All categories

3 years ago

What is the speed of sound at the atmospheric temperature of 30°C?

Physics

2 answers:

Rina8888 [55]3 years ago

5 0

Answer:

Option C is the correct answer.

Explanation:

The speed of sound in air is given approximately by

$v=331.4+0.6T_c$

Here $T_c$ is the temperature in degree Celsius.

Temperature $T_c=30^0C$

Substituting we will get

v = 331.4 + 0.6 x 30 = 349.4 m/s

Option C is the correct answer.

12345 [234]3 years ago

4 0

The parcel will undergo projectile motion, which means that it will have motion in both the horizontal and vertical direction.

First, we determine how long the parcel will fall using:

s = ut + 1/2 at²

where s will be the height, u is the initial vertical velocity of the parcel (0), t is the time of fall and a is the acceleration due to gravity.

5.5 = (0)(t) + 1/2 (9.81)(t)²

t = 1.06 seconds

You might be interested in

A cheetah can go from a state of resting to running at 20 m/s in just two seconds what is the cheetah's average acceleration?

Sunny_sXe [5.5K]

It's C i believe. To solve it we just take 20 and divide it by 2. Which gives us the average of 10 m/s

-Steel jelly

8 0

3 years ago

Read 2 more answers

A motorcyclist travels with an average speed of 6 m/s . If the cyclist is going to a friend's house 36 m away , how much time wi

vodka [1.7K]

Answer:

Explanation:

The time taken by the cyclist can be found by using the formula

$t = \frac{d}{v} \\$

v is the velocity

d is the distance

From the question we have

$t = \frac{36}{6} \\ = 6$

We have the final answer as

Hope this helps you

5 0

3 years ago

True or false Variables are those factors that can be changed in an experiment

tekilochka [14]

True

Variables is changing. It is any factor, trait or condition that can exist in differing amounts or types. An experiment has three types: Independent, Dependent and Controlled.

6 0

3 years ago

A block weighing 30kg is moved at a constant speed over a horizontal surface by a force of 100 N applied parallel

DENIUS [597]

30-100=70^3•É+ 30 =3^E

5 0

3 years ago

Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is

saveliy_v [14]

Answer:

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

Explanation:

a) The mass flow rate through the nozzle can be calculated with the following equation:

$\dot{m_{i}} = \rho_{i} v_{i}A_{i}$

Where:

$v_{i}$ : is the initial velocity = 20 m/s

$A_{i}$ : is the inlet area of the nozzle = 60 cm²

$\rho_{i}$ : is the density of entrance = 2.21 kg/m³

$\dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s$

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

$\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}$

$0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}$

$A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}$

Therefore, the exit area of the nozzle is 23.6 cm².

I hope it helps you!

5 0

3 years ago

Read 2 more answers