Answer:
period of oscillations is 0.695 second
Explanation:
given data
mass m = 0.350 kg
spring stretches x = 12 cm = 0.12 m
to find out
period of oscillations
solution
we know here that force
force = k × x .........1
so force = mg = 0.35 (9.8) = 3.43 N
3.43 = k × 0.12
k = 28.58 N/m
so period of oscillations is
period of oscillations = 2π × ................2
put here value
period of oscillations = 2π ×
period of oscillations = 0.6953
so period of oscillations is 0.695 second
Answer:
Failure rate = 20%
MTBF = 880 hours
Explanation:
given data
batteries = 10
tested = 200 hours
one failed = 20 hours
another fail at = 140 hours
solution
we know that Mean Time between Failures is express as = (Total up time) ÷ (number of breakdowns) ....................1
so here Total up time will be
Total up time = 200 × 10
Total up time = 2000
and here
Number of breakdown = 1 at 20 hour and another at 140 hour = 2
so it will be = (Total up time) ÷ (number of breakdowns) .......2
= = 1000
so here gap between occurrences is
gap between occurrences= 140 - 20
gap between occurrences = 120 hour
and
MTBF will be
MTBF = 1000 - 120
MTBF = 880 hours
and
Failure rate (FR) will be
Failure rate (FR) = 1 ÷ MTBF ................3
Failure rate (FR) = R÷T ......................4
as here R is the number of failures and T is total time
so Failure rate (FR) = 20%
Complete Question
The complete question is shown on the uploaded image
Answer:
The nuclei experience a force that will move it to the right of the conductor rod while the electrons experience a force that will move it to the left side of the conductor rod.
Explanation:
The force that act on the charges(both the positive and the negative charge ) is Mathematically expressed as
F = qE => E = F/q
where F is the force, q is the charge and E is the electric field
we can see that the force is directly proportional to the electric field,so an increase in the electric field would increase the force.
we can also see that the electric field is given as force per unit charge and generally the direction of this field is taken to be the direction of the force it would exert on a positive test charge
Now from the question we are being told that the external electric field is the direction of the positive x axis
Hence this field would drive the positive charge i.e the nuclei to the right.
In order to further explain let consider this
Generally the electric field is always radially outward originating from a positive point charge and radially in toward a negative point charge.
what this means for this question, is that the positive point charge is on the left side of the electric field while the negative point charge is at the right side of the field.
According to Coulomb's law which states that unlike term attract while like terms repel, it means that the electron would move to the left of the conductor rod while the nuclei would move to the right side of the conductor rod.
