Answer:
The charge on the third object is − 21.7nC
Explanation:
From Gauss's Law
Φ = Q/ε₀
where;
Φ is the total electric flux through the shell = − 533 N⋅m²/C
Q is the total charge Q in the shell = ?
ε₀ is the permittivity of free space = 8.85 x 10⁻¹²
From this equation; Φ = Q/ε₀
Q = Φ * ε₀ = − 533 * 8.85 x 10⁻¹²
Q = −4.7 X 10⁻⁹ C = -4.7nC
Q = q₁ + q₂ + q₃
− 4.7nC = − 14.0 nC + 31.0 nC + q₃
− 4.7nC − 17nC = q₃
− 21.7nC = q₃
Therefore, the charge on the third object is − 21.7nC
Answer: 14. 49 m
Explanation:
We can solve this problem with the following equations:
(1)
(2)
Where:
is the horizontal distance between the cannon and the ball
is the cannonball initial velocity
since the cannonball was shoot horizontally
is the time
is the final height of the cannonball
is the initial height of the cannonball
is the acceleration due gravity
Isolating
from (2):
(3)
(4)
(5)
Substituting (5) in (1):
(6)
Finally:
Answer:
Energy always involves. motion in some form. Matter is an object that that up space so it is. measured in units of space
Explanation:
Answer:
R1 = 5.13 Ω
Explanation:
From Ohm's law,
V = IR............... Equation 1
Where V = Voltage, I = current, R = resistance.
From the question,
I = 2 A, R = R1
Substitute into equation 1
V = 2R1................ Equation 2
When a resistance of 2.2Ω is added in series with R1,
assuming the voltage source remain constant
R = 2.2+R1, and I = 1.4 A
V = 1.4(2.2+R1)................. Equation 3
Substitute the value of V into equation 3
2R1 = 1.4(2.2+R1)
2R1 = 3.08+1.4R1
2R1-1.4R1 = 3.08
0.6R1 = 3.08
R1 = 3.08/0.6
R1 = 5.13 Ω