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
Answer:
The answer is C.
Explanation:
I guessed and it was right
The solution would be like
this for this specific problem:
<span>5.5 g = g + v^2/r </span><span>
<span>4.5 g =
v^2/r </span>
<span>v^2 = 4.5
g * r </span>
<span>v = sqrt
( 4.5 *9.81m/s^2 * 350 m) </span>
v = 124
m/s</span>
So the pilot will black out for this dive at 124
m/s. I am hoping that these answers have satisfied your query and it
will be able to help you in your endeavors, and if you would like, feel free to
ask another question.
Answer:
Option B
Solution:
As per the question:
Heat produced at the rate of 10 W
The resistor R and 2R are in series.
Also, in series, same current, I' passes through each element in the circuit.
Therefore, current is constant in series.
Also,
Power,
When current, I' is constant, then
P' ∝ R
Thus

P' = 20 W
Answer:
E = 420.9 N/C
Explanation:
According to the given condition:

where,
E = Magnitude of Electric Field = ?
v = speed of charge = 230 m/s
B = Magnitude of Magnetic Field = 0.61 T
θ = Angle between speed and magnetic field = 90°
Therefore,

<u>E = 420.9 N/C</u>