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3 years ago

A sound wave takes 0.2 seconds to travel 306 meters.

Physics

1 answer:

chubhunter [2.5K]3 years ago

6 0

Answer:

1530 m/s

Explanation:

Given that :

Speed = distance / time

Travel time take = 0.2 s

Distance covered = 306 metres

The speed of sound in the material :

Speed = distance / time = 306 m / 0.2

Speed = 1530 m/s

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If the evaporator outlet temperature on an r410A system is 50f and the evaporator superheat is 10f, what is the evaporating pres

Olenka [21]

Answer:

So, the evaporating pressure of the R410A = 118 psig

Explanation:

Solution:

For R410A system:

Data Given:

Evaporator Outlet Temperature = 50°F

Evaporator Superheat = 10°F

Required:

Evaporating Pressure in the system = ?

For this, first of all, we need to calculate inlet temperature on R410A system from the given value of outlet temperature.

Evaporator inlet temperature is the difference of outlet temperature and evaporator superheat.

Evaporator inlet temperature = Outlet Temperature - Evaporator Superheat

Evaporator inlet Temperature = 50°F - 10°F

Evaporator inlet Temperature = 40°F

Now, as we have the inlet temperature and the R410A system. We can consult the pressure temperature chart or PT chart, which I have attached and highlighted the value of evaporating pressure for 40°F inlet temperature.

So, the evaporating pressure of the R410A = 118 psig

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3 years ago

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boyakko [2]

Answer:

D 5m/s

Explanation:

6 0

3 years ago

4. A force of 5 N gives a mass m,, an acceleration of 10 m's, and a mass

Art [367]

Answer:

For mass m 1 newton 2nd law

F=m 1 a 1

5=m 1 ×10

m 1 = 2 1 kg

For mass m 2

F=m 2 a 2

5=m 2 ×20

m 2 = 4 1 kg

if tied together

Total mass =m 1 +m 2 = 1/2 +1/4=3/4kg

Now

F=M T Q T

a T = 5/m T = 5×4/3 = 3 /20m/s^2

Explanation:

8 0

3 years ago

A charge q of 1.3 × 10-16 coulombs moves from point A to a lower potential at point B in an electric field of 3.2 × 102 newtons/

LUCKY_DIMON [66]

Explanation :

It is given that,

Charge, $q=1.3\times 10^{-16}\ C$

Electric field, $E=3.2\times 10^2\ N/C$

Distance, $d=1.1\times 10^{-2}\ m$

The work done is stored in the form of potential energy.

$W=F.d$

$\because F=qE$

So, $W=qE\ d$

$W=1.3\times 10^{-16}\ C\times 3.2\times 10^2\ N/C\times 1.1\times 10^{-2}\ m$

$W=4.576\times 10^{-16}\ J$

Hence, this is the required solution.

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3 years ago

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A star's _______ brightness is the brightness seen by humans on Earth. A star's _______ brightness is its actual brightness and

baherus [9]

A star's apparent brightness is the brightness seen by humans on Earth. A star's absolute brightness is its actual brightness and does not depend on where the star is viewed from.

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3 years ago