Answer:
The force required to begin to lift the pole from the end 'A' is 240 N
Explanation:
The given parameters for the pole AB are;
The length of the pole, l = 10.0 m
The weight of the pole, W = 600 N ↓
The distance of the center of gravity of the pole from the side 'A' = 4.0 m
Let '
' represent the force required to begin to lift the pole from the end 'A' and let a force applied in the upwards direction be positive
For equilibrium, the sum of moment about the point 'B' = 0, therefore, taking moment about 'B', we have
× 10.0 m - W × 4.0 m = 0
∴ × 10.0 m = W × 4.0 m = 600 N × 4.0 m
× 10.0 m = 600 N × 4.0 m
∴ = 600 N × 4.0 m/(10.0 m) = 240 N
The force required to begin to lift the pole from the end 'A', = 240 N.
Answer:
13.78 mT
Explanation:
The peak voltage ε = ωNAB where ω = angular speed of coil = 1500 rpm = 1500 × 2π/60 rad/s = 50π rad/s = 157.08 rad/s, N = number of turns of coil = 250, A = area of coil = πr² where r = radius of coil = 10 cm = 0.10 m,
A = π(0.1 m)² = 0.03142 m² and B = magnetic field strength
So,
B = ε/ωNA
substituting the values of the variables into the equation given that ε = 17 V
So, B = ε/ωNA
B = 17 V/(157.08 rad/s × 250 turns × 0.03142 m²)
B = 17 V/(1233.8634 rad-turns-m²/s)
B = 0.01378 T
B = 13.78 mT
(a) -48.0 cm, diverging
We can use the lens equation:
where
f is the focal length
p = 16.0 cm is the object distance
q = -12.0 cm is the image distance (with a negative sign because the image is on the same side as the object, so it is virtual)
Solving for f, we find the focal length of the lens:
The lens is diverging, since the focal length is negative.
(b) 6.38 mm, erect
We can use the magnification equation:
where
y' is the size of the image
y = 8.50 mm is the size of the object
Substituting p and q that we used in the previous part of the problem, we find y':
and the image is erect, since the sign is positive.
(c)
See attached picture.
Answer:
The frequency of the signal is 2 GHz
Explanation:
Given;
period of the clock signal, T = 500 ps = 500 x 10⁻¹² s
the frequency of the signal is given by;
F = 2 GHz
Therefore, the frequency of the signal is 2 x 10⁹ Hz or 2 GHz
Answer:
7.3 newtons to the west
Explanation:
3.7kg × 11a - 3.7kg × ? = 3.7n
