Ask question

Ask question

All categories

3 years ago

7

Sound travels at a speed of about 340m/s in the air. How long will it take an explosion to be heard from a distance of 20 miles

away ?

1 answer:

Goryan [66]3 years ago

3 0

First convert the speed of sound from meters/second to miles/second:

$340\,\dfrac{\mathrm m}{\mathrm s}\cdot\dfrac{3.281\,\mathrm{ft}}{1\,\mathrm m}\cdot\dfrac{1\,\mathrm{mi}}{5280\,\mathrm{ft}}=0.21\,\dfrac{\mathrm{mi}}{\mathrm s}$

Then the time it takes for the sound to travel 20 miles is

$\dfrac{20\,\mathrm{mi}}{0.21\,\frac{\mathrm{mi}}{\mathrm s}}=95\,\mathrm s$

You might be interested in

a woman is swimming across a cold lake. her body temperature is 98 degrees fahrenheit , and the lake water is at 60 degrees fahr

sergiy2304 [10]

Answer:

the woman has more thermal energy becuase of her body temp.

Explanation:

none

5 0

3 years ago

Careful measurements have been made of Olympic sprinters in the 100-meter dash. A quite realistic model is that the sprinter's v

mihalych1998 [28]

Answer:

a.

$\displaystyle a(0 )=8.133\ m/s^2$

$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=0.52\ m/s^2$

b. $\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

c. $t=9.9 \ sec$

Explanation:

Modeling With Functions

Careful measurements have produced a model of one sprinter's velocity at a given t, and it's is given by

$\displaystyle V(t)=a(1-e^{bt})$

For Carl Lewis's run at the 1987 World Championships, the values of a and b are

$\displaystyle a=11.81\ ,\ b=-0.6887$

Please note we changed the value of b to negative to make the model have sense. Thus, the equation for the velocity is

$\displaystyle V(t)=11.81(1-e^{-0.6887t})$

a. What was Lewis's acceleration at t = 0 s, 2.00 s, and 4.00 s?

To compute the accelerations, we must find the function for a as the derivative of v

$\displaystyle a(t)=\frac{dv}{dt}=11.81(0.6887\ e^{0.6887t})$

$\displaystyle a(t)=8.133547\ e^{-0.6887t}$

For t=0

$\displaystyle a(0)=8.133547\ e^o$

$\displaystyle a(0 )=8.133\ m/s^2$

For t=2

$\displaystyle a(2)=8.133547\ e^{-0.6887\times 2}$

$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=8.133547\ e^{-0.6887\times 4}$

$\displaystyle a(4)=0.52\ m/s^2$

b. Find an expression for the distance traveled at time t.

The distance is the integral of the velocity, thus

$\displaystyle X(t)=\int v(t)dt \int 11.81(1-e^{-0.6887t})dt=11.81(t+\frac{e^{-0.6887t}}{0.6887})+C$

$\displaystyle X(t)=11.81(t+1.45201\ e^{-0.6887t})+C$

To find the value of C, we set X(0)=0, the sprinter starts from the origin of coordinates

$\displaystyle x(0)=0=>11.81\times1.45201+C=0$

Solving for C

$\displaystyle c=-17.1482\approx -17.15$

Now we complete the equation for the distance

$\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

c. Find the time Lewis needed to sprint 100.0 m.

The equation for the distance cannot be solved by algebraic procedures, but we can use approximations until we find a close value.

We are required to find the time at which the distance is 100 m, thus

$\displaystyle X(t)=100=>11.81(t+1.45\ e^{-0.6887t})-17.15=100$

Rearranging

$\displaystyle t+1.45\ e^{-0.6887t}=9.92$

We define an auxiliary function f(t) to help us find the value of t.

$\displaystyle f(t)=t+1.45\ e^{-0.687t}-9.92$

Let's try for t=9 sec

$\displaystyle f(9)=9+1.45\ e^{-0.687\times 9}-9.92=-0.92$

Now with t=9.9 sec

$\displaystyle f(9.9)=9.9+1.45\ e^{-0.687\times 9.9}-9.92=-0.0184$

That was a real close guess. One more to be sure for t=10 sec

$\displaystyle f(10)=10+1.45\ e^{-0.687\times 10}-9.92=0.081$

The change of sign tells us we are close enough to the solution. We choose the time that produces a smaller magnitude for f(t).

$At t\approx 9.9\ sec, \text{ Lewis sprinted 100 m}$

7 0

3 years ago

A boy is whirling a stone around his head by means of a string. The string makes one complete revolution every second; and the m

Damm [24]

Answer

given,

Tension of string is F

velocity is increased and the radius is not changed.

the string makes two complete revolutions every second

consider the centrifugal force acting on the stone

= $\dfrac{mv^2}{r}$

now centrifugal force is balanced by tension

T = $\dfrac{mv^2}{r}$

From the above expression we can clearly see that tension is directly proportional to velocity and inversely proportional to radius.

When radius is not changing velocity is increasing means tension will also increase in the string.

8 0

3 years ago

........................................................ waves are detected first because they move so fast .

wel

<u><em>PRIMARY </em></u>Waves Are Detected First Because They Move So Fast.

<u><em>RIGHT</em></u> Angles To The Direction of Movement.

A Kind Of Scale Used To Measure The Amount of Seismic Energy Released By An Earthquake <u><em>RICHTER SCALE</em></u>

7 0

3 years ago

The "atomic weight" of an atom reflects the average number of

Illusion [34]

E) Protons, neutrons, and electrons

8 0

3 years ago