What is the voltage drop across an alarm clock that is connected to a circuit with a current of 1.10A and a resistance of 90Ω?
1 answer:
You might be interested in
Since there is no friction between the ladder and the wall, there can be no vertical force component. That's the tricky part ;)
So to find the weight, divide the 100N <em>normal</em> force by earths gravitational acceleration, 9.8m/s^2
Then;
Draw an arrow at the base of the ladder pointing towards the wall with a value of 30N, to show the frictional force.
So to find the weight, divide the 100N <em>normal</em> force by earths gravitational acceleration, 9.8m/s^2
Then;
Draw an arrow at the base of the ladder pointing towards the wall with a value of 30N, to show the frictional force.
Explanation:
The electric field of an isolated charged parallel-plate capacitor is given by :
........(1)
Where
q is the electric charge
A is the area of cross section of parallel plate
It is clear from equation (1) that the electric field of a parallel plate capacitor is directly proportional to the charge on the plate and inversely proportional to the area of cross section of a plate.
So, the correct option is (E) i.e. "none of the above".
Answer
given,
F₁ = 15 lb
F₂ = 8 lb
θ₁ = 45°
θ₂ = 25°
Assuming the question's diagram is attached below.
now,
computing the horizontal component of the forces.
F_h = F₁ cos θ₁ - F₂ cos θ₂
F_h = 15 cos 45° - 8 cos 25°
F_h = 3.36 lb
now, vertical component of the forces
F_v = F₁ sin θ₁ + F₂ sin θ₂
F_v = 15 sin 45° + 8 sin 25°
F_v = 13.98 lb
resultant force would be equal to
F = 14.38 lb
the magnitude of resultant force is equal to 14.38 lb
direction of forces
θ = 76.48°
