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2 years ago

Which of the following statements correctly describe the various applications listed above?

Physics

1 answer:

fgiga [73]2 years ago

3 0

Answer:

The answer is "c,d,e, and g".

Explanation:

The correct choices can be defined as follows:

Higher-frequency microwaves aren't used in any of these systems.
Infrared waves aren't seen in each of these technologies.
Its shortest wavelength of all of the technologies listed is the above radiation generated by certain wireless networks.
These devices all produce waves of wavelengths ranging from 0.10 to 10.0 cm.

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Why is electric current scalar quantity with having direction ?

frez [133]

Explanation:

It represents the direction of flow of positive charge but is treated as a scalar quantity because current follows the laws of scalar addition and not the laws of vector addition. The angle between the wires carrying current does not affect the total current in the circuit.

8 0

1 year ago

Read 2 more answers

The sound from a bolt of lightning travelled 4.08 km in 12.0 s. What was the speed

LenaWriter [7]

Answer:

83.67 m/s

Explanation:

Set up a calculation to convert units of measure to what you need.

You have km/s and you need m/s.

4.08km 1000 m 83.67m

----------- X ---------- = --------------- the km will cancel out and you are left

12.0 s 1 km s with m/s

6 0

2 years ago

Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat

nikklg [1K]

Answers:

a) $T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) $a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds $V$ are given by:

$V=\omega L=\frac{2\pi}{T} L$ (1)

Where:

$\omega$ is the angular frequency

$L$ is the radius of the orbit of each satellite

$T$ is the period of the orbit of each satellite

Isolating $T$ :

$T=\frac{2 \pi L}{V}$ (2)

Applying this equation to each satellite:

$T_{1}=\frac{2 \pi L}{V_{1}}=261.79 s$ (3)

$T_{2}=\frac{2 \pi L}{V_{2}}=1570.79 s$ (4)

$T_{3}=\frac{2 \pi L}{V_{3}}=196.349 s$ (5)

$T_{4}=\frac{2 \pi L}{V_{4}}=98.174 s$ (6)

$T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s$ (7)

$T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s$ (8)

Ordering this periods from largest to smallest:

$T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

b) Acceleration $a$ is defined as the variation of velocity in time:

$a=\frac{V}{T}$ (9)

Applying this equation to each satellite:

$a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2}$ (10)

$a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2}$ (11)

$a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2}$ (12)

$a_{4}=\frac{V_{4}}{T_{4}}=1.629 m/s^{2}$ (13)

$a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2}$ (14)

$a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2}$ (15)

Ordering this acceerations from largest to smallest:

$a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

4 0

2 years ago

Objects P and Q are placed in a bowl of water. P floats and Q sinks. Which one or more of the following can be deduced from this

Lady bird [3.3K]

The correct option that can be deduced for both Object P and Q is Option b) I and II only

To solve this question correctly, we need to understand the concept of density and it relation to mass and volume.

<h3>What is Density?</h3>

Density is a physical property of an object and can be expressed by using the relation:

$\mathbf{Density = \dfrac{mass}{volume}}$

From the given parameters, we are being told that:

P → floats
Q → sinks

This implies that Q has a greater density that P. Since Q has a greater density than P, Q will be heavier since it will have greater mass.

However, Q will not be denser than water because if that happens, P will be have a greater density which is untrue in this scenario.

Therefore, we can conclude that:

1. Q is heavier than P
II. 1cm³ of Q has a greater mass than 1cm³ of P

Learn more about density here:

brainly.com/question/6838128

6 0

2 years ago

The Event Horizon Telescope needs a 22 micro-arcsecond resolution to view the event horizon regions around black holes. If the a

likoan [24]

Answer:

14869817.395 m

Explanation:

$\theta$ =22 microarcsecond

λ = Wavelength = 1.3 mm

Converting to radians we get

$22\times 10^{-6}\frac{\pi}{180\times 3600}\ radians$

From Rayleigh Criterion

$\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{\theta}\\\Rightarrow D=1.22\frac{1.3\times 10^{-3}}{22\times 10^{-6}\frac{\pi}{180\times 3600}}\\\Rightarrow D=14869817.395\ m$

Diameter of the effective primary objective is 14869817.395 m

It is not possible to build one telescope with a diameter of 14869817.395 m. But, we need this type of telescope. So, astronomers use an array of radio telescopes to achieve a virtual diameter in order to observe objects that are the size of supermassive black hole's event horizon.

7 0

2 years ago