Let us first find out the radioactive constant of Phosphorus 32.
Radioactive constant, λ =
Here, is half life of phosphorus 32 = 14.3 days
λ =
The amount of 4 mg of phosphorus remain after 71.5 days can be found using the formula,
m=m₀e⁻(λt)
=
=
=
=
=
= 0.125 mg
The mass of 4 mg of phosphorus 32 remains after 71.5 days will be 0.125 mg.
Simple, A. None of the other answers make sense so, A.
Answer
given,
mass of the boiled egg = 50 g = 0.05 Kg
spring constant, k = 25 N/m
initial Amplitude, A₁ = 0.3 m
final amplitude, A₂ = 0.1 m
time, t = 5 s
considering the amplitude of damped harmonic oscillation to calculate damping factor.
b is the damping factor and t is the time.
on solving
inserting all the given values
b = 0.0219 Kg/s
damping coefficient in three significant figure
b = 0.022 Kg/s
Answer:
U = 0.413 J
Explanation:
the potential energy between two charges q1 and q2 is given by the following formula:
(1)
k: Coulomb's constant = 8.98*10^9 NM^2/C^2
q1: first charge = 4.6 μC = 4.6*10^-6 C
q2: second charge = 1.0 μC*10^-6 C
r: distance between charges = 10.0 cm = 0.10 m
You replace the values of all variables in the equation (1):
Hence, the energy between charges is 0.413 J
Answer
given,
mass of base ball = 0.14 kg
speed before it made the contact with the ball (V i) = 42 m/s
speed after batter hit the ball(V f) = - 48 m/s
a)
impulse = change in momentum
=
=
= -12.6 Kg m/s
Magnitude of impulse = 12.6 Kg m/s
b)
Force =
=
Force = 2520 N