Answer: A
<u>Explanation:</u>
NOTES:
d = 650 meters
t = 10 seconds
**********************************
v = d/t
= 650 meters/10 seconds
= 65 meters/second
Answer:
x = 0.775m
Explanation:
Conceptual analysis
In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.
We apply Coulomb's law to calculate the electrical forces on q₃:
(Electric force of q₂ over q₃)
(Electric force of q₁ over q₃)
Known data
q₁ = 15 μC = 15*10⁻⁶ C
q₂ = 6 μC = 6*10⁻⁶ C
Problem development
F₂₃ = F₁₃
(We cancel k and q₃)
q₂(2-x)² = q₁x²
6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)
6(2-x)² = 15(x)²
6(4-4x+x²) = 15x²
24 - 24x + 6x² = 15x²
9x² + 24x - 24 = 0
The solution of the quadratic equation is:
x₁ = 0.775m
x₂ = -3.44m
x₁ meets the conditions for the forces to cancel in q₃
x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel
The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.
The molar mass of the sample is equal to the summation of the molar mass of the elementas multiplied by the abundance of the elements by mole. In this case, copper has an abundance of 93.69 percent while zinc has 6.31 percent. In this case, the average molecular weight is 63.67 g/mol
Answer:
The current through the resistor is 0.5 A
Explanation:
Given;
power of the light bulb = 60 W
voltage in the wall outlet across the plug terminals = 120 V
power of the light bulb is the product of voltage in the wall outlet across the plug terminals and the current passing through the resistor.
power = voltage x current
Therefore, for a 60 W light bulb powered by a connection to a wall outlet with 120 V across the plug terminals, the current passing through the resistor is 0.5 A