DE which is the differential equation represents the LRC series circuit where
L d²q/dt² + Rdq/dt +I/Cq = E(t) = 150V.
Initial condition is q(t) = 0 and i(0) =0.
To find the charge q(t) by using Laplace transformation by
Substituting known values for DE
L×d²q/dt² +20 ×dq/dt + 1/0.005× q = 150
d²q/dt² +20dq/dt + 200q =150
Answer:
A 1.0 min
Explanation:
The half-life of a radioisotope is defined as the time it takes for the mass of the isotope to halve compared to the initial value.
From the graph in the problem, we see that the initial mass of the isotope at time t=0 is
The half-life of the isotope is the time it takes for half the mass of the sample to decay, so it is the time t at which the mass will be halved:
We see that this occurs at t = 1.0 min, so the half-life of the isotope is exactly 1.0 min.
Answer:
Explanation:
Given
See attachment for the graph
Required
Determine the frequency
Frequency (F) is calculated as:
Where
T = Time to complete a period
From the attachment, the wave complete a cycle or period in 3 seconds..
So:
--- Approximated
Answer:
67.9 kg*m/s
Explanation:
Pi = 38 kgm/s
F = 88.3N and ∆t = 0.338s
Final momentum Pf = Pi + F∆t = 38 + (88.3)(0.338) = 38 + 29.8454
=) Pf = 67.8454 kgm/s = 67.85kg*m/s
Your answer is 67.9kg*m/s with three significant figures
hope this helps your troubles!
Answer:
following are the solution to this question:
Explanation:
When I stand at such a scale in an elevated that's already rising upwards, its scale would appear to also be 0 because of free fall and would often reveal that weight whenever the lift is stable.
In this, the free fall is also known as the object, that is influenced exclusively by gravity, and an object operating only through the influence of gravity is said to be in a free-fall state.
