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Archy [21]
4 years ago
12

When the temperature of the air is 50°C, the velocity of a sound wave traveling through the air is approximately?

Physics
2 answers:
noname [10]4 years ago
5 0

The answer is:

C. 361 m/s

The explanation:

To calculate  the speed of sound at a given temperature (50°C) we are going to use this formula:

v = 331 + 0.6T

when V is the velocity

and T is the temperature = 50°C

by substitution:

v = 331 + 0.6(50)

v = 361 m/s

So, The correct answer is C.

because of the variation of the motion of the molecules of air with change of temperature so, the velocity (V) of the  sound in the air is change with temperature.

Blababa [14]4 years ago
5 0
<span>When the temperature of the air is 50°C, the velocity of a sound wave traveling through the air is approximately Answer</span><span>: 361 m/s</span>
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When magma cools quickly, what kind of texture or
Ksivusya [100]

Answer:

a

Explanation:

when magma cools Crystal's form because the solution is super saturated with respect to some minerals if the magma cools quickly the crystals do not have much time to form hence they are small and also the resulting rock is fine grained

6 0
3 years ago
Two long, parallel wires are separated by 2.2 mm. Each wire has a 32-AA current, but the currents are in opposite directions. Pa
Alex

Answer:

B=1.1636*10^{-3}T

Explanation:

Given data

d_{wires}=2.2mm=0.022m\\ I_{current}=32A\\

To find

Magnitude of the net magnetic field B

Solution

The magnitude of the net magnetic field can be find as:

B=2*u\frac{I}{2\pi r}\\ B=2*(4\pi*10^{-7}  )\frac{32}{2\pi (0.022/2)} \\ B=1.1636*10^{-3}T

3 0
3 years ago
In which situations is elastic potential energy present? Check all that apply. A bow is drawn back from its equilibrium position
Simora [160]
1,4,6
a bow is drawn back
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6 0
3 years ago
Read 2 more answers
Two friction disks A and B are brought into contact when the angular velocity of disk A is 240 rpm counterclockwise and disk B i
mel-nik [20]

Answer:

a) αA = 4.35 rad/s²

αB = 1.84 rad/s²

b) t = 3.7 rad/s²

Explanation:

Given:

wA₀ = 240 rpm = 8π rad/s

wA₁ = 8π -αA*t₁

The angle in B is:

\theta _{B} =4\pi =\frac{1}{2} \alpha _{B} t_{1}^{2}  =\frac{1}{2} (\frac{r_{A} }{r_{B} } )^{3} \alpha _{A} t_{1}^{2}=\frac{1}{2} (\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}^{2}

\alpha _{A} =8\pi (\frac{0.2}{0.15} )^{3} =59.57rad

w_{B,1} =\alpha _{B} t_{1}=(\frac{0.15}{0.2} )^{3} \alpha _{A} t_{1}=0.422\alpha _{A} t_{1}

The velocity at the contact point is equal to:

v=r_{A} w_{A} =0.15*(8\pi -\alpha _{A} t_{1})=1.2\pi -0.15\alpha _{A} t_{1}

v=r_{B} w_{B} =0.2*(0.422\alpha _{A} t_{1})=0.0844\alpha _{A} t_{1}

Matching both expressions:

1.2\pi -0.15\alpha _{A} t_{1}=0.0844\alpha _{A} t_{1}\\\alpha _{A} t_{1}=16.09rad/s

b) The time during which the disks slip is:

t_{1} =\frac{\alpha _{A} t_{1}^{2}}{\alpha _{A} t_{1}} =\frac{59.574}{16.09} =3.7s

a) The angular acceleration of each disk is

\alpha _{A}=\frac{\alpha _{A} t_{1}}{t_{1} } =\frac{16.09}{3.7} =4.35rad/s^{2} (clockwise)

\alpha _{B}=(\frac{0.15}{0.2} )^{3} *4.35=1.84rad/s^{2} (clockwise)

6 0
3 years ago
A source produces 20 crests and 20 troughs in 4 seconds. The second crest is 3 cm away from the first crest.Calculate :
sineoko [7]

Answer:

Solution given:

No of waves[N] =20crests & 20 troughs

=20waves

Time[T]=4seconds

distance[d]=3cm=0.03m

Now

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<u>Frequency</u>=\frac{No of waves}{time}

=\frac{20}{4}=5Hertz

and

Wave speed:wave length×frequency=3 × {10}^{ - 2}m×5=1.5 × {10}^{ - 1} \tt{ {ms}^{ - 1}}.

3 0
3 years ago
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