Answer:
0.695s
Explanation:
From Hooke's law, the restoring force is given has
F = -ky .......1
Where F is the force, y is the spring displacement and k force constant of the spring.
Also recall,
F=mg ............ 2
Where m is the mass of object, g is the acceleration due to gravity.
Equating 1 and 2
Ky = mg
Given that g=9.8m/s2 , y is 3.4cm and g is 8g
K×3.4/100m =8/1000kg × 9.8m/s2
K= ( 0.008kg × 9.8m/s2 ) ÷ 0.034
K= 0.0784÷0.035
K=2.24N/m
Mass ofvthe second object is 25g =0.025kg
Period of oscillation T
T=2π√m/k
T=2×3.142√0.025/2.24
T=6.284√0.0111
T=0.659seconds
Answer:
0.8c and -0.14c
Explanation:
The first fragment will have a speed of +0.5c respect of a frame of reference moving at +0.5c
Lest name v the velocity of the frame of reference, and u' the velocity of the object respect of this moving frame of reference.
The Lorentz transform for velocity is:
u = (u' + v) / (1 + (u' * v) / c^2)
u = (0.5c + 0.5c) / (1 + (0.5c * 0.5c) / c^2) = 0.8c
The other fragment has a velocity of u' = -0.6c respect of the moving frame of reference.
u = (-0.6v + 0.5c) / (1 + (0.5c * 0.5c) / c^2) = -0.14c
Answer:
(a). Z = 54.54 ohm
(b). R = 36 ohm
(c). The circuit will be Capacitive.
Explanation:
Given data
I = 2.75 A
Voltage = 150 V
rad = 48.72°
(a). Impedance of the circuit is given by
Z = 54.54 ohm
(b). We know that resistance of the circuit is given by
Put the values of Z & 
in above formula we get
R = 36 ohm
(c). Since the phase angle is negative so the circuit will be Capacitive.
Answer:
Explanation:
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the electric conductivity of gold is very high
