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3 years ago

13

Two identical ambulances with loud sirens are driving directly towards you at a speed of 40 mph. One ambulance is 2 blocks away

and the other is 10 blocks away. Which of the following is true?  [Note that pitch = frequency.]a) The siren from the closer ambulance sounds higher pitched to you.b) The siren from the farther ambulance sounds higher pitched to you.c) The pitch of the two sirens sounds the same to you.d) The siren from the farther ambulance sounds higher pitched, until the  closer ambulance passes you.

2 answers:

Rasek [7]3 years ago

5 0

Answer:

c) The pitch of the two sirens sounds the same to you

Explanation:

The pitch does not depend on the distance of the object from the observer.

As per the given data

pitch = frequency

Frequency = $f_{0}$ $\frac{V +- V_{0}}{V +- V_{s}}$

$f^{'}$ = $f_{0}$ $\frac{V }{V - V_{s}}$

Hence, the pitch of the two sirens remains the same for the observer.

Jobisdone [24]3 years ago

5 0

Answer:

c) The pitch of the two sirens sounds the same to you

Explanation:

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A rocket has initail mass M begins to move from space , with an exhaust constant speed . Find the mass of the rocket while has t

irina1246 [14]

Answer:

$m=\frac{m_{0}}{e}$

Explanation:

Equation of the rocket is,

$m\frac{dv}{dt} =F-v'\frac{dm}{dt}$

Here, v' is the relative velocity of rocket.

In space F is zero.

So,

$m\frac{dv}{dt} =-v'\frac{dm}{dt}\\dv=-v'\frac{dm}{m} \\v=-v'ln\frac{m}{m_{0} }$

Now the momentum can be obtained by multiply by m on both sides.

$P=-v'mln\frac{m}{m_{0} }$

Now for maxima, $\frac{dP}{dm}=0$

$-v'ln\frac{m}{m_{0} }-v'm\frac{m_{0}}{m }m_{0=0$

Now,

$ln(\frac{m}{m_{0} } )=-1\\\frac{m}{m_{0} }=\frac{1}{e} \\m=\frac{m_{0}}{e}$

Therefore, the mass of the rocket while having maximum momentum is $\frac{m_{0}}{e}$

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4 years ago

A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.

Mama L [17]

Answer: $f_{r}$ = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

$W_{x}$ = horizontal component of the weight = mgsinФ

$W_{y}$ = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - $f_{r}$ = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

g = acceleration due to gravity = 9.8 $m/s^{2}$

$f_{r}$ = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

$v^{2} = u^{2} + 2aS$

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

$2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}$

we slot in a into the equation below to get frictional force

mgsinФ - $f_{r}$ = ma

3 * 9.8 * sin 37 - $f_{r}$ = 3* 0.4

17.9633 - $f_{r}$ = 1.2

$f_{r}$ = 17.9633 - 1.2

$f_{r}$ = 16.49N

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4 years ago

A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 40.0 kg. Initially open and at re

xxTIMURxx [149]

Answer:

The final angular speed is 0.223 rad/s

Explanation:

By the conservation of angular moment:

ΔL=0

L₁=L₂

L₁ is the initial angular moment

L₂ is the final angular moment

L₁ is given by:

$L_1=L_{door} + L_{mud}$

As the door is at rest its angular moment is zero and the angular moment of mud can be considered as a point object, then:

$L_1= L_{mud}= mvr$

where

r is the distance from the support point to the axis of rotation (the mud hits at the center of the door; r=0.5 m)

v is the speed

m is the mass of the mud

L₂ is given by:

$L_2= (I_{door} + I_{mud}) \omega_f$

ωf is the final angular speed

The moment of inertia of the door can be considered as a rectangular plate:

$I_{door}=\frac{1}{3}MW^2$

M is the mass of the door

W is the width of the door

The moment of inertia of the mud is:

$I_{mud}=mr^2$

Hence,

$L_1=L_2\\mvr= (I_{door} + I_{mud}) \omega_f\\\omega_f=\frac{mvr}{I_{door} + I_{mud}} \\\omega_f=\frac{mvr}{I_{door} + I_{mud}}$

$\omega_f=\frac{0.5kg \times 12m/s \times 0.5m}{\frac{1}{3}40kg(1m)^2+0.5kg \times (0.5m)^2}$

$\omega_f=0.223 \frac{rad}{s}$

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Answer:

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