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D. Gamma rays
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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.
Answer:
Angular acceleration, is 
Explanation:
Given that,
Initial speed of the drill, 
After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, 
We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

So, the drill's angular acceleration is
.
Answer:
r=6.05km/hr
z=59.1 degree to the horizontal
Explanation:
A bird is flying east at 5.2 kilometers/hour relative to the air. There's a crosswind blowing at 3.1 kilometers/hour toward the south relative to the ground. What is the bird's velocity relative to the ground? State your answer to one decimal place
can be solved using pythagoras theorem
r^2=o^2+a^2
r^2=5.2^2+3.1^2
r^2=36.65
r=6.1km/hr is te birds velocity relative to the ground
tanz=5.2/3.1
z=tan^-1(5,2/3.1)
z=59.1 degree to the horizontal