A wheel of mass M and radius R rolls on a level surface without slipping. If the angular velocity of the wheel about its center
1 answer:
The most important thing you should remember in order to get how is its linear momentum relative to the surface is that the<span> point on the circumference is moving at w rad/second.
Therefore, here is the solving formulae:
if </span>
you can easily get <span> linear momentum p :
</span>
I'm sure that helps.
Therefore, here is the solving formulae:
if </span>
</span>
I'm sure that helps.
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Answer:
460.52 s
Explanation:
Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that
dV/dt ∝ V
dV/dt = kV
separating the variables, we have
dV/V = kdt
integrating both sides, we have
∫dV/V = ∫kdt
㏑(V/V₀) = kt
V/V₀ =
Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V
Since dV/dt = kV
-0.01V = kV
k = -0.01
So, V/V₀ =
V = V₀
Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀
So, V = 0.1V₀
Thus
V = V₀
0.1V₀ = V₀
0.1V₀/V₀ =
0.1 =
to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have
㏑(0.01) = -0.01t
So, t = ㏑(0.01)/-0.01
t = -4.6052/-0.01
t = 460.52 s
Answer:
E) are almost circular, with low eccentricities.
Explanation:
Kepler's laws establish that:
All the planets revolve around the Sun in an elliptic orbit, with the Sun in one of the focus (Kepler's first law).
A planet describes equal areas in equal times (Kepler's second law).
The square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit (Kepler's third law).
Where T is the period of revolution and a is the semi-major axis.
Planets orbit around the Sun in an ellipse with the Sun in one of the focus. Because of that, it is not possible to the Sun to be at the center of the orbit, as the statement on option "C" says.
However, those orbits have low eccentricities (remember that an eccentricity = 0 corresponds to a circle)
In some moments of their orbit, planets will be closer to the Sun (known as perihelion). According with Kepler's second law to complete the same area in the same time, they have to speed up at their perihelion and slow down at their aphelion (point farther from the Sun in their orbit).
Therefore, option A and B can not be true.
In the celestial sphere, the path that the Sun moves in a period of a year is called ecliptic, and planets pass very closely to that path.
Explanation:
Let f is the frequency of an oscillation and T is the period of the oscillation. There exists an inverse relationship between the frequency and the time period of the oscillation. Mathematically, it is given by :
Also,
So,
The time taken to complete one oscillation is called the period of the oscillation and the number of oscillation is called the frequency if an oscillation.