What are tiny sacs at the end of the bronchioles filled with air called?
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Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)
Answer:
d= 0.46 m
Explanation:
The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.
For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:
V =
We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.
If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.
So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:
V = +
= 0
⇒:
Replacing by the values of q1, q2, and k, and solving for x, we get:
⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.
Explanation:
answer is attached, kindly see answer
Answer:
2.87 km/s
Explanation:
radius of planet, R = 1.74 x 10^6 m
Mass of planet, M = 7.35 x 10^22 kg
height, h = 2.55 x 10^6 m
G = 6.67 x 106-11 Nm^2/kg^2
Use teh formula for acceleration due to gravity
g = 1.62 m/s^2
initial velocity, u = ?, h = 2.55 x 10^6 m , final velocity, v = 0
Use third equation of motion
0 = v² - 2 x 1.62 x 2.55 x 10^6
v² = 8262000
v = 2874.37 m/s
v = 2.87 km/s
Thus, the initial speed should be 2.87 km/s.