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victus00 [196]
2 years ago
9

What are three types of electromagnetic waves that are used to transmit information? What types of devices are used to receive e

ach type of wave?
Type of electromagnetic wave:


Receiver devices:
Physics
1 answer:
MrMuchimi2 years ago
3 0

Answer:

Three types of electromagnetic waves, used to transmit various information

Explanation:

A form of energy waves having both electric & magnetic fields are Electromagnetic waves. Three types -

Radio Waves - These have longest wavelengths & transmit data through radio, satellites, radar .

Micro Waves - These have shorter wavelengths & are used in cooking appliances & predicting weather.

X rays - These have more short wavelength and can penerate soft tissues like skin & muscle, hence are used for medical examining

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Consider the following neutral electron configurations in which n has a constant value. Which configuration would belong to the
grandymaker [24]

Answer:

he configuration with the highest electronic affinity is 2s2 2p5

Explanation:

Electronic affinity is the variation of energy when we add an electron to a neutral atom to form an ion

When an electron is added, it must occupy a space is the sub-level of the atom, giving more stability when it approaches the configuration of a complete shell with eight electrons (noble gas), so the affinity must increase when moving in a period Group VIII noble gases)

Let's examine the given settings

In this case, when adding an electron, 2s2 is very far from a complete level configuration, so its affinity must be small.

2s2 2p2 when adding an electro the one has a little more affinity, but is still a long way from a full shell, it would be missing 3 electrons

2s2 2sp5 this is the atom with the highest electronic affinity, since i = that when adding an electron the ion has the configuration of a noble gas. This is the most stable on the list

2s2 2p6 already has a full shell making it very difficult to insert an electron into this atom.

In summary, the configuration with the highest electronic affinity is 2s2 2p5

3 0
3 years ago
Read 2 more answers
Can someone answer this please.?? I really need help
svetlana [45]

Answer:

1. is the age group 35 and 44

2. is 2006 i think its 2006 i cant really tell in the picture but its the one before the last one!

6 0
2 years ago
Certain rifles can fire a bullet with a speed of 950 m/s just as it leaves the muzzle (this speed is called the muzzle velocity)
TiliK225 [7]

Answer:

a) By v^2 = u^2 + 2as => a= 70291.70.

(b)By v = u + at => t= 1.58 ms.

(c)By v^2 = u^2 - 2gh => H = 46045.92 m.

Explanation:

a) By v^2 = u^2 + 2as

(950)^2 = 0 + 2 \times a \times 0.75\\a = 601666.67 m/s^2\\a/g = 688858.70/9.8 = 70291.70

(b)By v = u + at

950 = 0 + 601666.67 \times t\\t = 1.58 x 10^-3 sec = 1.58 ms

(c)By v^2 = u^2 - 2gh

0 = (950)^2 - 2 \times9.8 \times H\\H = 46045.92 m

4 0
2 years ago
Find its moment of inertia about an axis perpendicular to its plane and passing through the midpoint of the line connecting its
antoniya [11.8K]

A) Moment of inertia about an axis passing through the point where the two segments meet : $I_A=\frac{1}{12} M L^2$

B) Moment of inertia passing through the point where the midpoint of the line connects to its two ends: $I x=\frac{1}{3} M L^2$

What is Moment of inertia?

The term "moment of inertia" refers to a physical quantity that quantifies a body's resistance to having its speed of rotation along an axis changed by the application of a torque (turning force). The axis might be internal or exterior, fixed or not.

A) The moment of inertia about an axis passing through the point where the two segments meet is $I_A=\frac{1}{12} M L^2$given that the rod is bent at the center and distance from all the points to the axis remains the same, the moment of inertia about the center will remain the same.

B) Determine the moment of inertia about an axis passing through the point midpoint of the line which connects the two ends

First step: determine the distance between the ends ( d )

After applying Pythagoras theorem$\mathrm{d}=\frac{\sqrt{2}}{2} L$

Next step : determine distance between the two axis $(\mathrm{x})$

After applying Pythagoras theorem

\mathrm{x}=\frac{\sqrt{2}}{4} L$$

Final step : Calculate the value of $\mathrm{I}_{\mathrm{x}}$

applying Parallel Axis Theorem

$$I_x=I_8+M x^2$$

$$\begin{aligned}& =\frac{1}{12} M L^2+\frac{1}{4} M L^2 \\& \therefore \quad I x=\frac{1}{3} M L^2 \\&\end{aligned}$$

Hence we can conclude that Moment of inertia about an axis passing through the point where the two segments meet: $I_A=\frac{1}{12} M L^2$, Moment of inertia passing through the point where the midpoint of the line connects its two ends: $I x=\frac{1}{3} M L^2$

To learn more about moment of inertia visit:brainly.com/question/15246709

#SPJ4

5 0
1 year ago
HELP
dem82 [27]

Answer:

Explanation:

28 / 70 = 0.3857142...  = 0.39 hr

280 / 100 = 2.8 hrs.

(100 - 0) / 10 = 10 m/s²

(60 - 20) / 4 = 10 m/s²

3 0
3 years ago
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