Answer:
B. w=12.68rad/s
C. α=3.52rad/s^2
Explanation:
B)
We can solve this problem by taking into account that (as in the uniformly accelerated motion)
( 1 )
where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.
In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have
to calculate the angular speed w we can use
Thus, wf=12.68rad/s
C) We can use our result in B)
I hope this is useful for you
regards
The solution would be like
this for this specific problem:
<span>
The force on m is:</span>
<span>
GMm / x^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] ->
1
The force on 2m is:</span>
<span>
GM(2m) / (L - x)^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2]
-> 2
From (1), you’ll get M = 2mx^2 / L^2 and from
(2) you get M = m(L - x)^2 / L^2
Since the Ms are the same, then
2mx^2 / L^2 = m(L - x)^2 / L^2
2x^2 = (L - x)^2
xsqrt2 = L - x
x(1 + sqrt2) = L
x = L / (sqrt2 + 1) From here, we rationalize.
x = L(sqrt2 - 1) / (sqrt2 + 1)(sqrt2 - 1)
x = L(sqrt2 - 1) / (2 - 1)
x = L(sqrt2 - 1) </span>
= 0.414L
<span>Therefore, the third particle should be located the 0.414L x
axis so that the magnitude of the gravitational force on both particle 1 and
particle 2 doubles.</span>
When light passes from a faster medium into a slower medium, light will be refracted toward a line drawn perpendicular to the point of refraction. <em>(B)</em>
The statement “When
an object is in orbit, it is falling at the same rate at which the Earth is
curving” is true. The speed of a satellite orbiting the earth depends only on
the mass of the earth and the mass of the satellite.
