Answer:
(a)11.24 m/s
(b)7.44 m/s
(c)409 N
(d)
(e) 0
Explanation:
The period for 1 circle
of the merry go around is 9.5s. It means the angular speed is:

(a)The speed is

(b) Centripetal acceleration:

(c) Magnitude of the force that keeps you go around at this acceleration

(d) let the coefficient of friction by
. The frictional force shall be this coefficient multiplied by normal force reverting gravity of the man

It is the branch of science, in which we study different phenomena of atmosphere including climate and weather.
Magnitude of the force of tension: 139 N
Explanation:
The surface of the ramp here is assumed to be the positive x-direction.
To solve this problem and find the magnitude of the force of tension, we have to analyze only the situation along the x-direction, since the force of tension lie in this direction.
There are three forces acting along the x-direction:
- The force of tension,
, acting up along the plane - The force of friction,
, acting down along the plane - The component of the weight in the x-direction,
, acting down along the plane
We know that the magnitude of the weight is

So its x-component is

The net force along the x-direction can be written as

And therefore, since the net force is 98 N, we can find the magnitude of the force of tension:

Learn more about inclined planes:
brainly.com/question/5884009
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C would be the right answer edu
The clicker emits EM (electromagnetic) wave which travels at the speed of light, that is
v = 3 x 10⁸ m/s
The frequency is
f = 900mHz = 9 x 10⁸ Hz
Because velocity = frequency * wavelength, the wavelength, λ, is given by
fλ = v
λ = v/f
= (3 x 10⁸ m/s) / (9 x 10⁸ 1/s)
= 1/3 m
Answer: 1/3 m