Answer:
1). Average speed = 1.5 m per second
2). Average velocity = 1.5 m per second
Explanation:
1). Since, speed is a scalar quantity
Therefore, average speed of the trip =
From the graph attached,
Total distance covered = 10 + 10 + 20 + 0 + 20 + 30
= 90 meters
Total time taken = 60 seconds
Average speed =
= 1.5 meter per second
2). Velocity is a vector quantity.
Therefore, average velocity =
=
=
= 1.5 meter per second
243 rad is the angle through which the wheel turns during this time.
<h3>What is torque and angular speed?</h3>
Torque is a measure of the force that can cause an object to rotate about an axis.
Angular speed is define as the measuring angle of rotation over a given amount of time.
Mathematically , ω = θ t
From the given question ,
Torque = 19.8 rad/s
ω = 23.5 rad/s
t = 11.2 s
We get the equation,
= 243 rad
Learn more about Angular momentum here:
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Answer:
0.5
Explanation:
The energy of a charged capacitor is given by where
is the charge on it and
is its capacitance. The capacitance is defined, geometrically, as
where is a constant that is determined by the material between the plates,
is the area of the plates and
is the distance between them. It is then that the capacitance is linearly inversely proportional to the separation distance; as the distance increases, the capacitance reduces.
Because it is linear, when the separation distance is doubled, other factors remaining constant, the capacitance is halved. Because the capacitance is halved, the energy is halved.
Let's see the mathematics.
Initial capacitance, at initial separation,
If separation is doubled, separation becomes . Then the capacitance becomes
Initial energy,
Final energy,
Hence,
Answer:
Explanation:
The amplitude of the oscillation under SHM will be .5 m and the equation of
SHM can be written as follows
x = .5 sin(ωt + π/2) , here the initial phase is π/2 because when t = 0 , x = A ( amplitude) , ω is angular frequency.
x = .5 cosωt
given , when t = .2 s , x = .35 m
.35 = .5 cos ωt
ωt = .79
ω = .79 / .20
= 3.95 rad /s
period of oscillation
T = 2π / ω
= 2 x 3.14 / 3.95
= 1.6 s
b )
ω =
ω² = k / m
k = ω² x m
= 3.95² x .6
= 9.36 N/s
c )
v = ω
At t = .2 , x = .35
v = 3.95
= 3.95 x .357
= 1.41 m/ s
d )
Acceleration at x
a = ω² x
= 3.95 x .35
= 1.3825 m s⁻²