Answer:
0.34 sec
Explanation:
Low point of spring ( length of stretched spring ) = 5.8 cm
midpoint of spring = 5.8 / 2 = 2.9 cm
Determine the oscillation period
at equilibrum condition
Kx = Mg
g= 9.8 m/s^2
x = 2.9 * 10^-2 m
k / m = 9.8 / ( 2.9 * 10^-2 ) = 337.93
note : w = 
= 
= 18.38 rad/sec
Period of oscillation = 
= 0.34 sec
You can use fixture wires: For installation in luminaires where they are enclosed and protected and not subject to bending and twisting and also can be used to connect luminaires to their branch circuit conductors.
<h3>What are some uses of fixture wires?</h3>
Fixture wires are flexible conductors that are used for wiring fixtures and control circuits. There are some special uses and requirements for fixture wires and no fixture can be smaller than 18 AWG
In modern fixtures, neutral wire is white and the hot wire is red or black. In some types of fixtures, both wires will be of the same color.
To know more about fixture wires, refer
brainly.com/question/26098282
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Answer:
Option B. 5 nC
Explanation:
From the question given above, the following data were obtained:
Capicitance (C) = 100 pF
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:
1 pF = 1×10¯¹² F
Therefore,
100 pF = 100 pF × 1×10¯¹² F / 1 pF
100 pF = 1×10¯¹⁰ F
Next, we shall determine the quantity of charge. This can be obtained as follow:
Capicitance (C) = 1×10¯¹⁰ F
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Q = CV
Q = 1×10¯¹⁰ × 50
Q = 5×10¯⁹ C
Finally, we shall convert 5×10¯⁹ C to nano coulomb (nC). This can be obtained as follow:
1 C = 1×10⁹ nC
Therefore,
5×10¯⁹ C = 5×10¯⁹ C × 1×10⁹ nC / 1 C
5×10¯⁹ C = 5 nC
Thus, the quantity of charge is 5 nC
to your question is 54 cm
0.4823 m/s
The initial velocity u1 of the ball=0
From the law of conservation of linear momentum.
m1u1+m2u2=m1v1+m2v2
(160×0)+(170×u1)=(160×0.3)+(170×0.2)
u1=0.4823m/s
