Answer:
Part a)
T = 3.96 s
Part b)
T = 1.98 s
Part c)
T = 2.8 s
Explanation:
As we know that time period of spring block system is given as
T = 2.8 s
Part a)
If the mass of the block attached is doubled
then we will have
Part b)
If the spring constant is doubled
then we have
Part c)
If the amplitude is halved but mass and spring constant will remain the same
so here we know that time period does not depends on Amplitude
so we will have
T = 2.8 s
Answer:
1 by 3 units
Explanation:
The resistance (R) of a conductor is given by the formula:
R = ρL / A
where L is the length of the conductor, ρ is resistivity and A is the cross sectional area.
Let us assume that the metal bar has a resistivity of ρ.
a) If the leads is attached to the two opposite sides that have dimensions of 1 by 3 units.
The length of the bar would be 13 units and the cross sectional area (A) would be = 1 * 3 = 3 units²
R₁ = ρL / A = ρ(13) / 3 = 13ρ / 3
b) If the leads is attached to the two opposite sides that have dimensions of 3 by 13 units.
The length of the bar would be 1 units and the cross sectional area (A) would be = 3 * 13 = 39 units²
R₂ = ρL / A = ρ(1) / 39 = ρ / 39
c) If the leads is attached to the two opposite sides that have dimensions of 1 by 13 units.
The length of the bar would be 3 units and the cross sectional area (A) would be = 1 * 13 = 13 units²
R₃ = ρL / A = ρ(3) / 13 = 3ρ / 13
Therefore we can see that the largest resistance is gotten If the leads is attached to the two opposite sides that have dimensions of 1 by 3 units
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