Answer:
The final charges of each sphere are: q_A = 3/8 Q
, q_B = 3/8 Q
, q_C = 3/4 Q
Explanation:
This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.
Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point
q_A = Q / 2
q_B = Q / 2
Now sphere A touches sphere C, ending with half the charge
q_A = ½ (Q / 2) = ¼ Q
q_B = ¼ Q
Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge
q = Q / 4 + Q / 2 = ¾ Q
This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q
q_A = 3/8 Q
q_B = 3/8 Q
The final charges of each sphere are:
q_A = 3/8 Q
q_B = 3/8 Q
q_C = 3/4 Q
Answer:
the wind carries abrasive materials
Explanation:
such as sand and salt over time theses small particles slowly strip way at the land form sculpting it by eroding the softer layers first
Given that the function of the wave is f(x) = cos(π•t/2), we have;
a. The graph of the function is attached
b. 4 units of time
c. Even
d. 4.935 J/kg
e. 1.234 W/kg
<h3>How can the factors of the wave be found?</h3>
a. Please find attached the graph of the signal created with GeoGebra
b. The period of the signal, T = 2•π/(π/2) = <u>4</u>
c. The signal is <u>even</u>, given that it is symmetrical about the y-axis
d. The energy of the signal is given by the formula;

Which gives;
E = 0.5 × 1.571² × 1² × 4 = <u>4.935 J/kg</u>
e. The power of the wave is given by the formula;
E = 0.5 × 1.571² × 1² × 4 × 0.25 = <u>1.234 W/</u><u>kg</u>
Learn more about waves here:
brainly.com/question/14015797
Answer:
Gravity changes with altitude. as we know The gravitational force is proportional to 1/R2, where R is your distance from the center of the Earth.
eg. The radius of the Earth at the equator is 6400 kilometers.
Let's say you were in a jet at the equator that was 40 kilometers high above the earth's surface.
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