All categories

3 years ago

A herdsman yelling out to a fellow herdsman heard his voice reflected by a cliff 4s later.What is

Physics

1 answer:

Sloan [31]3 years ago

4 0

Answer:

v = 340 m/s

Explanation:

Given that,

A herdsman yelling out to a fellow herdsman heard his voice reflected by a cliff 4s later.

The cliff is 680 m away

We need to find the velocity of sound in air.

Velocity = distance/time

Distance = 2 × 680 = 1360 m

$v=\dfrac{1360\ m}{4\ s}\\\\=340\ m/s$

So, the velocity of sound in air is 340 m/s.

You might be interested in

200 pages about string thirory

joja [24]

Answer:

do not know the answer

Explanation:

7 0

3 years ago

Read 2 more answers

11. Trait theory claims that

svetoff [14.1K]

A. people from the same location share the same personality type.

3 0

2 years ago

You just calibrated a constant volume gas thermometer. The pressure of the gas inside the thermometer is 286.0 kPa when the ther

diamong [38]

Answer:

$T_{2} = 606.69 K$

Explanation:

In that the gas thermometer is a constant volume, it is satisfied that:

$\frac{P_{1} }{T_{1} } = \frac{P_{2} }{T_{2} }$

How the boiling water is under regular atmospheric pressure, then

$T_{1} = 373 .15 K$

Thus

$\frac{286000}{373.15} = \frac{465000}{T_{2} }$

$T_{2} = 606.69 K$

5 0

3 years ago

At the Indianapolis 500, you can measure the speed of cars just by listening to the difference in pitch of the engine noise betw

allsm [11]

To develop this problem it is necessary to apply the concepts related to the Dopler effect.

The equation is defined by

$f_i = f_0 \frac{c}{c+v}$

Where

$f_h$ = Approaching velocities

$f_i$ = Receding velocities

c = Speed of sound

v = Emitter speed

And

$f_h = f_0 \frac{c}{c+v}$

Therefore using the values given we can find the velocity through,

$\frac{f_h}{f_0}=\frac{c-v}{c+v}$

$v = c(\frac{f_h-f_i}{f_h+f_i})$

Assuming the ratio above, we can use any f_h and f_i with the ratio 2.4 to 1

$v = 353(\frac{2.4-1}{2.4+1})$

$v = 145.35m/s$

Therefore the cars goes to 145.3m/s

7 0

3 years ago

1. A piece of metal weighs 50.0 N in air, 36.0 N in water, and 41.0 N in an unknown

denis23 [38]

Answer:

a) 3.37 x $10^{3} kg/m^3$

b) 6.42kg/ $m^{3}$

Explanation:

a) Firstly we would calculate the volume of the metal using it`s weight in air and water , after finding the weight we would find the density .

Weight of metal in air = 50N = mg implies the mass of metal is 5kg.

Now the difference of weight of the metal in air and water = upthrust acting on it = volume (metal) p (liquid) g = V (1000)(10) = 14N. So volume of metal piece = 14 x $10^{-4}$ $kg/m^{3}$ . So density of metal = mass of metal / volume of metal = 5 / 14 x $10^{-4}$ $kg/m^{3}$ = 3.37 x $10^{3} kg/m^3$

b) Water exerts a buoyant force to the metal which is 50−36 = 14N, which equals the weight of water displaced. The mass of water displaced is 14/10 = 1.4kg Since the density of water is 1kg/L, the volume displaced is 1.4L. Hence, we end up with 3.57kg/l. Moreover, the unknown liquid exerts a buoyant force of 9N. So the density of this liquid is 6.42kg/ $m^{3}$

3 0

3 years ago