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Marizza181 [45]

3 years ago

6

zybooks One AA battery in a flashlight stores 9400 J. The three LED flashlight bulbs consume 0.5 W. How many hours will the flas

hlight last?

1 answer:

Rom4ik [11]3 years ago

7 0

Answer:

The time for which the flashlight last is 5.23 hours.

Explanation:

Given that,

One AA battery in a flashlight stores 9400 J, E = 9400 J

Power consumed by the LED flashlight bulbs, P = 0.5 W

We need to find the time for which the flashlight last. The power of a device is given by the energy stored per unit time as :

$P=\dfrac{E}{t}$

$t=\dfrac{E}{P}$

$t=\dfrac{9400\ J}{0.5\ W}$

t = 18800 seconds

or

Since, 1 hour = 3600 seconds

t = 5.23 hours

So, the time for which the flashlight last is 5.23 hours. Hence, this is the required solution.

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Explain the advantages of using parallel circuits to series circuits in wiring rooms

irina [24]

The major advantage of parallel circuit which makes it advantageous in the application of wiring rooms are the supply of equal voltage across each circuit, which is not present in series circuit.

<u> Explanation:</u>

The use of parallel circuit is preferred over series circuit while wiring the room as the parallel gives the same voltage across every circuit which keeps it safe from damaging or burning as they give consistent voltage.

Apart from this, the parallel circuit allows you for independent operation without letting use of all the equipment present in the room. Then there is scope for additional components without changing the voltage.

5 0

2 years ago

A centrifuge is a common laboratory instrument that separates components of differing densities in solution. This is accomplishe

melamori03 [73]

Answer:

$a_r=389\ rev.s^{-2}$

$n=58350 rev$

Explanation:

Given:

time of constant deceleration, $t=10\ s$

A.

initial angular speed, $N_i=3890\ rpm\$

<u>Using equation of motion:</u>

$N_f=N_i+a_r.t$

$0=3890+a_r\times 10$

$a_r=389\ rev.s^{-2}$

B.

Using eq. of motion for no. of revolutions, we have:

$n=N_i.t+\frac{1}{2} a_r.t^2$

$n=3890\times 10+0.5\times 389\times 100$

$n=58350\ rev$

8 0

3 years ago

Which of the following statements does not represent a benefit of peer relationships? A. Jason and Sean argued over whose turn i

Wewaii [24]

C does not represent a benefit, so that seems most logical.

A) they comprised.
B) at least ones listening
D) nothings wring with that

3 0

2 years ago

Read 2 more answers

A normal mode of a closed system is an oscillation of the system in which all parts oscillate at a single frequency. In general

valentina_108 [34]

Answer:

Explanation:

(A)

The string has set of normal modes and the string is oscillating in one of its modes.

The resonant frequencies of a physical object depend on its material, structure and boundary conditions.

The free motion described by the normal modes take place at the fixed frequencies and these frequencies is called resonant frequencies.

Given below are the incorrect options about the wave in the string.

• The wave is travelling in the +x direction

• The wave is travelling in the -x direction

• The wave will satisfy the given boundary conditions for any arbitrary wavelength $\lambda_i$

• The wave does not satisfy the boundary conditions $y_i(0;t)=0
$

Here, the string of length L held fixed at both ends, located at x=0 and x=L

The key constraint with normal modes is that there are two spatial boundary conditions, $y(0,1)=0
$

and $y(L,t)=0$

.The spring is fixed at its two ends.

The correct options about the wave in the string is

• The wavelength $\lambda_i$ can have only certain specific values if the boundary conditions are to be satisfied.

(B)

The key factors producing the normal mode is that there are two spatial boundary conditions, $y_i(0;t)=0$ and $y_i(L;t)=0$ , that are satisfied only for particular value of $\lambda_i$ .

Given below are the incorrect options about the wave in the string.

• $A_i$ must be chosen so that the wave fits exactly o the string.

• Any one of $A_i$ or $\lambda_i$ or $f_i$ can be chosen to make the solution a normal mode.

Hence, the correct option is that the system can resonate at only certain resonance frequencies $f_i$ and the wavelength $\lambda_i$ must be such that $y_i(0;t) = y_i(L;t)=0
$

(C)

Expression for the wavelength of the various normal modes for a string is,

$\lambda_n=\frac{2L}{n}$ (1)

When $n=1$ , this is the longest wavelength mode.

Substitute 1 for n in equation (1).

$\lambda_n=\frac{2L}{1}\\\\2L$

When $n=2$ , this is the second longest wavelength mode.

Substitute 2 for n in equation (1).

$\lambda_n=\frac{2L}{2}\\\\L$

When $n=3$ , this is the third longest wavelength mode.

Substitute 3 for n in equation (1).

$\lambda_n=\frac{2L}{3}$

Therefore, the three longest wavelengths are $2L$ , $L$ and $\frac{2L}{3}$ .

(D)

Expression for the frequency of the various normal modes for a string is,

$f_n=\frac{v}{\lambda_n}$

For the case of frequency of the $i^{th}$ normal mode the above equation becomes.

$f_i=\frac{v}{\lambda_i}$

Here, $f_i$ is the frequency of the $i^{th}$ normal mode, v is wave speed, and $\lambda_i$ is the wavelength of $i^{th}$ normal mode.

Therefore, the frequency of $i^{th}$ normal mode is $f_i=\frac{v}{\lambda_i}$

.

6 0

2 years ago

A light platform is suspended from the ceiling by a spring. A student with a mass of 90 kg climbs onto the platform. When it sto

Ilya [14]

Refer to the diagram shown.

When the student climbs onto the platform, the spring stretches by 0.82 m to reach the equilibrium position.
The mass of the student is m = 90 kg, so his weight is
mg = (90 kg)*(9.8 m/s²) = 882 N

By definition, the spring constant is
k = (882 N)/(0.82 m) = 1075.6 N/m

When the spring is stretched by x from the equilibrium position, the restoring force is
F = - k*x.

If damping is ignored, the equation of motion is
F = m * acceleration
or
$m \frac{d^{2}x}{dt^{2}} = -kx \\ \frac{d^{2}x}{dt^{2}} + \frac{k}{m} x = 0$

Define ω² = k/m = 11.751 => ω = 3.457.
Then the solution of the ODE is
x(t) = c₁ cos(ωt) + c₂ sin(ωt)

x'(t) = -c₁ω sin(ωwt) + c₂ω cos(ωt)
When t=0, x' =0, therefore c₂ = 0

The solution is of the form
x(t) = c₁ cos(ωt)
When t = 0, x = 0.32 m. Therefore c₁ = 0.32

The motion is
x(t) = 0.32 cos(3.457t)
The single amplitude is 0.32 m, and the double amplitude is 0.64 m.

Answer:
0.32 m (single amplitude), or
0.64 m (double amplitude)

6 0

2 years ago