Answer:
0.37sec
Explanation:
Period of oscillation of a simple pendulum of length L is:
T
=
2
π
×
√
(L
/g)
L=length of string 0.54m
g=acceleration due to gravity
T-period
T = 2 x 3.14 x √[0.54/9.8]
T = 1.47sec
An oscillating pendulum, or anything else in nature that involves "simple harmonic" (sinusoidal) motion, spends 1/4 of its period going from zero speed to maximum speed, and another 1/4 going from maximum speed to zero speed again, etc. After four quarter-periods it is back where it started.
The ball will first have V(max) at T/4,
=>V(max) = 1.47/4 = 0.37 sec
Answer:
L = 8694 Kg.m²/s
Explanation:
r = 270 ĵ m
v = 14 î m/s
m = 2.3 kg
θ = 90º
L = ?
We can apply the equation
L = m*v*r*Sin θ
L = (2.3 kg)*(14 m/s)*(270 m)*Sin 90º = 8694 Kg.m²/s
Answer:
Resistance of the second wire is twice the first wire.
Explanation:
Let us first see the formula of resistance;
R = pxL/A
Here L is the lenght of the wire, A the area and p is the resistivity of wire.
As we are given that the length of second wire is double than that of the first wire, hence the resistance of second wire would be double.
Since we have two loop in second case, inducing double voltage but as resistance is doubled so the current would remain same according to ohms law
I = V/R
The answer is C, as there is not increase or decrease in speed during that time frame.
