Sphere A travels a horizontal distance of (10 m/s) <em>t</em> after time <em>t</em>, while sphere B travels a distance of (5 m/s) <em>t</em>. So sphere B lands on the floor some point between the table's edge and the point X (A).
The second object, the one that had twice the force applied to it, would move twice as far, I believe.
The critical angle is the angle of incident in the optically dense media that produces the angle of refraction as 90°.
sin c =1/n, where n is the refractive index and c is the critical angle.
Also, n1sinФ1=n2sinФ2
In this case we have,
1.33×sinc = 1.309×sin90° Where c is the critical angle.
sin c =1.309/1.33
sin c = 0.981995498
c = 79.11°
Answer:
The bird's average velocity is 4.00 m/s.
Explanation:
Given that,
Distance = 5190 km
Time
We need to calculate the bird's average velocity for the return flight
Using formula of average velocity
Where, d = position
t = total time
Put the value into the formula
Negative sign shows the bird is traveling in negative x -axis.
Hence, The bird's average velocity is 4.00 m/s.
Answer:
500÷25=20
so 20 coulombs per second
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