Since carbon has 4 valence electrons, a total of 4 bonds will be drawn withing methane, one between each H and the center C
Hope this helps
Answer:
13.23J
Explanation:
PE = m*g*h
PE = (3 kg ) * (9.8 m/s/s) * (0.45 m)
Answer:
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Explanation:
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Answer:
a) force between them is attraction, b) F = 1.125 10⁻² N
Explanation:
In this case the electric force is given by Coulomb's law
F =
In the exercise they give us the values of the loads
q1 = - 10 mC = -10 10⁻³ C
q2 = 5 mC = 5 10⁻³ C
d = 20 cm = 0.20 m
let's calculate
F = 9 10⁹
F = 1.125 10⁻² N
To find the direction of the force we use that charges of the same sign repel each other, as in this case there is a positive and a negative charge, the force between them is attraction
Answer:
Explanation:
Given that:
mass of stone (M) = 0.100 kg
mass of bullet (m) = 2.50 g = 2.5 ×10 ⁻³ kg
initial velocity of stone (
) = 0 m/s
Initial velocity of bullet (
) = (500 m/s)i
Speed of the bullet after collision (
) = (300 m/s) j
Suppose we represent
to be the velocity of the stone after the truck, then:
From linear momentum, the law of conservation can be applied which is expressed as:





∴
The magnitude now is:


Using the tangent of an angle to determine the direction of the velocity after the struck;
Let θ represent the direction:

