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Ulleksa [173]
4 years ago
7

What frequency fapproach is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction

opposite to the first train and approaching it?
Physics
1 answer:
Sergio039 [100]4 years ago
7 0

Answer:

The frequency is 302.05 Hz.

Explanation:

Given that,

Speed = 18.0 m/s

Suppose a train is traveling at 30.0 m/s relative to the ground in still air. The frequency of the note emitted by the train whistle is 262 Hz .

We need to calculate the frequency

Using formula of frequency

f'=f(\dfrac{v+v_{p}}{v-v_{s}})

Where, f = frequency

v = speed of sound

v_{p} = speed of passenger

v_{s} = speed of source

Put the value into the formula

f'=262\times(\dfrac{344+18}{344-30})

f'=302.05\ Hz

Hence, The frequency is 302.05 Hz.

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Which of the following will cause an increase in gas pressure in a closed container?
Ludmilka [50]

C. Both A. and B.

Explanation:

Statement A. Reducing the volume is true because of Boyle's law, which states that for a gas at fixed temperature, the pressure p and the volume V are inversely proportional:

pV=const.

Therefore, when the volume V is reduced, the pressure p increases.

Statement B. Adding more gas is also true: in fact, if we add gas into the container, we will have more molecules of the gas hitting the wall of the container. But the pressure of a gas is exactly given by this: by the collision of the molecules against the wall of the container, so the more the molecules of gas, the greater the pressure.

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A container of gas is at a pressure of 1.3x10^5 Pa and a volume of 6 m^3. How much work is done by the gas if it expands at a co
Ivenika [448]
In thermodynamics, work of a system at constant pressure conditions is equal to the product of the pressure and the change in volume. It is expressed as follows:

W = P(V2 - V1)
W = 1.3x10^5 (2x6 - 6 )
<span>W = 780000 J
</span>
Hope this answers the question. Have a nice day.
5 0
3 years ago
8. A car moved 20 km East and 70 km West.<br> What is the distance?
SCORPION-xisa [38]
Your answer is.......a car moved 60 km East and 90 km in west.
7 0
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The density of a material in CGS system of units is 4g cm-³. In a system of units in which unit of length is 10 cm and unit of m
butalik [34]

\sf\underline{Solution:}

Here , the density of the material is 4g cm³ but it is not given in CGS system.

$\sf{As\:we\:know\:that:}$

$\sf\bold{Density=}$ $\sf\dfrac{Mass}{Volume}$

$\space$

\sf{Now,according \: to \:the\:question:}

$\sf\small{Density\:of\:the\:material=4}$ $\sf\dfrac{g}{cm^2}$

$\space$

$\sf{It\:is\:given\:that:}$

In the system of units the mass is 100gram.

$\space$

Hence,

$\sf{The\:mass\:unit\:for\:4g=}$ $\sf\dfrac{4}{100}$ $\sf{units}$

$\space$

In the system of units,the length is 10cm.

Henceforth,

$\sf\small{The\:length \:for\:1cm\:units=}$ $\sf\dfrac{1}{10}$ $\sf{units}$

$\space$

<u>☆</u><u> </u><u>Substitute</u><u> </u><u>the</u><u> </u><u>required</u><u> </u><u>values</u><u> </u><u>in</u><u> </u><u>the</u><u> </u><u>given</u><u> </u><u>formula</u><u>-</u>

$\sf\purple{Density=}$ $\sf\dfrac\purple{Mass}\purple{volume}$

$\space$

$\sf\underline\bold{Density\:of\:the\:material:}$

= $\sf\dfrac{4/100}{1/10^3}$ $\sf\bold{units}$

$\space$

= $\sf\dfrac{4/100}{1/1000}$ $\sf\bold{units}$

$\space$

= $\sf\dfrac{4000}{100}$ $\sf\bold{units}$

$\space$

$\sf\underline\bold\blue{=40\:units}$

$\sf\small{Therefore,option\:2nd\:is\:correct!}$

_______________________________

6 0
3 years ago
You wad up a piece of paper and throw it into the wastebasket. How far will
astraxan [27]

Answer:

Since the paper is wadded up tight, and if there's any

air resistance left we assume there isn't any, it might

just as well be a stone that's tossed.  This is just a

stripped down projectile situation.

You said "an angle of 36 degrees", but you didn't say relative

to what.  I'll assume that it's 36 degrees above horizontal, and

now I'll proceed to answer the question with the information that

I just gave myself.

-- The vertical component of the velocity is  1.4 sin(36)

                                                                        = 0.823 m/s up.

-- The projectile rises for (0.823/9.8) second, runs out of gas,

and then falls for another (0.823/9.8) second to its original height.

So it's in the air for

                                  2 (0.823/9.8) = 0.168 second

                                                            (not very long at all)

-- The horizontal component of the velocity is  1.4 cos(36)

                                                                           = 1.133 m/s  

                                                             and it doesn't change.

-- During the 0.168 second that it's in the air,

the wad travels horizontally

                                              (0.168 s) x (1.133 m/s)

                                          =            0.19 meter

                                              (19 cm, ~ 7.5 inches)

If you find my mistake on this one, please please tell me.  

As of now, it looks like with that velocity at that angle, your

paper wad only makes it 7.5 inches from your hand into the can.

Explanation:

6 0
3 years ago
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