Answer:
the direction is towards upward.i think.
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
<h2>Answer: Francium
</h2>
Let's start by explaining that electronegativity is a term coined by Linus Pauling and is determined by the <em>ability of an atom of a certain element to attract electrons when chemically combined with another atom.
</em>
So, the more electronegative an element is, the more electrons it will attract.
It should be noted that this value can not be measured directly by experiments, but it can be determined indirectly by means of calculations from other atomic or molecular properties of the element. That is why the scale created by Pauling is an arbitrary scale, where the maximum value of electronegativity is 4, assigned to Fluorine (F) and the <u>lowest is 0.7, assigned to Francium (Fr).</u>
Thank you for posting your question here at brainly. Below is Yoland's study:
<span>Yolanda is studying two waves. The first wave has an amplitude of 2 m, and the second has an amplitude of 3 m.
</span>
I think the answer is "She can use constructive interference to generate a wave with an amplitude of 1.5 m."
1. 0.16 N
The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:
where
G is the gravitational constant
is the mass of the asteroid
m = 100 kg is the mass of the man
r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid
Substituting, we find
2. 1.7 m/s
In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force
where v is the minimum speed required to stay in orbit.
Re-arranging the equation and solving for v, we find:
