Question: A loader sack of total mass
is l000 grams falls down from
the floor of a lorry 200 cm high
Calculate the workdone by the
gravity of the load.
Answer:
19.6 Joules
Explanation:
Applying
W = mgh........................ Equation 1
Where W = Workdone by gravity on the load, m = mass of the loader sack, h = height, g = acceleration due to gravity
From the question,
Given: m = 1000 grams = (1000/1000) kilogram = 1 kg, h = 200 cm = 2 m
Constant: g = 9.8 m/s²
Substitute these values into equation 1
W = (1×2×9.8)
W = 19.6 Joules
Hence the work done by gravity on the load is 19.6 Joules
The length of a 2 sec pendulum is 1 m.
Given that, initial length of the simple pendulum L₁ = 1 m
Initial time period T₁ = 2 sec
We need to find the length of the pendulum whose time period is 2 sec
T₂ = 2 sec
L₂ = ?
We know that the time period of the simple pendulum is given by the formula,
T = 2π√(L/g)
From the above relation, we can write T ∝ √L
T₁ / T₂ = √(L₁/L₂)
Making L₂ from the above relation, we have,
L₂ = (T₂² * L₁)/ T₁² = 2² * 1/ 2² = 1 m
Thus, the length of a 2 sec pendulum is 1 m.
To know more about time period:
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Answer:
B. An electric current into a magnetic field
Explanation:
The generation of electrical power requires relative motion between a magnetic field and a conductor. In a generator, mechanical energy is converted into electrical energy. The electricity produced by most generators is in the form of alternating current.
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