The lytic cycle is one of the two phases of reproduction in viruses. In this part of reproduction, the viral RNA joins with the DNA of the host and copies itself into the host's DNA pattern. This will cause the host cells to reproduce the viral cells instead of the original host cells. Overtime as the viral cells continue to reproduce they will overload and eventually cause the host cell to explode. When this occurs, then the produced viruses will continue on to infect the next cell.
The energy stored in motion is called kinetic energy.
To solve the problem we will apply the concepts related to the Intensity as a function of the power and the area, as well as the electric field as a function of the current, the speed of light and the permeability in free space, as shown below.
The intensity of the wave at the receiver is




The amplitude of electric field at the receiver is


The amplitude of induced emf by this signal between the ends of the receiving antenna is


Here,
I = Current
= Permeability at free space
c = Light speed
d = Distance
Replacing,


Thus, the amplitude of induced emf by this signal between the ends of the receiving antenna is 0.0543V
Answer: V=IR
Explanation: for a series circuit connected to a battery supply, the total emf across the circuit is given as
E = I(R + r) and by expanding, we have that E =IR + It
Where r is the internal resistance of the battery
I is the total current flowing in the circuit
R total load resistance in the circuit.
E is the total emf of the circuit.
The total emf is the sum of 2 separate voltages.
"IR" which is the terminal voltage and "Ir" which is the loss voltage.
The teenila voltage is the voltage flowing in the circuit based on the equivalent resistance of the circuit while the loss voltage is the wasted voltage based on the internal resistance of the battery source.
Answer:
Explanation:
Given that,
5J work is done by stretching a spring
e = 19cm = 0.19m
Assuming the spring is ideal, then we can apply Hooke's law
F = kx
To calculate k, we can apply the Workdone by a spring formula
W=∫F.dx
Since F=kx
W = ∫kx dx from x = 0 to x = 0.19
W = ½kx² from x = 0 to x = 0.19
W = ½k (0.19²-0²)
5 = ½k(0.0361-0)
5×2 = 0.0361k
Then, k = 10/0.0361
k = 277.008 N/m
The spring constant is 277.008N/m
Then, applying Hooke's law to find the applied force
F = kx
F = 277.008 × 0.19
F = 52.63 N
The applied force is 52.63N