = 2.76 ft/s^2</span>
