Answer: 45.3°
Explanation:
Given,
Length of ladder = l
Weight of ladder = w
Coefficient of friction = μs = 0.495
Smallest angle the ladder makes = θ
If we assume the forces in the vertical direction to be N1, and the forces in the horizontal direction to be N2, then,
N1 = mg and
N2 = μmg
Moment at a point A in the clockwise direction is
N2 Lsinθ - mg.(L/2).cosθ = 0
μmgLsinθ - mg.(L/2).cosθ = 0
μmgLsinθ = mg.(L/2).cosθ
μsinθ = cosθ/2
sin θ / cos θ = 1 / 2μ
Tan θ = 1 / 2μ
Substituting the value of μ = 0.495, we have
Tan θ = 1 / 2 * 0.495
Tan θ = 1 / 0.99
Tan θ = 1.01
θ = tan^-1(1.01)
θ = 45.3°
Answer:
Q' = 115.2 KJ
Explanation:
Given that
Current I = 10 A
Voltage = 120 V
Time t= 120 s
The energy supplied Q= V I t
Q= 10 x 120 x 120 J
Q= 144 x 1000 J
Q= 144 KJ
The change in the energy ΔEngg = 28.8 kJ
By using energy conservation
Q= Q' + ΔEngg
Q'=Wasted energy
Now by putting the values in the above equation
144 = Q' + 28.8 KJ
Q' = 115.2 KJ
Therefore the waste energy will be 115 kJ.
Answer:
The length traveled by an object moving in any direction or even changing direction is called distance. The location of an object in a frame of reference is called position. For straight line motion, positions can be shown using a number line. The separation between original and final position is called displacement
Explanation:
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Answer
Given,
Magnetic field, B = 0.0000193 T
speed, v = 121 m/s
mass of electron, m = 9.11 x 10⁻³¹ Kg
charge of electron, q = 1.6 x 10⁻¹⁹ C
radius of the electron path, r = ?


r = 3.64 x 10⁻⁵ m
We know frequency is inverse of time period
d = v t



t = 1.889 x 10⁻⁶ s.
now, frequency



Explanation:
For this problem, use the first law of thermodynamics. The change in energy equals the increase in heat energy minus the work done.
ΔU=Q−W
We are not given a value for work, but we can solve for it using the force and distance. Work is the product of force and displacement.
W=FΔx
W=3N×2m
W=6J
Now that we have the value of work done and the value for heat added, we can solve for the total change in energy.
ΔU=Q−W
ΔU=10J−6J
ΔU=4J
Answer is 4J
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