Doppler radars work by using radio waves that bounce off objects and thus the object can be tracked.
Doppler radars work by emitting a beam of energy from an antenna known as radio waves. The energy they release when they collide with airborne objects scatters in all directions, with some of it returning directly to the radar.
The quantity of energy returned to the radar increases with object size. We can now perceive raindrops in the atmosphere because of this.
The amount of time it takes for the energy beam to be delivered and returned to the radar also gives us the object's distance.
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Particles q1 = +8.0 μC, q2= +3.5 μC, and
q3=-2.5 μC are in a line. Particles q1 and q2 are separated by 0.10 m and particles q2 and q3 are separated by 0.15 m. What is the net force on particle q2?
Remember: Negative forces (-F) will point Left Positive forces (+F) will point Right
Please answer bro I will do anything
The net force on particle q2 is a positive force pointing to the right.
To determine the net force on particle q2, we need to calculate the individual forces exerted on q2 by q1 and q3 and then add them vectorially.
The force between two charged particles can be calculated using Coulomb's law, which states that the force (F) between two charged particles is directly proportional to the product of their charges (q1 and q2) and inversely proportional to the square of the distance (r) between them:
F = k * (|q1| * |q2|) / r^2
where k is the electrostatic constant.
Given:
q1 = +8.0 μC
q2 = +3.5 μC
q3 = -2.5 μC
Distance between q1 and q2 (r12) = 0.10 m
Distance between q2 and q3 (r23) = 0.15 m
First, let's calculate the force between q1 and q2:
F12 = k * (|q1| * |q2|) / r12^2
Substituting the values:
[tex]F12 = (9 \times 10^9 N m^2/C^2) * ((8.0 \times 10^-6 C) * (3.5 \times 10^-6 C)) / (0.10 m)^2[/tex]
Calculating the force F12 will give us the magnitude of the force between q1 and q2. However, since q1 and q2 have the same charge sign, the force will be repulsive, pointing to the right.
Next, let's calculate the force between q2 and q3:
[tex]F23 = k * (|q2| * |q3|) / r_{23}^2[/tex]
Substituting the values:
F23 = (9 x 10^9 N m^2/C^2) * ((3.5 x 10^-6 C) * (2.5 x 10^-6 C)) / (0.15 m)^2
Calculating the force F23 will give us the magnitude of the force between q2 and q3. Since q2 and q3 have opposite charge signs, the force will be attractive, pointing to the left.
To find the net force on q2, we need to subtract the force F23 from F12 since they act in opposite directions:
Net force on q2 = F12 - F23
Finally, we need to consider the direction of the forces. Since F12 is repulsive (positive) and F23 is attractive (negative), the net force on q2 will be positive, pointing to the right.
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a negative charge Q1 = -5.50 is located at a point x1=-6.00 abd. positive chsrge Q2= 6.50 uc is located at point x2= 4.00 m find the magnitude and direction of the electric force between rhe charges
a negative charge Q1 = -5.50 is located at a point x1=-6.00 abd. positive charge Q2= 6.50 uc is located at point x2= 4.00 m then the magnitude and direction of the electric force between the charges is 3.56 mN and from left to right.
Coulomb's Law, which says that the electric force between two charges is directly proportional to the product of their charges and inversely proportional to the square of their distance, may be used to determine the size and direction of the electric force between the provided charges.
Given:
Q1 = -5.50 μC (negative charge)
Q2 = 6.50 μC (positive charge)
x1 = -6.00 m (position of Q1)
x2 = 4.00 m (position of Q2)
distance between the charges is
Distance (d) = |x2 - x1| = |4.00 m - (-6.00 m)| = 10.00 m
electric force (F) by using Coulomb's Law:
F = (k * |Q1| * |Q2|) / (d^2)
Where k is the electrostatic constant, which is equal to 9 × 10^9 N·m^2/C^2.
Substituting the given values:
F = (9 × 10⁹ N·m²/C² * |(-5.50 μC)| * |6.50 μC|) / (10.00 m)^2
F ≈ 3.2 mN
The magnitude of the electric force between the charges is approximately 3.56 mN.
Since Q1 is negative and Q2 is positive, the electric force acts in the direction from Q1 towards Q2. Therefore, the direction of the electric force is from left to right.
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