Ask question

Ask question

All categories

3 years ago

10

You adjust the temperature so that a sound wave travels more quickly through the air. You increase the temperature from 30°C to

36°C. What's the velocity of the sound wave at this new temperature? A. 344.0 m/s B. 335.6 m/s C. 350.0 m/s D. 352.6 m/s

2 answers:

zhuklara [117]3 years ago

6 0

yep you were right the answer is D

earnstyle [38]3 years ago

5 0

The correct answer to the question is : D) 352.6 m/s.

CALCULATION :

As per the question, the temperature is increased from 30 degree celsius to 36 degree celsius.

We are asked to calculate the velocity of sound at 36 degree celsius.

Velocity of sound is dependent on temperature. More is the temperature, more is velocity of sound.

The velocity at this temperature is calculated as -

V = 331 + 0.6T m/s

= 331 + 0.6 × 36 m/s

= 331 + 21.6 m/s

= 352.6 m/s.

Here, T denotes the temperature of the surrounding.

Hence, velocity of the sound will be 352.6 m/s.

You might be interested in

Timed! I would really appreciate some help! thank you!

GenaCL600 [577]

Answer:

x = 5[km]

Explanation:

We must convert the time from minutes to hours.

$t=30[min]*\frac{1h}{60min}= 0.5[h]\\$

We know that speed is defined as the relationship between space and time.

$v=x/t$

where:

x = space [m]

t = time = 0.5 [h]

v = velocity [m/s]

Now replacing:

$x = 10[\frac{km}{h} ]*0.5[h]\\x=5[km]$

4 0

2 years ago

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

5 0

3 years ago

What needed to be present in marine mud to form fossils fuels?

garri49 [273]

It would be oraganic matter I think.

6 0

3 years ago

Read 2 more answers

What is the numerical value, in meters per second squared, of the acceleration of an object experiencing true free fall?

Nuetrik [128]

Answer:

9.8 m/s/s

Explanation:

The numerical value, in meters per second squared, of the acceleration of an object experiencing true free fall is 9.8 m/s/s. This is called the acceleration due to gravity.

8 0

3 years ago

Read 2 more answers

A (B + 25.0) g mass is hung on a spring. As a result, the spring stretches (8.50 A) cm. If the object is then pulled an addition

Morgarella [4.7K]

Answer:

Time period of the osculation will be 2.1371 sec

Explanation:

We have given mass m = (B+25)

And the spring is stretched by (8.5 A )

Here A = 13 and B = 427

So mass m = 427+25 = 452 gram = 0.452 kg

Spring stretched x= 8.5×13 = 110.5 cm

As there is additional streching of spring by 3 cm

So new x = 110.5+3 = 113.5 = 1.135 m

Now we know that force is given by F = mg

And we also know that F = Kx

So $mg=Kx$

$K=\frac{mg}{x}=\frac{0.452\times 9.8}{1.135}=3.90N/m$

Now we know that $\omega =\sqrt{\frac{K}{m}}$

So $\frac{2\pi }{T} =\sqrt{\frac{K}{m}}$

$\frac{2\times 3.14 }{T} =\sqrt{\frac{3.90}{0.452}}$

$T=2.1371sec$

8 0

3 years ago