Answers
Answer:
.487 s⁻¹
Explanation:
Let damping constant be τ . The equation of decreasing amplitude can be written as
A = A₀
A / A₀ =
At t = 9.43 s , A / A₀ = .01
.01 =
ln.01 = - 9.43 τ
-4.6 = -9.43τ
τ = .487 s⁻¹
Answer:
0.05508 kg/sec
Explanation:
mass of the oscillator m= 0.231 Kg
amplitude of oscillation given by
Ao= maximum amplitude
t= time and 1.00% of its initial value in t= 9.43 s.
A= 0.01Ao
⇒0.01=e^(-I×9.43)
ln100= 9.43×l
l=0.4883
we know that l= c/2m
c= damping constant
c= 2ml
=2×0.231×0.4883
=0.05508 kg/sec
Related Questions
How much longer will it taketo travel a distance of 6ookm ata speed of 50kmh than itwould at aspeed of 6okmh?
25% part (c) assume that d is the distance the cheetah is away from the gazelle when it reaches full speed. Derive an expression in terms of the variables d, vcmax and vg for the time, tc, it takes the cheetah to catch the gazelle.
A particular string resonates in four loops at a frequency of 320 Hz . Name at least three other (smaller) frequencies at which it will resonate. Express your answers using two significant figures separated by commas.
A circular loop of wire has radius 7.80cm . A sinusoidal electromagnetic plane wave traveling in air passes through the loop, with the direction of the magnetic field of the wave perpendicular to the plane of the loop. The intensity of the wave at the location of the loop is 2.03�10?2W/m2 , and the wavelength of the wave is 6.20m .What is the maximum emf induced in the loop? Express your answer with the appropriate units.I stumbled through the formulas I do know for EMF but cant seem to figure out how to get the right answer. Please help and provide explanation! Thanks
Answers
The wasted chemical energy be "8.64 × 10⁸ J" and the equivalent meals could be cooked be "144".
Chemical energy
According to the question,
Bulb power, P = 100 W
Time, t = 1 month or,
= 1 × 30 × 24
= 720 h
Efficiency, η = 30% or,
= 0.30
Fuel consumption, E = 6 MJ or,
= 6 × 10⁶ J
Energy consumed be:
→ = P × t
By substituting the values,
= 100 × 720
= 72 kWh
Wasted energy be:
→ =
=
= 240 kWh or,
= 240 × 3.6 × 10⁶
= 8.64 × 10⁸ J
and,
The no. of meals be:
→ N =
= 144 meals
Thus the answers above are correct.
Find out more information about chemical energy here:
Answer:
a
b
Explanation:
From the question we are told that
The power rating of the bulb is P = 100 W
The duration is t = 1 month = 1 * 30 * 24 = 720 h
The efficiency is
The fuel consumption for one meal is
Generally the energy consumed by the bulb is mathematically represented as
=>
=>
Generally the energy generated at the power plant that was wasted by the bulb is mathematically represented as
=>
=>
Converting this value to Joules
Generally the number of means that would be cooked is
=>
Answers
Answer:
(A). The time is 5.47 sec.
(B). The speed of the rock just before it strikes the ground is 39.59 m/s.
Explanation:
Given that,
Initial velocity = 14.0 m/s
Height = 70.0 m
(A). We need to calculate the time
Using second equation of motion
Put the value into the formula
(B). We need to calculate the speed of the rock just before it strikes the ground
Using third equation of motion
Put the value into the formula
Hence, (A). The time is 5.47 sec.
(B). The speed of the rock just before it strikes the ground is 39.59 m/s.
Answers
Answer:
Explanation:
Cart A is moving to the right with constant speed i.e. net acceleration is zero
because acceleration is change in velocity in given time
Cart B is moving towards right with gradually speed up so there is net acceleration which helps to increase the velocity s
This indicates the net force acting on the cart towards right
For cart C there is gradual slow down of cart which indicates cart is decelerating and a net force is acting towards which opposes its motion.
Answers
Answer:
The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C, but the direction is still to the right.
Explanation:
From coulomb's law, F = Eq
Thus,
F = E₁q₁
F = E₂q₂
Then
E₂q₂ = E₁q₁
where;
E₂ is the external electric field due to second test charge = ?
E₁ is the external electric field due to first test charge = 4 x 10⁶ N/C
q₁ is the first test charge = 13 mC
q₂ is the second test charge = 23 mC
Substitute in these values in the equation above and calculate E₂.
The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C when 13 mC test charge is replaced with another test charge of 23 mC.
However, the direction of the external field is still to the right.
Answers
Answer:
3.27
Explanation:
Electric Power: This can be defined as the rate at which electric energy is consumed. The unit of power is Watt (W).
Mathematically, electric power is represented as
P = VI ..................................... Equation 1.
Where P = power, V = voltage, I = Current.
For Circuit A,
P₁ = V₁I₁ ................................... Equation 2
Where P₁ = maximum power delivered by circuit A, V₁ = Voltage of circuit A, I₁ = circuit breaker rating of circuit A.
Given: V₁ = 218 V, I₁ = 45 A.
Substituting into equation 2
P₁ = 218×45
P₁ = 9810 W.
For Circuit B,
P₂ = V₂I₂............................. Equation 3
Where P₂ = maximum power delivered by the circuit B, V₂ = voltage of circuit B, I₂ = circuit breaker rating of circuit B
Given: V₂ = 120 V, I₂ = 25 A.
Substitute into equation 3
P₂ = 120(25)
P₂ = 3000 W.
Ratio of maximum power delivered by circuit A to that delivered by circuit B = 9810/3000
= 3.27.
Thus the ratio of maximum power delivered by circuit A to circuit B = 3.27
Answers
Answer: i think it is B
Explanation: