A driver can exceed the postage maximum speed limit in a work zone T or F?

A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t

spin [16.1K]

Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

$f = f_{0}\frac{v + v_{o}}{v - v_{s}}$

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer

v: is the speed of the sound = 343 m/s

$v_{o}$ : is the speed of the observer = 0 (it is heard in the town)

$v_{s}$ : is the speed of the source =?

The frequency of the train before slowing down is given by:

$f_{b} = f_{0}\frac{v}{v - v_{s_{b}}}$ (1)

Now, the frequency of the train after slowing down is:

$f_{a} = f_{0}\frac{v}{v - v_{s_{a}}}$ (2)

Dividing equation (1) by (2) we have:

$\frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}}$

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}}$ (3)

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

$v_{s_{b}} = 2v_{s_{a}}$ (4)

Now, by entering equation (4) into (3) we have:

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}}$

$\frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}}$

By solving the above equation for $v_{s_{a}}$ we can find the speed of the train after slowing down:

$v_{s_{a}} = 11.06 m/s$

Finally, the speed of the train before slowing down is:

$v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s$

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

I hope it helps you!

7 0

2 years ago

if a ball is thrown straight up into the air with an initial velocity of 8080 ft/s, its height in feet after tt second is given

yanalaym [24]

The average velocity for the time period beginning when t=1 and lasting

(i) 0.01 seconds = 63.84 ft/s

(ii) 0.001 seconds = 63.984 ft/s

Given that a ball is thrown with an initial velocity = 80 ft/s

Let 'y' be the height in feet after 't' seconds.

Given, $y=80t-16t^2$ gives the height in 't' seconds.

Average velocity = Rate of change of distance

= Change in distance/Change in time.

The initial time can be taken as 0 s.

When t =1 s, y = 80 - 16 = 64 ft

(1) t = 0.01 s

y = 80 x 0.01 - 16 x 0.01 x 0.01 = 0.7984 ft

Average velocity = (64 - 0.7984) / (1 -0.01) = 63.84 ft/s

(2) t = 0.001 s

y = 80 x 0.001 - 16 x 0.001 x 0.001 = 0.079984 ft

Average velocity = (64 - 0.079984) / (1 -0.001) = 63.984 ft/s

The question is incomplete. Find out the complete question below:

If a ball is thrown straight up into the air with an initial velocity of 80 ft/s, it height in feet after t second is given by $y=80t-16t^2$ .Find the average velocity for the time period beginning when t=1 and lasting

(i) 0.01 seconds

(ii) 0.001 seconds

Learn more about average velocity at brainly.com/question/6504879

#SPJ4

7 0

2 years ago

Rick shoots a basketball at an angle of 35' from the horizontal. It leaves his hands 7 feet from the ground with a velocity of 2

Korvikt [17]

Given:

The angle of projection of the basketball, θ=35°

The height at which the ball leaves the hand, h=7 ft

The initial velocity of the basketball, v=20 ft/s

To find:

The parametric equations describing the shot.

Explanation:

The range, x of the basketball is given by,

$x=v\cos\theta t$

On substituting the known values,

$\begin{gathered} x=20\times\cos35\degree\times t \\ \implies x=16.4t \end{gathered}$

The change in the height, y of the basketball is given by,

$y=-v\sin\theta t+\frac{1}{2}gt^2$

Where g is the acceleration due to gravity.

On substituting the known values,

$\begin{gathered} y=-20\times\sin35\degree\times t+\frac{1}{2}\times32\times t^2 \\ \implies y=-11.5t+16t^2 \end{gathered}$

Final answer:

The parametric equations describing the shot are

$\begin{gathered} \begin{equation*} x=16.4t \end{equation*} \\ \begin{equation*} y=-11.5t+16t^2 \end{equation*} \end{gathered}$

8 0

1 year ago

Suppose a cart with no fans has a starting velocity of 2 m/s. What will be the velocity of the cart when it reaches the wall?

Roman55 [17]

Answer:

less than stating velocity due to friction and air resistance.

Explanation:

3 0

2 years ago

Why might you want to use a single fixed pulley?

hjlf

A single fixed pulley can be used to raise or lower lightweight objects.

Option b

<u>Explanation:</u>

A pulley is a simple machine tool which is used to make lifting or lowering tasks easy. A single fixed pulley is a system involving only one pulley fixed on a constant rigid support with a rope wrapped around the wheel. Such a system can be used only to change the direction of applied force in raising or lowering small, lightweight objects which need minimal work force.

A single fixed pulley system helps only in redirecting the applied force direction by using a rope and wheel assembly. The work done in such a case remains the same and hence it is not preferred to use it in lifting heavy objects. Neither is the required force reduced in case of a single fixed pulley system. A movable pulley helps in achieving (A) and (C).

4 0

3 years ago