It depends on the circuit. Sometimes it becomes a bit weaker, sometimes it stays the same.
Answer:
6 days.
Explanation:
From radioactivity, The expression for half life is given as,
R/R' = 2⁽ᵃ/ᵇ)................... Equation 1
Where R = original mass of the radioactive substance, R' = Remaining mass of the radioactive substance after decay, a = Total time taken to decay, b = half life.
Given: R = 80 g, R' = 10 g, b = 2 days.
Substitute into equation 1
80/10 = 2⁽ᵃ/²⁾
8 = 2⁽ᵃ/²⁾
2³ = 2⁽ᵃ/²)
Equating the base and solving for a
3 = a/2
a = 2×3
a = 6 days.
Answer:
d=0.137 m ⇒13.7 cm
Explanation:
Given data
m (Mass)=3.0 kg
α(incline) =34°
Spring Constant (force constant)=120 N/m
d (distance)=?
Solution
F=mg
F=(3.0)(9.8)
F=29.4 N
As we also know that
Force parallel to the incline=FSinα
F=29.4×Sin(34)
F=16.44 N
d(distance)=F/Spring Constant
d(distance)=16.44/120
d(distance)=0.137 m ⇒13.7 cm
Frequency = 1 / (period)
Frequency = 1 / (10 seconds) = (1/10) ( / second) = 0.1 per second = <em>0.1 Hz</em>.