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Lorico [155]
3 years ago
5

Does sound travel outside earth's atmosphere in space explain

Physics
1 answer:
Yuliya22 [10]3 years ago
8 0
On a large scale, no. Sound travels through space by the compression and relaxation of molecules. When you speak to someone on Earth, each inflection of voice compresses local molecules, and relaxes the molecules around the compressed ones. In space, however, the large lack of molecules means that the compression waves can’t be transmitted from the sound to the listener.
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On a nice winter day at the South Pole, the temperature rises to −54°F. What is the approximate temperature in degrees Celsius?
Evgesh-ka [11]

Answer:

Its going to be about -47.78°C

3 0
3 years ago
Read 2 more answers
A glass flask whose volume is 1000 cm^3 at a temperature of 1.00°C is completely filled with mercury at the same temperature. W
Marat540 [252]

Answer:

the coefficient of volume expansion of the glass is \mathbf{  ( \beta_{glass} )= 1.333 *10^{-5} / K}

Explanation:

Given that:

Initial volume of the glass flask = 1000 cm³ = 10⁻³ m³

temperature of the glass flask and mercury= 1.00° C

After heat is applied ; the final temperature = 52.00° C

Temperature change ΔT = 52.00° C - 1.00° C = 51.00° C

Volume of the mercury overflow = 8.50 cm^3 = 8.50 ×  10⁻⁶ m³

the coefficient of volume expansion of mercury is 1.80 × 10⁻⁴ / K

The increase in the volume of the mercury =  10⁻³ m³ ×  51.00 × 1.80 × 10⁻⁴

The increase in the volume of the mercury = 9.18*10^{-6} \ m^3

Increase in volume of the glass =  10⁻³ × 51.00 × \beta _{glass}

Now; the mercury overflow = Increase in volume of the mercury - increase in the volume of the flask

the mercury overflow = (9.18*10^{-6}  -  51.00* \beta_{glass}*10^{-3})\ m^3

8.50*10^{-6} = (9.18*10^{-6}  -51.00* \beta_{glass}* 10^{-3} )\ m^3

8.50*10^{-6} - 9.18*10^{-6} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3

-6.8*10^{-7} =  ( -51.00* \beta_{glass}* 10^{-3} )\ m^3

6.8*10^{-7} =  ( 51.00* \beta_{glass}* 10^{-3} )\ m^3

\dfrac{6.8*10^{-7}}{51.00 * 10^{-3}}=  ( \beta_{glass} )

\mathbf{  ( \beta_{glass} )= 1.333 *10^{-5} / K}

Thus; the coefficient of volume expansion of the glass is \mathbf{  ( \beta_{glass} )= 1.333 *10^{-5} / K}

5 0
2 years ago
A body with mass 4 kg moves with a velocity of 108km/h <br> Calculate it kinetic<br> energy
Fed [463]

Explanation:

The answer is in the pic above

4 0
3 years ago
Which part of the microscope is the circular area on the stage that light passes through?
patriot [66]

Answer: The part of the microscope that is the circular area is the APERTURE

I hope this helped!

6 0
1 year ago
An automobile with a tangential speed of 54.1 km/h follows a circular road that has a radius of 41.6 m. The automobile has a mas
Viefleur [7K]

1) Available force of friction: 6174 N

2) No

Explanation:

1)

The magnitude of the frictional force between the car's tires and the pavement of the road is given by

F_f=\mu mg

where

\mu is the coefficient of friction

m is the mass of the car

g is the acceleration of gravity

For the car in this problem, we have:

\mu=0.500 (coefficient of friction)

m = 1260 kg (mass of the car)

g=9.8 m/s^2

Therefore, the force of friction is

F_f=(0.500)(1260)(9.8)=6174 N

2)

In order to mantain the car in circular motion, the force of friction must be at least equal to the centripetal force.

The centripetal force is given by

F=m\frac{v^2}{r}

where

m is the mass of the car

v is the tangential speed

r is the radius of the curve

In this problem, we have

m = 1260 kg

v=54.1 km/h =15.0 m/s is the tangential speed

r = 41.6 m is the radius of the curve

Therefore, the centripetal force is

F=(1260)\frac{15.0^2}{41.6}=6814 N

Therefore, the force of friction is not enough to keep the car in the curve, since F_f

4 0
3 years ago
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