Answer:
1.99×10^-4coulombs
Explanation:
The charge (Q) across the resistor the directly proportional to the voltage (V) where capacitance of the capacitor(C) is the proportionality constant. Mathematically, Q = CV
If V is the voltage across the resistor, V = IR (according to ohm's law) where I is the current in the resistor and R is the resistance.
We need to calculate the voltage on the resistor first when 0.18A current is passed through it.
V = 0.18 × 185
V = 33.3Volts
The charge Q on the resistor will be;
Q = CV
Were C = 6.00 μF, V = 33.3
Q= 6×10^-6 ×33.3
Q = 0.0001998
Q= 1.99×10^-4Coulombs
Say the initial point is (0,0)
The final point is
x = 200 + 135*cos(30) = 200 + 135*sqrt(3)/2 = 316.91 ft
y = 135*sin(30) = 135/2 = 67.5 ft
Resultant vector = (316.91, 67.5) - (0,0) = 316.91, 67.5) ft
Answer:
The correct answer is c. When the balloon hits the ground, the rubber envelope stretches, storing elastic potential energy; this elastic potential energy is converted to the gravitational potentiaL
Explanation:
Let's analyze the situation of the globe
When it touches the ground, the part that is in contact decreases its velocity to zero, but the upper part of the ball continues to move, which creates that the molecules approach slightly, if we approximate the spring links, a repulsive force is created that after all the particles reach zero speed. The force of the springs moves the ball up until the force decreases to zero.
We can relate this force of Hooke with an elastic energy
This energy can be stored in the deformation of the system, as elastic potential energy, which is subsequently transformed into gravitational potential energy when the balloon is lifted.
The correct answer is c
Answer:
884Hz
Explanation:
Beats is the absolute difference between two frequencies therefore
Beats = f1-f2
4=f1-880
F1=880+4
F1=884Hz
