Two objects have charges 2.0 C and 1.0 C. If the objects are placed 2 meters apart, what is the magnitude of the force that the
2 C object exerts on the 1 C obejct?
1 answer:
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Answer:
n = 756.25 giga electrons
Explanation:
It is given that,
If the charge on the negative plate of the capacitor,
Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :
e is the charge on electron
or
n = 756.25 giga electrons
So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.
Answer: the constant angular velocity of the arms is 86.1883 rad/sec
Explanation:
First we calculate the linear velocity of the single sprinkler;
Area of the nozzle = π/4 × d²
given that d = 8mm = 8 × 10⁻³
Area of the nozzle = π/4 × (8 × 10⁻³)²
A = 5.024 × 10⁻⁵ m²
Now total discharge is dived into 4 jets so discharge for single jet will be;
Q_single = Q / n = 0.006 / 4 = 1.5 × 10⁻³ m³/sec
So using continuity equation ;
Q_single = A × V_single
V_single = Q_single/A
we substitute
V_single = (1.5 × 10⁻³) / (5.024 × 10⁻⁵)
V_single = 29.8566 m/s
Now resolving the forces as shown in the second image,
Vt = Vcos30°
Vt = 29.8566 × cos30°
Vt = 25.8565 m/s
Finally we calculate the angular velocity;
Vt = rω
ω_single = Vt / r
from the given diagram, radius is 300mm = 0.3m
so we substitute
ω_single = 25.8565 / 0.3
ω_single = 86.1883 rad/sec
Therefore the constant angular velocity of the arms is 86.1883 rad/sec
