All categories

3 years ago

The velocity of sound in air at 300C is approximately :

Physics

1 answer:

kumpel [21]3 years ago

6 0

The velocity of sound in at 300C is 511.3 m/s.

Explanation:

The equation that gives the speed of sound in ar as a function of the air temperature is the following:

$v=(331.3+0.6T) m/s$

where

T is the temperature of the air, measured in Celsius degrees

In this problem, we want to find the speed of sound in ar for a temperature of

$T=300^{\circ}C$

Substituting into the equation, we find:

$v=331.3 + 0.6(300)=511.3 m/s$

So, the velocity of sound in at 300C is 511.3 m/s.

Learn more about sound waves:

brainly.com/question/4899681

#LearnwithBrainly

You might be interested in

A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co

algol [13]

Answer:

$\displaystyle X_T=66.6\ km$

Explanation:

<u>Accelerated Motion </u>

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

$\displaystyle V_f=V_o+a\ t$

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of $0.987 m/s^2$ for 182 seconds. Then it brakes with a constant acceleration of $-0.321 m/s^2$ until it comes to a stop. We need to find the total distance traveled.

The equation for the distance is

$\displaystyle X=V_o\ t+\frac{a\ t^2}{2}$

Our data is

$\displaystyle V_o=0,a=0.987m/s^2,\ t=182\ sec$

Let's compute the first distance X1

$\displaystyle X_1=0+\frac{0.987\times 182^2}{2}$

$\displaystyle X_1=16,346.7\ m$

Now, we find the speed at the end of the first period of time

$\displaystyle V_{f1}=0+0.987\times 182$

$\displaystyle V_{f1}=179.6\ m/s$

That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

$\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0$

$\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}$

$\displaystyle t=559.5\ sec$

Computing the second distance

$\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}$

$\displaystyle X_2=50,243.2\ m$

The total distance is

$\displaystyle X_t=x_1+x_2=16,346.7+50,243.2$

$\displaystyle X_t=66,589.9\ m$

$\displaystyle \boxed{X_T=66.6\ km}$

8 0

4 years ago

Two balls move away from each other, both traveling at 7 m/s. One has a mass of 2 kg and the other has a mass of 3 kg

zavuch27 [327]

A 300-kg bear grasping a vertical tree slides down at constant velocity. The friction force between the
tree and the bear is

5 0

3 years ago

the two rotating systems shown in the figure differ only in that the two identical movable masses are positioned at different di

svetoff [14.1K]

Answer: the block at the right lands first

Explanation:

7 0

3 years ago

What must be the pressure difference between the two ends of a 2.6 km section of pipe, 36 cm in diameter, if it is to transport

ad-work [718]

Answer: $1.13(10)^{3} Pa$

Explanation:

This problem can be solved by the following equation:

$\Delta P=\frac{8 \eta L Q}{\pi r^{4}}$

Where:

$\Delta P$ is the pressure difference between the two ends of the pipe

$\eta=0.20 Pa.s$ is the viscosity of oil

$L=2.6 km=2600 m$ is the length of the pipe

$Q=900 \frac{cm^{3}}{s} \frac{1 m^{3}}{(100 cm)^{3}}=0.0009 \frac{m^{3}}{s}$ is the Rate of flow of the fluid

$d=36 cm=0.36 m$ is the diameter of the pipe

$r=\frac{d}{2}=0.18 m$ is the radius of the pipe

Soving for $\Delta P$ :

$\Delta P=\frac{8 (0.20 Pa.s)(2600 m)(0.0009 \frac{m^{3}}{s})}{\pi (0.18 m)^{4}}$

Finally:

$\Delta P=1135.26 Pa \approx 1.13(10)^{3} Pa$

7 0

3 years ago

An ice skater starts with a velocity of 2.25 m/s in a 50.0 direction. after 8.33 m/s in a 120 direction. what is the y-component

Bond [772]

The y-component of the acceleration is $0.33 m/s^2$

Explanation:

The y-component of the acceleration is given by:

$a_y = \frac{v_y-u_y}{t}$

where

$v_y$ is the y-component of the final velocity

$u_y$ is the y-component of the initial velocity

t is the time elapsed

For the ice skater in this problem, we have:

$u_y = u sin \theta_1 = (2.25 m/s)(sin 50.0^{\circ})=1.72 m/s$

where

u = 2.25 m/s is the initial velocity

$\theta_1 = 50.0^{\circ}$ is the initial direction

$v_y = v sin \theta_2 = (4.65)(sin 120^{\circ})=4.03 m/s$ , where

v = 4.65 m/s is the final velocity

$\theta_2 = 120.0^{\circ}$ is the final direction

The time elapsed is

t = 8.33 s

Therefore, we can find the y-component of the acceleration:

$a_y=\frac{4.03-1.72}{8.33}=0.33 m/s^2$

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

3 0

3 years ago