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;
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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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Answer:
2.019 seconds.
Explanation:
Here we use the equation:
Distance covered = (initial speed x time) + 0.5 x (acceleration x time^2)
The vertical distance to cover is 20 meters.
Since the cannonball is fired horizontally, the initial vertical speed is 0 m/s.
Acceleration due to gravity is 9.81 m/s^2.
Substituting these values into the equation we get:
20 = (0 x time) + 0.5 x (9.81 x time^2)
Solving this equation for time gives us 2.019 seconds as the answer.
Answer:
A.
Explanation:
I remember this one from my science class.
The flattened rotatiing disj of gas adn dushdhakafna
Answer:
A. bent
Explanation:
Water molecule (H2O) is said to contain four valence electron pairs (2 bond pairs and 2 lone pairs). The presence of lone pair of electrons in the water molecule influences its molecular geometry or shape.
Since water has two lone pairs of electrons, which repel each other according to the VSEPR theory, water molecule is said to have a BENT molecular geometry.