Answer: 6.47m/s
Explanation:
The tangential speed can be defined in terms of linear speed. The linear speed is the distance traveled with respect to time taken. The tangential speed is basically, the linear speed across a circular path.
The time taken for 1 revolution is, 1/3.33 = 0.30s
velocity of the wheel = d/t
Since d is not given, we find d by using formula for the circumference of a circle. 2πr. Thus, V = 2πr/t
V = 2π * 0.309 / 0.3
V = 1.94/0.3
V = 6.47m/s
The tangential speed of the tack is 6.47m/s
Answer:
(a): a = 0.4m/s²
(b): α = 8 radians/s²
Explanation:
First we propose an equation to determine the linear acceleration and an equation to determine the space traveled in the ramp (5m):
a= (Vf-Vi)/t = (2m/s)/t
a: linear acceleration.
Vf: speed at the end of the ramp.
Vi: speed at the beginning of the ramp (zero).
d= (1/2)×a×t² = 5m
d: distance of the ramp (5m).
We replace the first equation in the second to determine the travel time on the ramp:
d = 5m = (1/2)×( (2m/s)/t)×t² = (1m/s)×t ⇒ t = 5s
And the linear acceleration will be:
a = (2m/s)/5s = 0.4m/s²
Now we determine the perimeter of the cylinder to know the linear distance traveled on the ramp in a revolution:
perimeter = π×diameter = π×0.1m = 0.3142m
To determine the angular acceleration we divide the linear acceleration by the radius of the cylinder:
α = (0.4m/s²)/(0.05m) = 8 radians/s²
α: angular aceleration.
Answer:

Explanation:
The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.
Also, we know that the centripetal force of an object describing a circular motion is given by:

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.
Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So
and
(Since
). Then, we get:

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).
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