Each corner of a right-angled triangle is occupied by identical point charges "A", "B", and "C" respectively. Draw a sketch of t
his arrangement. "A" exerts force F on "B". An equal force F is exerted by "C" on "B" (/_ ABC= 90 degrees). Determine an expression for the net force on "B".
1 answer:
Answer:
Fnet = F√2
Fnet = kq²/r² √2
Explanation:
A exerts a force F on B, and C exerts an equal force F on B perpendicular to that. The net force can be found with Pythagorean theorem:
Fnet = √(F² + F²)
Fnet = F√2
The force between two charges particles is:
F = k q₁ q₂ / r²
where
k is Coulomb's constant, q₁ and q₂ are the charges, and r is the distance between the charges.
If we say the charge of each particle is q, then:
F = kq²/r²
Substituting:
Fnet = kq²/r² √2
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Answer:
v_max = (1/6)e^-1 a
Explanation:
You have the following equation for the instantaneous speed of a particle:
(1)
To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:
(2)
where you have use the derivative of a product.
Next, you equal the expression (2) to zero in order to calculate t:
For t = 1/6 you obtain the maximum speed.
Then, you replace that value of t in the expression (1):
hence, the maximum speed is v_max = ((1/6)e^-1)a