You may look at what group they are in
Group
1A=Group 1
2A = Group 2
3A = Group 13
4A= Group 14
5A=Group 15
6A=Group 16
7A=Group 17
The #A tells you how many valence electrons there are by the # before A. Such as Chlorine, which is in 7A, so therefore has 7 valence electrons.
To solve this problem, we use this formula:
s = rθ
where
s is the arc length or in this case, the distance traveled by a point of a wheel
r is the radius of the circle or the wheel
θ is the subtended angle or the angle of rotation of the wheel
Plugging in the given values and converting degrees to radians
s = 22 *128 (π/180)
s = 49.15 cm
Who created the theory of general relativity?:
The answer would be: Albert Einstein.
Albert Einstein developed the general relativity theory (gravitation).
The year he developed the general relativity theory or (GR) was back in 1907 and 1915. Then many other contributions after 1915.
Thanks,
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Using the Equation:
v² = vi² + 2 · a · s → Eq.1
where,
v = final velocity
vi = initial velocity
a = acceleration
s = distance
<span><span>We know that vi = 0 because the ball was at rest initially.
</span><span>
Therefore,
Solving Eq.1 for acceleration,
</span></span> v² = vi² + 2 · a · s
v² = 0 + 2 · a · s
v² = 2 · a · s
Rearranging for a,
a = v ²/2·<span>s
Substituting the values,
a = 46</span>²/2×1<span>
a = 1058 m/s</span>²
<span>Now applying Newton's 2nd law of motion,
</span>
<span>F = ma
= 0.145</span>×<span>1058
F = 153.4 N</span>
Answer:
Hello your question is incomplete attached below is the missing part of the question
In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these situations, two objects of masses m1 and m2 are attached to a seesaw. The seesaw is made of a bar that has length l and is pivoted so that it is free to rotate in the vertical plane without friction. Assume that the pivot is attached tot he center of the bar.
You are to find the angular acceleration of the seesaw when it is set in motion from the horizontal position. In all cases, assume that m1>m2.I
answer : part A = 2*[(M1 - M2)/(M1 + M2)]*g/L
part A = attached below
Explanation:
Part A :
Assuming that mass of swing is negligible
α = T/I
where ; T = torque, I = inertia,
hence T = L/2*9*(M1 - M2)
also; I = 
= ( M1 + M2) * (L/2)^2
Finally the magnitude of the angular acceleration α
α = 2*[(M1 - M2)/(M1 + M2)]*g/L
Part B attached below
