Answer:
D. Uranium
Explanation:
I just got the answer right on my quiz.
Answer:
V = 0.0806 m/s
Explanation:
given data
mass quarterback = 80 kg
mass football = 0.43 kg
velocity = 15 m/s
solution
we consider here momentum conservation is in horizontal direction.
so that here no initial momentum of the quarterback
so that final momentum of the system will be 0
so we can say
M(quarterback) × V = m(football) × v (football) ........................1
put here value we get
80 × V = 0.43 × 15
V = 0.0806 m/s
The distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.
<h3>Distance from the center of the meter rule</h3>
The distance from the centre of the rule at which a 2N weight must be suspend from A is calculated as follows;
-----------------------------------------------------------------
20 A (30 - x)↓ x ↓ 20 cm B 30 cm
2N 0.9N
Let the center of the meter rule = 50 cm
take moment about the center;
2(30 - x) + 0.9(x)(30 - x) = 0.9(20)
(30 - x)(2 + 0.9x) = 18
60 + 27x - 2x - 0.9x² = 18
60 + 25x - 0.9x² = 18
0.9x² - 25x - 42 = 0
x = 29.3 cm
Thus, the distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.
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Newton’s fifth law says so i’m sorry it’s just logic
