Answer:
Focal length of the convex lens is 2.72 cm
Explanation:
As we know that the convex lens produce 1.2 times magnified image
Now we know that the object distance from the lens is 5 cm
so we have
now we have
now by the formula of lens
A different group of students decided to do this experiment by
1 answer:
Answer:
Using an object of less mass can reduce the energy lost due to friction.
Explanation:
let the mass of the object be M.
acceleration due to gravity be g, coefficient of kinetic friction between object and slider be , angle between horizntal and slider be ∅ and distance slided by the object be d.
work done by friction of the ramp will be the energy lost by ramp due to friction.
work done = force×displacement
force = ×M×g×cos∅
displacement= d
therefore work done = ×M×g×cos∅×d
therefore on decreasing the mass M we can see that work done by friction decreases so the energy lost due to friction also decreases.
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Answer:
The new value of g
Explanation:
Mass =290 g
Spring size = 40 cm
The size of spring after streches = 23 cm
Pulling the mass down at = 11.7 cm
The stopwatch shows the number of oscillations in 17.7 s = 9 osscillations.
The oscillation frequency is independent to g.
The amount of spring stretch due to the hanging mass, x, is dependent on g.
Therefore, the new value of g
The maximum voltage generated in the coil is 3.14 V.
According to the question, we are provided that a bicycle turns 1000 of the area of 0.10 m2, where the coil rotates half a revolution per second, which is placed in a uniform magnetic field of 0.01 T.
Therefore, no .of turns the coil took, N = 1000
Area of the coil, A= 0.10 m²
Frequency of the system =0.5 revolutions per second
= 0.5 Hz
The intensity of the Magnetic field , B =0.01 T
So the Equation we refer to for the calculating the desired value
E(ind) =
= -d(NBAcosωt) / dt
= − NBA d(cosωt) /dt
= NBAωsinωt
if sinωt=1,
maximum E(ind)= NBAω
Since we know, ω=2πf
Therefore, (Eind)max = NBA(2πf)
= (1000)(0.01)(0.1)(2π)(0.5)
(E(ind))max= 3.14v.
the value for the voltage we got is 3.14 V
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