Answer:
8 m/s to the left.
Explanation:
Applying,
V = d/t...................... Equation 1
Where V = Velocity of the car, d = distance, t = time
From the question,
Given: d = 24 meters, t = 3 seconds
Substitute these values into equation 1
V = 24/3
V = 8 m/s to the left.
Hence the velocity of the car is 8 m/s to the left.
Given Information:
Resistance of circular loop = R = 0.235 Ω
Radius of circular loop = r = 0.241 m
Number of turns = n = 10
Voltage = V = 13.1 V
Required Information:
Magnetic field = B = ?
Answer:
Magnetic field = 0.00145 T
Explanation:
In a circular loop of wire with n number of turns and radius r and carrying a current I induces a magnetic field B
B = μ₀nI/2r
Where μ₀= 4πx10⁻⁷ is the permeability of free space and current in the loop is given by
I = V/R
I = 13.1/0.235
I = 55.74 A
B = 4πx10⁻⁷*10*55.74/2*0.241
B = 0.00145 T
Therefore, the magnetic field at the center of this circular loop is 0.00145 T
Answer:
Block A will have a final charge of 3.5nC.
Explanation:
This is because at the point of contact with Block B, which is electrically positive, the electrons in Block A will be attracted to the excess 'unpaired' protons in block B. Hence, the electrons will flow into Block B causing unpaired protons to remain in Block A.
This process is called Charging by Conduction.
This charging process will continue until the charges are evenly distributed between both objects.
In case you're wondering, "<em>how's all this possible within a few seconds</em>?", remember that electrons travel very fast and so, this process is a rather rapid one.
Answer:
a) ω = 9.86 rad/s
b) ac = 194. 4 m/s²
c) minimum coefficient of static friction, µs = 19.8
Explanation:
a) angular speed, ω = 2πf, where f is frequency of revolution
1 rps = 6.283 rad/s, π = 3.142
ω = 2 * 3.14 * 0.25 * 6.28
ω = 9.86 rad/s
b) centripetal acceleration, a = rω²
where r is radius in meters; r = 200 cm or 2 m
a = 2 * 9.86²
a = 194. 4 m/s²
c) µs = frictional force/ normal force
frictional force = centripetal force = ma; where a is centripetal acceleration
normal force = mg; where g = 9.8 m/s²
µs = ma/mg = a/g
µs = 194.4 ms⁻²/9.8 ms⁻²
c) minimum coefficient of static friction, µs = 19.8
