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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?
Physics
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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An object of mass 6.00 kg falls with an aceleration of 8.00 m/s2. The magnitude of air resitance must be ____ N
allsm [11]
Using Newton's Second Law, we can find the air resistance. We know the net force is equal to mass times acceleration.F_{net} = m*a = (6.00kg)(8.00 \frac{m}{s^2}) = 48N 

F_{g} - F_{d} = 48N

48N = (6.00kg)(9.81m/s^2) - F_{d} 

F_{d} = 10.86N


7 0
3 years ago
A ball rolls horizontally off a table of height 0.6 m with a speed of 9 m/s. How long does it take the ball to reach the ground?
denis23 [38]

Answer: 0.067 s

Explanation:s = Ut + 1/2at^2

0.6 = 9t + 0.5 *10 *t^2

Where a = g =10m/s/s

Solving the quadratic equation

5t^2 + 9t - 0.6=0,

t= 0.067 s and - 1.7 s

Of which 0.067 s is a valid time

4 0
3 years ago
A child has an ear canal that is 1.3 cm long. Assume the speed of sound is v = 344 m/s.
kap26 [50]

Answer:

The  frequencies are (f, f_1) =  (6615.4 \ Hz , 19846.2\ Hz)

Explanation:

From the question we are told that

  The  length of the ear canal is  l = 1.3 \ cm  =\frac{1.3}{100}  =  0.013 \ m

   The  speed of sound is assumed to be  v_s  =  344 \ m/s

Now  taking look at a typical  ear canal  we see that we assume it is  a  closed pipe

   Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as

            f = \frac{v_s}{4 * l }

 substituting values  

          f = \frac{344}{4 * 0.013 }

         f = 6615.4 \ Hz

Also the the second harmonic for the pipe (ear canal) is mathematically represented as

        f_1 =  \frac{3v_s}{4 * l}

 substituting values  

       f_1 =  \frac{3 *  344}{4 * 0.013}

       f_1 =   19846.2 \ Hz

Given that sound would be loudest in the pipe at the frequency, it implies that the child  will have an increased audible sensitivity at this  frequencies

6 0
3 years ago
Each plate of an air-filled parallel-plate capacitor has an area of 45.0 cm2, and the separation of the plates is 0.080 mm. A ba
maw [93]

Answer:

Option (e)

Explanation:

A = 45 cm^2 = 0.0045 m^2, d = 0.080 mm = 0.080 x 10^-3 m,

Energy density = 100 J/m

Let Q be the charge on the plates.

Energy density = 1/2 x ε0 x E^2

100 = 0.5 x 8.854 x 10^-12 x E^2

E = 4.75 x 10^6 V/m

V = E x d

V = 4.75 x 10^6 x 0.080 x 10^-3 = 380.22 V

C = ε0 A / d

C = 8.854 x 10^-12 x 45 x 10^-4 / (0.080 x 10^-3) = 4.98 x 10^-10 F

Q = C x V = 4.98 x 10^-10 x 380.22 = 1.9 x 10^-7 C

Q =  190 nC

3 0
3 years ago
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