The velocity of sound wave in air at 35°c is 353m/s, calculate the velocity of sound at 46°c

What part of the electromagnetic spectrum do you think is most important and why?

KonstantinChe [14]

Radio waves, gamma-rays, visible light, and all the other parts of the electromagnetic spectrum are electromagnetic radiation. Electromagnetic radiation can be described in terms of a stream of mass-less particles, ...

The electromagnetic spectrum is a map of all the types of light that we can identify. It separates all the types of light by wavelength because that directly relates to how energetic the wave is. More energetic wave

For most of history, visible light was the only known part of the electromagnetic spectrum. The ancient Greeks recognized that light traveled in straight lines and studied some of its properties, including reflection

4 0

3 years ago

Read 2 more answers

Find the minor measurement of the vernier scale by taking 49, 1mm divisions of the main scale and dividing it into 50 vernier di

olga nikolaevna [1]

<h2>♨ANSWER♥</h2>

length of V-50 = 49mm

length of V-1 = 49/50mm

= 0.98mm

so,

minor measurement = (M-1) - (V-1)

= 1mm -0.98mm

= 0.02mm

☆ Therefore,

The minor measurement of the vernier scale is 0.02mm.

☆...hope this helps...☆

_♡_mashi_♡_

6 0

2 years ago

What is difference between kilowatt and kilowatt hour?

Scrat [10]

Those two units can be compared to a 'mile per hour' and a 'mile per hour - hour'.
One is a rate. The other is a quantity, after maintaining a rate for some time.

-- 'Joule' is a unit of energy. It's the amount of work (energy) you do
when you push with a force of 1 newton though a distance of 1 meter.
Lifting 10 pound of beans 3 feet off the floor takes about 40.7 joules of energy.

-- 'Watt' is a rate of using energy . . . 1 joule per second.
If you lift 10 pounds 3 feet off the floor in 1 second, your power is 40.7 watts.

-- 'Watt-second' is the amount of energy used in one second,
at the rate of 1 joule per second . . . 1 joule.

-- 'Watt-hour' is the amount of energy used in one hour,
at the rate of 1 joule per second . . . 3,600 joules.

-- 'Kilowatt' is a bigger rate of using energy . . . 1,000 joules per second.

-- 'Kilowatt - second' is the amount of energy used in one second,
at the rate of 1,000 joules per second . . . 1,000 joules .

-- 'Kilowatt - hour' is the amount of energy used in one hour,
at the rate of 1,000 joules per second . . . 3,600,000 joules .

Depending on where you live, 3,600,000 joules of energy bought
from the electric company costs something between 5¢ and 25¢.

6 0

3 years ago

Read 2 more answers

Ls -2 a solution of 4x +3= -5?.

torisob [31]

Answer:

yes

Explanation:

Let's solve your equation step-by-step.

4x+3=−5

Step 1: Subtract 3 from both sides.

4x+3−3=−5−3

4x=−8

Step 2: Divide both sides by 4.

4x / 4 = −8 / 4

x=−2

Hope it helps,

Please mark me as the brainliest

Thank you

5 0

3 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:

$v_{2}=\sqrt{\left(15.5\,\frac{m}{s} \right)^{2}+2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (-3.5\,m)}$

$v_{2} \approx 13.100\,\frac{m}{s}$

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

4 0

3 years ago