All categories

3 years ago

I need the answer asap plz!!

Physics

1 answer:

nikdorinn [45]3 years ago

8 0

Answer:

266.67 K

Explanation:

Recall the formula for ideal gasses:

P * V = n * R * T

from which a proportion can be obtained if the volume of the gas and its number of moles remain constant. This is:

T / P = constant and therefore T1 / P1 = T2/ P2

Then for our case:

300 / 36 = T2 / 32

solving for T2:

T2 = 300 * 32 / 36 = 266.67 K

You might be interested in

A 300 gg ball on a 70-cmcm-long string is swung in a vertical circle about a point 200 cmcm above the floor. The string suddenly

Ratling [72]

Answer:

the tension in the string an instant before it broke = 34 N

Explanation:

Given that :

mass of the ball m = 300 g = 0.300 kg

length of the string r = 70 cm = 0.7 m

At highest point, law of conservation of energy can be expressed as :

$\frac{1}{2} mv^2 = mgh\\\\v = \sqrt{2gh}\\\\v = \sqrt{2*(9.8 \ m/s^2)*(6.00 \ m - 2.00 \ m)}\\\\$

$v = 8.854 \ m/s$

The tension in the string is:

$T = \frac{mv^2}{r}\\\\T = \frac{(0.300 \ kg)*(8.854 \ m/s^2)}{0.70 \ m}\\\\T = 33.59 N\\\\T = 34 \ N$

Thus, the tension in the string an instant before it broke = 34 N

6 0

3 years ago

Use this table of a school bus during morning pickups to calculate its average speed between 0 h and 2.340 h.

Gelneren [198K]

The average speed between 0 h and 2.340 h is 6.97 Km/h

Average speed is defined as the total distance travelled divided by the total time taken to cover the distance.

$Average \: speed = \frac{total \: distance}{total \: time} \\ \\$

With the above formula, we can obtain the average speed between 0 h and 2.340 h as illustrated below:

Total time = 2.340 – 0 = 2.340 h
Total distance = 16.3 – 0 = 16.3 Km
Average speed =?

$Average \: speed = \frac{total \: distance}{total \: time} \\ \\Average \: speed = \frac{16.3}{2.340} \\ \\ Average \: speed = 6.97 \: Km/h \\ \\$

Learn more about average speed: brainly.com/question/24884027

8 0

2 years ago

Water (density = 1x10^3 kg/m^3) flows at 15.5 m/s through a pipe with radius 0.040 m. The pipe goes up to the second floor of th

RUDIKE [14]

Answer:

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

Explanation:

By assuming that fluid is incompressible and there are no heat and work interaction through the line of current corresponding to the pipe, we can calculate the speed of the water floor in the pipe on the second floor by Bernoulli's Principle, whose model is:

$P_{1} + \frac{\rho\cdot v_{1}^{2}}{2}+\rho\cdot g\cdot z_{1} = P_{2} + \frac{\rho\cdot v_{2}^{2}}{2}+\rho\cdot g\cdot z_{2}$ (1)

Where:

$P_{1}$ , $P_{2}$ - Pressures of the water on the first and second floors, measured in pascals.

$\rho$ - Density of water, measured in kilograms per cubic meter.

$v_{1}$ , $v_{2}$ - Speed of the water on the first and second floors, measured in meters per second.

$z_{1}$ , $z_{2}$ - Heights of the water on the first and second floors, measured in meters.

Now we clear the final speed of the water flow:

$\frac{\rho\cdot v_{2}^{2}}{2} = P_{1}-P_{2}+\rho \cdot \left[\frac{v_{1}^{2}}{2}+g\cdot (z_{1}-z_{2}) \right]$

$\rho\cdot v_{2}^{2} = 2\cdot (P_{1}-P_{2})+\rho\cdot [v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})]$

$v_{2}^{2}= \frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})$

$v_{2} = \sqrt{\frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2}) }$ (2)

If we know that $P_{1}-P_{2} = 0\,Pa$ , $\rho=1000\,\frac{kg}{m^{3}}$ , $v_{1} = 15.5\,\frac{m}{s}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $z_{1}-z_{2} = -3.5\,m$ , then the speed of the water flow in the pipe on the second floor is: