Answer:
Current in the loop is 0.063 A
Explanation:
Number of turns in the coil N = 150
Radius of the circular loop r = 7.5 cm = 0.075 m
So area 
Magnetic field B = 1.5 T
Maximum torque is given 
We have to find current in the coil
Torque on circular coil in magnetic field is equal to
I = 0.063 A
Answer:
6 voltage is applied by the batteries.
Explanation:
To solve this sort of problem involving current, resistance and voltage, we use the relation:<em> </em><em>Voltage</em><em>= </em><em>Current x Resistance</em>.
From the problem, the following have been given:
Resistance= 2.4 ohms.
Current= 2.5 amps.
Required: Voltage?
Fix the values of current and resistance into the relation:
Voltage= <em>2.5 x 2.4</em>
=6 volts.
Answer:
Internal resistance = 0.545 ohm
Explanation:
As per ohm's law we know that
here we know that
i = electric current = 2.75 A
V = potential difference = 1.50 Volts
now from above equation we have
now we have
Because their is nothing at the geographical poles that attracts the magnet
Answer:
k = 1 700.7 N/m
v0 = 9.8 m/s^2
Explanation:
Hello!
We can answer this question using conservation of energy.
The potential energy of the spring (PS) will transform to kinetic energy (KE) of the ball, and eventually, when the velocity of the ball is zero, all that energy will be potential gravitational (PG) energy.
When the kinetic energy of the ball is zero, that is, when it has reached its maximum heigh, all the potential energy of the spring will be equal to the potential energy of the gravitational field.
PS = (1/2) k x^2 <em>where x is the compresion or elongation of the spring</em>
PG = mgh
a)
Since energy must be conserved and we are neglecting any energy loss:
PS = PG
Solving for k
k = (2mgh)/(x^2) = ( 2 * 1.7 * 9.81 * 4.9 Nm)/(0.31^2 m^2)
k = 1 700.7 N/m
b)
Since the potential energy of the spring transfors to kinetic energy of the ball we have that:
PS = KE
that is:
(1/2) k x^2 = (1/2) m v0^2
Solving for v0
v0 = x √(k/m) = (0.31 m ) √( 1 700.7 N/m / 1.7kg)
v0 = 9.8 m/s^2
