In an atomic clock there are approximately 9.193 × 109oscillations of the specified light emitted by cesium-133 atoms. The text
1 answer:
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Answer:
v = 6.06 m/s
Explanation:
In order for the rider to pass the top of the loop without falling, his weight must be equal to the centripetal force:
where,
v = minimum speed of motorcycle at top of the loop = ?
g = acceleration due to gravity = 9.8 m/s²
r = radius of the loop = diameter/2 = 7.5 m/2 = 3.75 m
Therefore, using these values in equation, we get:
<u>v = 6.06 m/s</u>
Answer:
W = 2.3 10²
Explanation:
The force of the weight is
W = m g
let's use the concept of density
ρ= m / v
the volume of a sphere is
V = π r³
V = π (1.0 10⁻³)³
V = 4.1887 10⁻⁹ m³
the density of water ρ = 1000 kg / m³
m = ρ V
m = 1000 4.1887 10⁻⁹
m = 4.1887 10⁻⁶ kg
therefore the out of gravity is
W = 4.1887 10⁻⁶ 9.8
W = 41.05 10⁻⁶ N
now let's look for the electric force
F_e = q E
F_e = 12 10⁻¹² 15000
F_e = 1.8 10⁻⁷ N
the relationship between these two quantities is
= 41.05 10⁻⁶ / 1.8 10⁻⁷
\frac{W}{F_e} = 2,281 10²
W = 2.3 10²
therefore the weight of the drop is much greater than the electric force