The period of oscillation is T = 2 * pi * sqrt ( ( m2/3 + m1) / k )
<h3>What is period of oscillation?</h3>
This is the time in seconds it takes to complete one oscillation. where an oscillation is a repetitive to and fro motion. period if the inverse of frequency and both are basic when calculation motion in simple harmonic motion.
The period of oscillation is given as T
T = 2 * pi * sqrt ( m / k )
where
m = mass on this case mass of the spring will be inclusive to the mass of the block such that we have:
m1 = mass of the block
m2 = mass pf the spring
k = force constant of the spring
including the two masses to the period gives
T = 2 * pi * sqrt ( ( m2/3 + m1) / k )
Read more on period of oscillation here: brainly.com/question/22499336
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Answer:
It would take
time for the capacitor to discharge from
to
.
It would take
time for the capacitor to discharge from
to
.
Note that
, and that
.
Explanation:
In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is
, and the initial charge of the capacitor be
. Then at time
, the charge stored in the capacitor would be:
.
<h3>a)</h3>
.
Apply the equation
:
.
The goal is to solve for
in terms of
. Rearrange the equation:
.
Take the natural logarithm of both sides:
.
.
.
<h3>b)</h3>
.
Apply the equation
:
.
The goal is to solve for
in terms of
. Rearrange the equation:
.
Take the natural logarithm of both sides:
.
.
.
Answer:
the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh
Explanation:
Given that;
weight of vehicle = 4000 lbs
we know that 1 kg = 2.20462
so
m = 4000 / 2.20462 = 1814.37 kg
Initial velocity
= 60 mph = 26.8224 m/s
Final velocity
= 30 mph = 13.4112 m/s
now we determine change in kinetic energy
Δk =
m(
² -
² )
we substitute
Δk =
×1814.37( (26.8224)² - (13.4112)² )
Δk =
× 1814.37 × 539.5808
Δk = 489500 Joules
we know that; 1 kilowatt hour = 3.6 × 10⁶ Joule
so
Δk = 489500 / 3.6 × 10⁶
Δk = 0.13597 ≈ 0.136 kWh
Therefore, the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh
Answer:
A) F=-20.16×10⁹N
B) if the distance doubles, force is 4 times smaller.
Explanation:
q1=-28C
q2=5mC=0.005C
d=25cm=0.25m
Electrostatic force between charges: F=k×q1×q2/d², where k is a coefficient that has the value k=9 × 10⁹ N⋅m²⋅C^(-2) for air.
Thus:
F=9×10⁹×(-28)×0.005/0.25²
F=-20.16×10⁹N
The minus sign indicates attraction.
If distance doubles, d1=2×d, then we have 4d² at the denominator and the force is 4 times smaller.
Answer:

Explanation:
Electric field strength= Force/unit charge
E= (kQq/r²)/q ₓ r
where r is the unit vector in the direction of unit charge
E= 