How much work is required to push a 2.0-kg object up a frictionless inclines plane whose length is 2.0m and whose height is 1.0m
1 answer:
Answer:
20 Joules.
Explanation:
As , the incline is frictionless , only work done to push a block up the incline is only due to the gravitation resisting the upward movement .
Also, gravitational work only depends upon the height upto which the block is taken .
W = mgh .
Here, mass m of the object = 2.0 kg
g = 10m/
h = 1.0 m
Thus, W = 2×10×1 = 20 Joules.
So, 20 Joules of work is required to push an object of 2.0 kg mass up the frictionless incline .
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Answer:
Explanation:
The resistance of a wire can be given by the following expression
where R is resistance , ρ is specific resistance , L is length of wire and A is cross sectional area
specific resistance of metals are almost the same . So in the present case ρ and l are same . Hence the formula becomes
R = k / A where k is a constant .
The diameter of wire becomes two times hence area of cross section becomes 4 times or 4A .
Resistance becomes 1/4 times . Hence if resistance of metal wire is R , resistance of silver wire will be R / 4 .
current = voltage / resistance
In case of metal wire
8 x 10⁻³ = V / R
In case of silver wire
I = V / (R / 4 ) , I is current , V is potential difference .
I = 4 x V/R
= 4 x 8 x 10⁻³ A
= 32 mA.