Answer:
F = 5.33 10⁻⁶ N , directed to the left
Explanation:
For this exercise we will use the force addition law where each force is electric given by Coulomb's law
F = ∑ F
F = F₁₃ + F₂₃
where the force is given by
F = k q₁ q₂ / r₁₂²
in our case
F₁₃ = k q₁ q₃ / r₁₃²
F₂₃ = k q₂ q₃ / r₂₃²
F = k q₃ (q₁ / r₁₃² + q₂ / r₂₃²)
as the charges are of the same sign the electric force is repulsive and due to the position of the charges it is directed to the left
let's calculate
F = 9 10⁹ 8 10⁻⁹ (3 10⁻⁹ / (0.26 - 0) 2 + 6 10⁻⁹ / (0.45-0) 2)
F = 72 (44,379 + 29,630) 10⁻⁹
F = 5.33 10⁻⁶ N
directed to the left
Answer:
5.5 km
Explanation:
First, we convert the distance from km/h to m/s
910 * 1000/3600
= 252.78 m/s
Now, we use the formula v²/r = gtanθ to get our needed radius
making r the subject of the formula, we have
r = v²/gtanθ, where
r = radius of curvature needed
g = acceleration due to gravity
θ = angle of banking
r = 252.78² / (9.8 * tan 50)
r = 63897.73 / (9.8 * 1.19)
r = 63897.73 / 11.662
r = 5479 m or 5.5 km
Thus, we conclude that the minimum curvature radius needed for the turn is 5.5 km
Answer:
Thus, the energy stored by a 50 Ampere minute battery is found to be 36 KJ.
Explanation:
The power delivered by a battery is given by the formula:
P = VI
where,
P = Power Delivered by battery in 1 second
V = Voltage of battery = 12 volt
I = Current stored in battery
But, if we multiply both sides of equation by time (t), then:
Pt = VIt
where,
Pt = Power x Time = E = Energy Stored = ?
It = Rating of Battery = (50 A.min)(60 sec/min) = 3000 A.sec
Therefore,
E = (12 volt)(3000 A.sec)
<u>E = 36000 J = 36 KJ</u>
<u>Thus, the energy stored by a 50 Ampere minute battery is found to be 36 KJ.</u>
Answer:
40watts
Explanation:
Given parameters:
Work done = 3000J
Time = 75s
Unknown:
Power generated = ?
Solution:
Power is the rate at which work is done;
Power =
Now, insert the parameters and solve;
Power = = 40watts
Answer:
I would probably say C to be completely honest
Explanation:
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