I just took it 100% 11/11
1.D
2.A
3.A
4.A
5.B
6.C
7.D
8C
9A
10B
11C
Answer:
The current is reduced to half of its original value.
Explanation:
- Assuming we can apply Ohm's Law to the circuit, as the internal resistance and the load resistor are in series, we can find the current I₁ as follows:

- where Rint = r and RL = r
- Replacing these values in I₁, we have:

- When the battery ages, if the internal resistance triples, the new current can be found using Ohm's Law again:

- We can find the relationship between I₂, and I₁, dividing both sides, as follows:

- The current when the internal resistance triples, is half of the original value, when the internal resistance was r, equal to the resistance of the load.
C and D are units of length or distance.
A is a measured angle.
B is a unit of angular measurement.
Answer:
Explanation:
Given that,
A point charge is placed between two charges
Q1 = 4 μC
Q2 = -1 μC
Distance between the two charges is 1m
We want to find the point when the electric field will be zero.
Electric field can be calculated using
E = kQ/r²
Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.
Then, the magnitude of the electric at point x is zero.
E = kQ1 / r² + kQ2 / r²
0 = kQ1 / x² - kQ2 / (1-x)²
kQ1 / x² = kQ2 / (1-x)²
Divide through by k
Q1 / x² = Q2 / (1-x)²
4μ / x² = 1μ / (1 - x)²
Divide through by μ
4 / x² = 1 / (1-x)²
Cross multiply
4(1-x)² = x²
4(1-2x+x²) = x²
4 - 8x + 4x² = x²
4x² - 8x + 4 - x² = 0
3x² - 8x + 4 = 0
Check attachment for solution of quadratic equation
We found that,
x = 2m or x = ⅔m
So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.
At its maximum height, the ball will have zero vertical velocity, so the ball's velocity at this point is exactly equal to its horizontal velocity.
At any time <em>t</em>, the horizontal component of its velocity is
<em>v</em> = (15 m/s) cos(40°) ≈ 11.49 m/s
so at the highest point of its trajectory, the ball has a velocity of about 11.49 m/s pointed in the positive horizontal direction.