Sound waves are known to be the one that's not considered as a type of electromagnetic energy. As for microwaves and x-rays, they tend to share the same frequencies that can be considered as electromagnetic, and sound waves have a different frequency than them.

5 0

3 years ago

The speed of sound in room temperature (20°C) air is 343 m/s; in room temperature helium, it is 1010 m/s. The fundamental freque

Lera25 [3.4K]

Answer: f = 927.55Hz

Explanation: Since the the tube is open-closed, the length of air and the wavelength of sound passing through the tube is given below

L = λ/4 where λ = wavelength.

speed of sound in air = v = 343m/s.

fundamental frequency of open closed tube = 315Hz

λ = 4L.

v = fλ

343 = 315 * 4L

343 = 1260 * L

L = 343/ 1260

L = 0.27m

In the same tube of length L = 0.27m but different medium ( helium), the speed of sound is 1010m/s.

The length of tube and wavelength are related by the formulae below

L = λ/4, λ=4L

λ = 4 * 0.27

λ = 1.087m.

v = fλ

1010 = f * 1.087

f = 1010/1.807

f = 927.55Hz

4 0

3 years ago

A car starts from rest and accelerates at 2.5m/s^2 for 2s. What is the final velocity of the car

Westkost [7]

Answer:

5m/s ez

Explanation:

plz make me brainliest

5 0

3 years ago

An RC car is carrying a tiny slingshot with a spring constant of 85 N/m at 0.2 m off the ground at 5.6 m/s. The sling shot is pu

ololo11 [35]

Answer:

The velocity of the steel pellet relative to the ground when it leaves the sling shot is approximately 5.960 meters per second.

Explanation:

Let suppose that RC car-slingshot-steelpellet is a conservative system, that is, that non-conservative forces (i.e. friction, air viscosity) can be neglected. The velocity of the steel pellet can be found by means of the Principle of Energy Conservation and under the consideration that change in gravitational potential energy is negligible and that the RC car travels at constant velocity:

$\frac{1}{2}\cdot (m_{C}+m_{P})\cdot v_{o}^{2} + \frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m_{C}\cdot v_{o}^{2} + \frac{1}{2}\cdot m_{P}\cdot v^{2}$

$\frac{1}{2}\cdot m_{P}\cdot v_{o}^{2} + \frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m_{P}\cdot v^{2}$

$m_{P}\cdot v_{o}^{2} + k\cdot x^{2} = m_{P}\cdot v^{2}$

$v^{2} = v_{o}^{2} + \frac{k}{m_{P}}\cdot x^{2}$

$v = \sqrt{v_{o}^{2}+\frac{k}{m_{P}}\cdot x^{2} }$ (1)

Where:

$v_{o}$ - Initial velocity of the steel pellet, measured in meters per second.

$v$ - Final velocity of the steel pellet, measured in meters per second.

$k$ - Spring constant, measured in newtons per meter.

$m_{P}$ - Mass of the steel pellet, measured in kilograms.

$m_{C}$ - Mass of the RC car, measured in kilograms.

$x$ - Initial deformation of the spring, measured in meters.

If we know that $v_{o} = 5.6\,\frac{m}{s}$ , $k = 85\,\frac{N}{m}$ , $m_{P} = 0.025\,kg$ and $x = 0.035\,m$ , then the velocity of the steel pellet relative to the ground when it leaves the sling shot is:

$v = \sqrt{\left(5.6\,\frac{m}{s} \right)^{2}+\frac{\left(85\,\frac{N}{m} \right)\cdot (0.035\,m)^{2}}{0.025\,kg} }$

$v \approx 5.960\,\frac{m}{s}$

The velocity of the steel pellet relative to the ground when it leaves the sling shot is approximately 5.960 meters per second.

5 0

3 years ago