Answer:
n=6.56×10¹⁵Hz
Explanation:
Given Data
Mass=9.1×10⁻³¹ kg
Radius distance=5.3×10⁻¹¹m
Electric Force=8.2×10⁻⁸N
To find
Revolutions per second
Solution
Let F be the force of attraction
let n be the number of revolutions per sec made by the electron around the nucleus then the centripetal force is given by
F=mω²r......................where ω=2π n
F=m4π²n²r...............eq(i)
as the values given where
Mass=9.1×10⁻³¹ kg
Radius distance=5.3×10⁻¹¹m
Electric Force=8.2×10⁻⁸N
we have to find n from eq(i)
n²=F/(m4π²r)
Answer:
the force between the building and the ball is non-conservative (friction-type force)
Explanation
Explanation:For this exercise the student must create an impulse to move the ball towards the building, in this part he performs positive work since the applied force and the displacement are in the same direction.
When the ball moves it has a kinetic energy and if its height increases or decreases its potential energy also changes, but the sum of being must be equal to the initial work.
When the ball arrives and collides with the building, non-conservative forces, of various kinds; rubbing, breaking, etc. It transforms this energy into a part of heat and another in mechanical energy that the building must absorb, let us destroy its wall
Consequently, the force between the building and the ball is non-conservative (friction-type force
Answer:
serie Ceq=0.678 10⁻⁶ F and the charge Q = 9.49 10⁻⁶ C
Explanation:
Let's calculate all capacity values
a) The equivalent capacitance of series capacitors
1 / Ceq = 1 / C1 + 1 / C2 + 1 / C3 + 1 / C4 + 1 / C5
1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 1 / 6.2 + 1 / 6.2
1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 2 / 6.2
1 / Ceq = 0.666 + 0.3030 +0.1818 +0.3225
1 / Ceq = 1,147
Ceq = 0.678 10⁻⁶ F
b) Let's calculate the total system load
Dv = Q / Ceq
Q = DV Ceq
Q = 14 0.678 10⁻⁶
Q = 9.49 10⁻⁶ C
In a series system the load is constant in all capacitors, therefore, the load in capacitor 5.5 is Q = 9.49 10⁻⁶ C
c) The potential difference
ΔV = Q / C5
ΔV = 9.49 10⁻⁶ / 5.5 10⁻⁶
ΔV = 1,725 V
d) The energy stores is
U = ½ C V²
U = ½ 0.678 10-6 14²
U = 66.4 10⁻⁶ J
e) Parallel system
Ceq = C1 + C2 + C3 + C4 + C5
Ceq = (1.5 +3.3 +5.5 +6.2 +6.2) 10⁻⁶
Ceq = 22.7 10⁻⁶ F
f) In the parallel system the voltage is maintained
Q5 = C5 V
Q5 = 5.5 10⁻⁶ 14
Q5 = 77 10⁻⁶ C
g) The voltage is constant V5 = 14 V
h) Energy stores
U = ½ C V²
U = ½ 22.7 10-6 14²
U = 2.2 10⁻³ J
Answer:
The answer is D.
Explanation:
There is no gravity in Space so that means that it will decrease your weight but not your mass.
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Magnitudes are measured by intensity so a 3.4 earthquake is much less stronger than a 4.5 earthquake it’s very literally when measuring them the higher the number the stronger it is
