Describe how switching the desk lamp on and off shows that light waves transfer energy
2 answers:
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Answer:
A) P.E = 138.44 J
B) The velocity of swing at bottom, v = 3.33 m/s
C) The work done, W = -138.44 J
Explanation:
Given,
The mass of the child, m = 25 Kg
The length of the swing rope, L = 2.2 m
The angle of the swing to the vertical position, ∅ = 42°
A) The potential energy at the initial position ∅ = 42° is given by the relation
P.E = mgh joule
Considering h = 0 for the vertical position
The h at ∅ = 42° is h = L (1 - cos∅)
P.E = mgL (1 - cos∅)
Substituting the given values in the above equation
P.E = 25 x 9.8 x 2.2 (1 - cos42°)
= 138.44 J
The potential energy for the child just as she is released, compared to the potential energy at the bottom of the swing is, P.E = 138.44 J
B) The velocity of the swing at the bottom.
At bottom of the swing the P.E is completely transformed into the K.E
∴ K.E = P.E
1/2 mv² = 138.44
1/2 x 25 x v² 138.44
v² = 11.0752
v = 3.33 m/s
The velocity of the swing at the bottom is, v = 3.33 m/s
C) The work done by the tension in the rope from initial position to the bottom
Tension on string, T = Force acting on the swing, F
=
= - 2.2 x 25 x 9.8 [cos0 - cos 42°]
= - 138.44 J
The negative sign in the in energy is that the work done is towards the gravitational force of attraction.
The work done by the tension in the ropes as the child swings from the initial position to the bottom of the swing, W = - 138.44 J
Answer:
The charge resides on the outer surface = C
Explanation:
Surface area of cell
Separation between two plate
Dielectric constant
Potential difference
The capacitance of parallel plate capacitor in free space is given by,
Where permittivity of free space =
The Capacitance of capacitor is increase by times when it placed in dielectric medium.
And we know that,
So charge on the outer surface is given by,
The magnitude of the force of friction is 40 N
Explanation:
To solve the problem, we just have to analyze the forces acting on the student and the scooter along the horizontal direction. We have:
- The constant pushing force forward, of magnitude F = 40 N
- The frictional force, acting backward,
Since the two forces are in opposite direction, the equation of motion is
where
m is the mass of the student+scooter
a is the acceleration
However, here the scooter is moving at constant speed: this means that its acceleration is zero, so
a = 0
And therefore,
which means that the magnitude of the force of friction is also equal to 40 N.
Learn more about force of friction here:
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