Answer:
a) (95.4 i^ + 282.6 j^) N
, b) 298.27 N 71.3º and c) F' = 298.27 N θ = 251.4º
Explanation:
a) Let's use trigonometry to break down Jennifer's strength
sin θ = Fjy / Fj
cos θ = Fjx / Fj
Analyze the angle is 32º east of the north measuring from the positive side of the x-axis would be
T = 90 -32 = 58º
Fjy = Fj sin 58
Fjx = FJ cos 58
Fjx = 180 cos 58 = 95.4 N
Fjy = 180 sin 58 = 152.6 N
Andrea's force is
Fa = 130.0 j ^
We perform the summary of force on each axis
X axis
Fx = Fjx
Fx = 95.4 N
Axis y
Fy = Fjy + Fa
Fy = 152.6 + 130
Fy = 282.6 N
F = (95.4 i ^ + 282.6 j ^) N
b) Let's use the Pythagorean theorem and trigonometry
F² = Fx² + Fy²
F = √ (95.4² + 282.6²)
F = √ (88963)
F = 298.27 N
tan θ = Fy / Fx
θ = tan-1 (282.6 / 95.4)
θ = tan-1 (2,962)
θ = 71.3º
c) To avoid the movement they must apply a force of equal magnitude, but opposite direction
F' = 298.27 N
θ' = 180 + 71.3
θ = 251.4º
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(a) 907.5 N/m
The force applied to the spring is equal to the weight of the object suspended on it, so:
The spring obeys Hook's law:
where k is the spring constant and 
is the stretching of the spring. Since we know 
, we can re-arrange the equation to find the spring constant:
(b) 1.45 cm
In this second case, the force applied to the spring will be different, since the weight of the new object is different:
So, by applying Hook's law again, we can find the new stretching of the spring (using the value of the spring constant that we found in the previous part):
(c) 3.5 J
The amount of work that must be done to stretch the string by a distance is equal to the elastic potential energy stored by the spring, given by:
Substituting k=907.5 N/m and 
, we find the amount of work that must be done:
DE which is the differential equation represents the LRC series circuit where
L d²q/dt² + Rdq/dt +I/Cq = E(t) = 150V.
Initial condition is q(t) = 0 and i(0) =0.
To find the charge q(t) by using Laplace transformation by
Substituting known values for DE
L×d²q/dt² +20 ×dq/dt + 1/0.005× q = 150
d²q/dt² +20dq/dt + 200q =150
