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yarga [219]
3 years ago
7

EM waves consist of changing electric and magnetic fields moving perpendicular with respect to each other. What kind of wave is

it?
-Transverse waves
-Longitudinal waves
-Mechanical waves
-Compressional waves
Physics
1 answer:
Serhud [2]3 years ago
5 0

Answer:

Transverse

Explanation:

Electromagnetic waves don't depend on the medium they travel through like a mechanical wave does, so they aren't mechanical. They don't oscillate (move back in forth) in the direction they travel either, ruling out compressional and longitudinal waves.

That leaves tranverse waves, the ones we're most used to, since they look very "wavelike," with smooth peaks and valleys. Electromagnic waves behave like these, oscillating in a plane perpendicular to the direction they're traveling in.

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Answer:

the product of mass and velocity

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3 years ago
A 3.0-kg block moves up a 40o incline with constant speed under the action of a 26-N force acting up and parallel to the incline
CaHeK987 [17]

Answer:

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3 0
3 years ago
What is the element with the lowest electronegativity value?
Lerok [7]
<h2>Answer: Francium </h2>

Let's start by explaining that electronegativity is a term coined by Linus Pauling and is determined by the <em>ability of an atom of a certain element to attract electrons when chemically combined with another atom. </em>

So, the more electronegative an element is, the more electrons it will attract.

It should be noted that this value can not be measured directly by experiments, but it can be determined indirectly by means of calculations from other atomic or molecular properties of the element. That is why the scale created by Pauling is an arbitrary scale, where the maximum value of electronegativity is 4, assigned to Fluorine (F) and the <u>lowest is 0.7, assigned to Francium (Fr).</u>

7 0
4 years ago
During electrical storms, a bolt of lightning can transfer 10 C of charge in 2.0 µs (the amount of time can vary considerably).
nydimaria [60]

Answer:

Explanation:

charge, q = 10 C

time, t = 2 micro second

Current, i = q / t

i = 10 / (2 x 10^-6) = 5 x 10^6 A

(a)

distance, d = 1 m

the formula for the magnetic field is given by

B = \frac{\mu_{0}}{4\pi}\frac{2i}{d}

B = 10^{-7}\frac{2\times 5\times 10^{6}}{1}

B = 1 Tesla

Now the distance is d' = 1 km = 1000 m

B' = \frac{\mu_{0}}{4\pi}\frac{2i}{d}

B' = 10^{-7}\frac{2\times 5\times 10^{6}}{1000}

B' = 0.001 Tesla

(b) The magnetic field of earth is Bo = 3 x 10^-5 tesla

B / Bo = 3.3 x 10^4

B'/Bo = 33.3

4 0
3 years ago
A fireboat is to fight fires at coastal areas by drawing seawater with a density of 1030 kg/m3 through a 10-cm-diameter pipe at
GaryK [48]

Answer:

50.93 m/s

199.5 kW

Explanation:

From the question, the nozzle exit diameter = 5 cm, Radius= diameter/2= 5cm/2= 2.5cm. we can convert it to metre for unit consistency= (2.5×0.01)=

0.025m

We can calculate the The cross sectional area of the nozzle as

A= πr^2

A= π ×0.025^2

= 1.9635 ×10^- ³ m²

From the question, the water is moving through the pipe at a rate of 0.1 m /s , then for the water to move through it at a seconds, it must move at

(0.1 / 1.9635 ×10^- ³ m²)

= 50.93 m/s

During the Operation of the pump, the Dynamic energy of the water= potential energy provided there is no loss during the Operation

mgh = 1/2mv²

We can make "h" subject of the formula, which is the height of required head of water

h = (1/2mv²)/mg

h= v² / 2g

h = 50.93² / (2 ×9.81)

h = 132.21m

From the question;

The total irreversible head loss of the system = 3 m,

the given position of nozzle = 3 m

the total head the pump needed=(The total irreversible head loss of the system + the position of the nozzle + required head of water )

=(3 + 3 + 132.21m)

=138.21m

mass of water pumped in a seconds can be calculated since we know that mass is a product of volume and density

Volume= 0.1m³

Density of sea water=1030 kg/m

(0.1 m^3× 1030)

= 103kg

We can calculate the Potential enegry, which is = mgh

= (103 ×9.81 × 138.21)

= 139651.5 Watts

= 139.65kW

To determine required shaft power input to the pump and the water discharge velocity

Energy= efficiency × power

But we are given efficiency of 70 percent, then

139651.5 Watts = 0.7P

=199502.18 Watts

P=199.5 kW

Therefore, the required shaft power input to the pump and the water discharge velocity is 199.5 kW

5 0
3 years ago
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