Besides the force due to gravity or g, what else determines the period of a pendulum?

2 answers:

5 0

For a simple pendulum, the mathematical equation that describes the period (T) is T=√(l/g) where l is the length of the string and g is the acceleration due to gravity (9.81 metres per second). Therefore it can be seen that both gravity and the length of the string within a pendulum effects its period.

Sidana [21]2 years ago

4 0

Length of the wire affects the period of a pendulem.

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A car travels forward with constant velocity. It goes over a small stone, which gets stuck in the groove of a tire. The initial

kolbaska11 [484]

Answer:

a) vertically up

Explanation:

A car travels forward with constant velocity. It goes over a small stone, which gets stuck in the groove of a tire. The initial acceleration of the stone, as it leaves the surface of the road is,

a) vertically up

b) horizontally forward

c) horizontally backwards

d) zero

e) greater than zero but less than 45 degrees below the horizontal

A vertically upward since that is the direction of centripetal acceleration at that time. as the stone enters the circular path of the tire, it does so with a centripetal acceleration which is directed upward

8 0

2 years ago

Calculate the total resistance in a parallel circuit made up of resistances of 2Ω, 3Ω, and 4Ω. (Hint: When converting to a decim

natta225 [31]

Answer:

Total resistance = 0.92Ω

Explanation:

For parallel connected resistors we have effective resistance

$\frac{1}{R_{eff}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+.....$

Here parallel circuit made up of resistances of 2Ω, 3Ω, and 4Ω.

That is

R₁ = 2Ω

R₂ = 3Ω

R₃ = 4Ω

Substituting

$\frac{1}{R_{eff}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\\\\\frac{1}{R_{eff}}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\\\\\frac{1}{R_{eff}}=\frac{6+4+3}{12}\\\\R_{eff}=\frac{12}{13}=0.92ohm$