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2 years ago

What do radio waves and micowaves have in common

1 answer:

8 0

Both are ...

-- transverse
-- electromagnetic
-- waves
-- have the same speed
-- used for communication
-- regulated by the FCC
-- carry energy from place to place

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What are the physical characteristics of sound waves

OlgaM077 [116]

•THAT THE PROPAGATION OF SOUND WAVES NEED MEDIUM TO TRAVEL
•THE MEDIUM SHOULD POSSES ELASTICITY
•FOR THE FASTER PROPAGATION OF SOUND THE PARTICLES SHOULD BE VERY CLOSE TO EACH OTHER

6 0

3 years ago

Find the pressure exerted by a 3000 N crate that has an area of 2m squared

ipn [44]

Pressure = Force/ Area = 3000/2 = 1500 pascal.

8 0

2 years ago

How many neutrons does element X have if its atomic number is 23 and its mass number is 90?

givi [52]

Answer:

Explanation:

- The atomic number (Z) of an atom is equal to the number of protons in the nucleus

- The mass number (A) of an atom is equal to the sum of protons and neutrons in the nucleus

Therefore, calling p the number of protons and n the number of neutrons, for element X we have:

Z = p = 23

A = p + n = 90

Substituting p=23 into the second equation, we find the number of neutrons:

n = 90 - p = 90 - 23 = 67

4 0

3 years ago

Read 2 more answers

A 24 liter tank contains ideal helium gas at 27 degrees C anda pressure of 22 atm. How many moles of gas are in the tank?

frozen [14]

Answer:

d)21.5 moles

Explanation:

Given that

L = 24 L

T = 27 °C = 300 K

P = 22 atm

We know that ideal gas equation

P V = n R T

P=Pressure ,V= Volume ,n=Moles ,R=Universal gas constant ,T=Temperature

Now by putting the values

22 x 24 = n x 0.08206 x 300

n= 21.447 moles

n= 21.5 moles

Therefore the number of moles will be 21.5 moles.

The answer is "d".

5 0

2 years ago

At t=0 a grinding wheel has an angular velocity of 25.0 rad/s. It has a constant angular acceleration of 26.0 rad/s2 until a cir

Agata [3.3K]

Answer:

a) The total angle of the grinding wheel is 569.88 radians, b) The grinding wheel stop at t = 12.354 seconds, c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

Explanation:

Since the grinding wheel accelerates and decelerates at constant rate, motion can be represented by the following kinematic equations:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

$\omega = \omega_{o} + \alpha \cdot t$

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})$

Where:

$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

$\omega_{o}$ , $\omega$ - Initial and final angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

Likewise, the grinding wheel experiments two different regimes:

1) The grinding wheel accelerates during 2.40 seconds.

2) The grinding wheel decelerates until rest is reached.

a) The change in angular position during the Acceleration Stage can be obtained of the following expression:

$\theta - \theta_{o} = \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

If $\omega_{o} = 25\,\frac{rad}{s}$ , $t = 2.40\,s$ and $\alpha = 26\,\frac{rad}{s^{2}}$ , then:

$\theta-\theta_{o} = \left(25\,\frac{rad}{s} \right)\cdot (2.40\,s) + \frac{1}{2}\cdot \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)^{2}$

$\theta-\theta_{o} = 134.88\,rad$

The final angular angular speed can be found by the equation:

$\omega = \omega_{o} + \alpha \cdot t$

If $\omega_{o} = 25\,\frac{rad}{s}$ , $t = 2.40\,s$ and $\alpha = 26\,\frac{rad}{s^{2}}$ , then:

$\omega = 25\,\frac{rad}{s} + \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)$

$\omega = 87.4\,\frac{rad}{s}$

The total angle that grinding wheel did from t = 0 s and the time it stopped is:

$\Delta \theta = 134.88\,rad + 435\,rad$

$\Delta \theta = 569.88\,rad$

The total angle of the grinding wheel is 569.88 radians.

b) Before finding the instant when the grinding wheel stops, it is needed to find the value of angular deceleration, which can be determined from the following kinematic expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})$

The angular acceleration is now cleared:

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}$