How much energy will be transferred to your eardrum while listening to this sound for 1.0 min?

Describe how an oscilloscope should be used to measure the frequency of the sound wave from the sonometer

ratelena [41]

Answer:

T = reading (cm) time base (s / cm)

f = 1 / T

Explanation:

An oscilloscope is a piece of equipment that allows you to visualize and measure a wave that reaches you, in the case of having a sonometer this transforms the sound wave into an electrical signal to be introduced through one of the voltage channels of the equipment, on the screen we will see the oscillating alternating signal, if it is fixed we can make the reading, if it is moving the time base and the trigger must be adjusted to stop it.

In the oscilloscope we can read the period of the signal, this is the time it takes for the signal to repeat itself with this value, we can calculate the frequency with the formula, for the reading of the period the distance is measured on the labeled screen and multiplied by the time base

T = reading (cm) time base (s / cm)

f = 1 / T

6 0

3 years ago

Can someone please give me the answer by 10 minutes

Dmitry [639]

24N is the answer
You would need to multiply 8 until you get to a number.

8 0

3 years ago

A student in gym class swings from a rope and they are moving 5 m/s at the bottom of their swing. What is the height they reach

vaieri [72.5K]

Answer:

A

Explanation:

4 0

3 years ago

A competitive go-cart driver is traveling at a speed of 32m/s. He sees a caution flag go up and slows down at a rate of -1.5 m/s

djyliett [7]

Answer:

His final velocity is 15.8 m/s.

Step-by-step explanation:

Given:

Initial velocity of the driver is, $u=32$ m/s

Acceleration of the driver is, $a=-1.5$ m/s²

Time taken to reach final velocity is, $t=10.8$ s.

The final velocity is given using the Newton's equations of motion as:

$v=u+at$ , where, $v$ is the final velocity.

Now, plug in the given values and solve for $v$ .

$v=32-1.5(10.8)\\v=32-16.2=15.8\textrm{ m/s}$

Therefore, his final velocity is 15.8 m/s.

5 0

3 years ago

Usain Bolt's world-record 100 m sprint on August 16, 2009, has been analyzed in detail. At the start of the race, the 94.0 kg Bo

ZanzabumX [31]

a) 893 N

b) 8.5 m/s

c) 3816 W

d) 69780 J

e) 8030 W

Explanation:

a)

The net force acting on Bolt during the acceleration phase can be written using Newton's second law of motion:

$F_{net}=ma$

where

m is Bolt's mass

a is the acceleration

In the first 0.890 s of motion, we have

m = 94.0 kg (Bolt's mass)

$a=9.50 m/s^2$ (acceleration)

So, the net force is

$F_{net}=(94.0)(9.50)=893 N$

And according to Newton's third law of motion, this force is equivalent to the force exerted by Bolt on the ground (because they form an action-reaction pair).

b)

Since Bolt's motion is a uniformly accelerated motion, we can find his final speed by using the following suvat equation:

$v=u+at$

where

v is the final speed

u is the initial speed

a is the acceleration

t is the time

In the first phase of Bolt's race we have:

u = 0 m/s (he starts from rest)

$a=9.50 m/s^2$ (acceleration)

t = 0.890 s (duration of the first phase)

Solving for v,

$v=0+(9.50)(0.890)=8.5 m/s$

c)

First of all, we can calculate the work done by Bolt to accelerate to a speed of

v = 8.5 m/s

According to the work-energy theorem, the work done is equal to the change in kinetic energy, so

$W=K_f - K_i = \frac{1}{2}mv^2-0$

where

m = 94.0 kg is Bolt's mass

v = 8.5 m/s is Bolt's final speed after the first phase

$K_i = 0 J$ is the initial kinetic energy

So the work done is

$W=\frac{1}{2}(94.0)(8.5)^2=3396 J$

The power expended is given by

$P=\frac{W}{t}$

where

t = 0.890 s is the time elapsed

Substituting,

$P=\frac{3396}{0.890}=3816 W$

d)

First of all, we need to find what is the average force exerted by Bolt during the remaining 8.69 s of motion.

In the first 0.890 s, the force exerted was

$F_1=893 N$

We know that the average force for the whole race is

$F_{avg}=820 N$

Which can be rewritten as

$F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}$

And solving for $F_2$ , we find the average force exerted by Bolt on the ground during the second phase:

$F_{avg}=\frac{0.890 F_1 + 8.69 F_2}{0.890+8.69}\\F_2=\frac{(0.890+8.69)F_{avg}-0.890F_1}{8.69}=812.5 N$

The net force exerted by Bolt during the second phase can be written as

$F_{net}=F_2-D$ (1)

where D is the air drag.

The net force can also be rewritten as

$F_{net}=ma$

where

$a=\frac{v-u}{t}$ is the acceleration in the second phase, with

u = 8.5 m/s is the initial speed

v = 12.4 m/s is the final speed

t = 8.69 t is the time elapsed

Substituting,

$a=\frac{12.4-8.5}{8.69}=0.45 m/s^2$

So we can now find the average drag force from (1):

$D=F_2-F_{net}=F_2-ma=812.5 - (94.0)(0.45)=770.2 N$

So the increase in Bolt's internal energy is just equal to the work done by the drag force, so:

$\Delta E=W=Ds$

where

d is Bolt's displacement in the second part, which can be found by using suvat equation:

$s=\frac{v^2-u^2}{2a}=\frac{12.4^2-8.5^2}{2(0.45)}=90.6 m$

And so,

$\Delta E=Ds=(770.2)(90.6)=69780 J$

e)

The power that Bolt must expend just to voercome the drag force is given by

$P=\frac{\Delta E}{t}$

where

$\Delta E$ is the increase in internal energy due to the air drag

t is the time elapsed

Here we have:

$\Delta E=69780 J$

t = 8.69 s is the time elapsed

Substituting,

$P=\frac{69780}{8.69}=8030 W$

And we see that it is about twice larger than the power calculated in part c.

3 0

3 years ago