Answer:
The height of the bridge is 78.4 m.
Explanation:
Given;
time of the stone motion off the bridge, t = 4.0 s
acceleration due to gravity, g = 9.8 m/s²
The height of the bridge is given by;
h = ut + ¹/₂gt²
where;
u is the initial velocity of the stone, u = 0
h = ¹/₂gt²
h = ¹/₂(9.8)(4)²
h = 78.4 m
Therefore, the height of the bridge is 78.4 m.
Work is done when spring is extended or compressed. Elastic potential energy is stored in the spring. Provided inelastic deformation has not happened, the work done is equal to the elastic potential energy stored.
Answer: 2940 J
Explanation: solution attached:
PE= mgh
Substitute the values:
PE= 10kg x 9.8 m/s² x 30 m
= 2940 J
Answer: 580 N
Refer to attached figure.
The angle of inclination is 22 degrees
weight (gravitational force) acts downwards.
Normal force is a contact force which acts perpendicular to the point of contact.
The horizontal component (mg cos 22 ) balances the normal force and the vertical component balances the frictional force.
Gravitational force on an object = mg
The normal force 

Answer:
R1 = 5.13 Ω
Explanation:
From Ohm's law,
V = IR............... Equation 1
Where V = Voltage, I = current, R = resistance.
From the question,
I = 2 A, R = R1
Substitute into equation 1
V = 2R1................ Equation 2
When a resistance of 2.2Ω is added in series with R1,
assuming the voltage source remain constant
R = 2.2+R1, and I = 1.4 A
V = 1.4(2.2+R1)................. Equation 3
Substitute the value of V into equation 3
2R1 = 1.4(2.2+R1)
2R1 = 3.08+1.4R1
2R1-1.4R1 = 3.08
0.6R1 = 3.08
R1 = 3.08/0.6
R1 = 5.13 Ω