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Bumek [7]
3 years ago
12

A 58-dB sound wave strikes an eardrum whose area is 5.0Ã10â5m^2. The intensity of the reference level required to determine the

sound level is 1.0Ã10^â12W/m^2. 1yr=3.156Ã10^7s.
a. How much energy is absorbed by the eardrum per second?
b. At this rate, how long would it take the eardrum to receive a total energy of 1.0 J?
Physics
1 answer:
emmasim [6.3K]3 years ago
7 0

Answer: (a) Energy = 3.15 * 10-11 W

(b) time = 1000 years

Explanation:

ok to begin,

(a )  for the output intensity

58 = 10 * log(I/(1 *10^-12))

I = 6.31 *10^-7 W/m^2

energy absorbed = I * Area

energy absorbed = 6.31 *10^-7 * 5 *10^-5

energy absorbed = 3.15 * 10^-11 W

the energy absorbed is 3.15 * 10-11 W

part B)

let the time taken be given as t

1 = 3.15 * 10^-11 * t

t = 3.17 *10^10 s = 1000 years

the time taken is 3.17 *1010 s or 1000 years

i hope this helps

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007 (part 1 of 2) 1.0 points
levacccp [35]

Answer:

a) The angle of refraction is approximately 34.7

b) The angle the light have to be incident to give an angle of refraction of 90° is approximately 53.42°

Explanation:

According to Snell's law, we have;

\dfrac{n_1}{n_2} = \dfrac{sin (\theta_2)}{sin (\theta_1)}

The refractive index of the glass, n₁ = 1.66

The angle of incident of the light as it moves into water, θ₁ = 27.2°

a) The refractive index of water, n₂ = 1.333

Let θ₂ represent the angle of refraction of the light in water

By plugging in the values of the variables in Snell's Law equation gives;

\dfrac{1.66}{1.333} = \dfrac{sin (\theta_2)}{sin (27.2^{\circ})}

sin (\theta_2) = sin (27.2^{\circ}) \times \dfrac{1.66}{1.333} \approx 0.5692292265

θ₂ = arcsin(0.5692292265) ≈ 34.7°

The angle of refraction of the light in water, θ₂ ≈ 34.7°

b) When the angle of refraction, θ₂ = 90°, we have;

\dfrac{1.66}{1.333} = \dfrac{sin (90^{\circ})}{sin (\theta_1)}

sin (\theta_1) = \dfrac{sin (90^{\circ})}{\left( \dfrac{1.66}{1.333}\right)} = sin (90^{\circ}) \times \dfrac{1.333}{1.66} \approx 0.803

θ₁ ≈ arcsin(0.803) ≈ 53.42°

The angle of incident, θ₁, that would give an angle of refraction of 90° is θ₁ ≈ 53.42°

3 0
3 years ago
Identify the energy transformation in the image
Yuliya22 [10]

Answer:

d one is correct

Explanation:

as the electrical energy in the socket is transferred to the electric tea pot

4 0
2 years ago
A wheel has a rotational inertia of 16 kgm2. Over an interval of 2.0 s its angular velocity increases from 7.0 rad/s to 9.0 rad/
german

Answer:

<h2>128.61 Watts</h2>

Explanation:

Average power done by the torque is expressed as the ratio of the workdone by the toque to time.

Power = Workdone by torque/time

Workdone by the torque = \tau \theta = I\alpha * \theta

I is the rotational inertia = 16kgm²

\theta = angular\ displacement

\theta = 2 rev = 12.56 rad

\alpha \ is \ the\ angular\ acceleration

To get the angular acceleration, we will use the formula;

\alpha = \frac{\omega_f^2- \omega_i^2}{2\theta}

\alpha = \frac{9.0^2- 7.0^2}{2(12.54)}\\\alpha = 1.28\ rad/s^{2}

Workdone by the torque = 16 * 1.28 * 12.56

Workdone by the torque = 257.23 Joules

Average power done by the torque = Workdone by torque/time

=  257.23/2.0

= 128.61 Watts

8 0
3 years ago
You are presented with several wires made of the same conducting material. The radius and drift speed are given for each wire in
Zigmanuir [339]

Answer:

d > a > b > c

Explanation:

Given that

a) radius = 3r and drift speed = 1v

b) radius = 4r and drift speed = 0.5v

c) radius = 1r and drift speed = 5v

d) radius = 2r and  drift speed = 2.5v

Based on the above information, the ranking of the wires for reducing the electron current is

As we can see that the radius i.e to be less and the drift speed that is highest so it should be rank one

And, According to that, other options are ranked

Therefore, the ranking would be d > a > b > c

4 0
3 years ago
A 1.0 kg ball at the end of a 2.0 m string swings in a vertical plane. At its lowest point the ball is moving with a speed of 10
Trava [24]

Answer:

a) 4.65m/s

b) 59.8 N , 1.01125 N

Explanation:

a)

m = mass of the ball = 1 kg

r = length of the string = 2.0 m

h = height gained by the ball as it moves from lowest to topmost position = 2r = 2 x 2 = 4 m

v = speed at the lowest position = 10 m/s

v' = speed at the topmost position = ?

Using conservation of energy

Kinetic energy at topmost position + Potential energy at topmost position = Kinetic energy at lowest position

(0.5) m v'² + m g h = (0.5) m v²

(0.5) v'² + g h = (0.5) v²

(0.5) v'² + (9.8 x 4) = (0.5) (10)²

v' = 4.65m/s

b)

T' = Tension force in the string when the ball is at topmost position

T = Tension force in the string when the ball is at lowest position

At the topmost position:

force equation is given as

mg + T' = \frac{m v'^{2}}{r}

(1)(9.8) + T' = \frac{(1) (4.65)^{2}}{2}

T' = 1.01125 N

At the lowest position:

force equation is given as

T - mg = \frac{m v^{2}}{r}

T - (1) (9.8) = \frac{(1) (10)^{2}}{2}

T = 59.8 N

8 0
4 years ago
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