Torque on a closed current carrying loop is given by formula
here
N = number of turns = 1
i = current = 4 A
A = area of the loop
B = magnetic field = 0.60 T
= angle between normal of the plane and magnetic field = 90 - 30 = 60 degree
Now we will have
So area of the loop is 0.53 m^2
Answer:
the curve inclination is increased so that a weight component helps keep the car on track
Explanation:
In the sledging competition these devices go at quite high speeds over 100 km/h, so when reaching the curves the friction force is not enough to keep the car on the track. For this reason, the curve inclination is increased so that a weight component helps keep the car on track.
In general we can solve Newton's second law for this case, with the condition of no friction, it is found that
V² = r g tan θ
Where V is the maximum velocity, r is the radius of the curve a, θ is the angle of the inclination
Answer:
0.018 J
Explanation:
The work done to bring the charge from infinity to point P is equal to the change in electric potential energy of the charge - so it is given by
where
is the magnitude of the charge
is the potential difference between point P and infinity
Substituting into the equation, we find
Answer:
35°C
Explanation:
q = mCΔT
2130 J = (0.200 kg) (710 J/kg/°C) (T − 20.0°C)
T = 35°C
Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
fo = 5408 HZ
