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Strike441 [17]
3 years ago
7

Can someone help me with this I'm-- never mind I would just like help..

Physics
2 answers:
Alik [6]3 years ago
6 0
Silicon: Si
Iron: Fe
K: Potassium
Hg: Mercury
I: Iodine
Copper: Cu

A. Bromine
B.Beryllium
C. Gold
D. Tin

A. Alkaline Metals
B. Transition Metals
C. Reactive Nonmetals
D. Alkaline Earth Metals
serg [7]3 years ago
6 0
Dang it I love elements :(
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Một vật dao động điều hòa có phương trình là x = 4sin(πt + π/3) (cm; s). Lúc t = 0,5s vật có li độ và vận tốc là
nydimaria [60]

Answer:

0,10s

Explanation:

7 0
2 years ago
Which set of shapes contains members that are always similar to one another?
Colt1911 [192]
<span>The correct answer is option C. i.e.Equilateral triangles. All equilateral triangle have equal sides. Due to this these set of shapes will always be similar. Similarity in the sense that only the size of the triangles can change to large or smaller but the shape will be be similar always.</span>
3 0
3 years ago
A +0.05 C charge is placed in a uniform electric field pointing downward with a strength of 100 Newtons over Coulombs.. Determin
shtirl [24]

The magnitude of the electric force on the charge is 5 N.

<h3>Magnitude of force on the charge</h3>

The magnitude of force on the charge is calculated as follows;

F = Eq

where;

  • E is electric field
  • q is magnitude of the charge

F = 100 N/C  x 0.05 C

F = 5 N

Thus, the magnitude of the electric force on the charge is 5 N.

Learn more about electric force here: brainly.com/question/20880591

#SPJ1

8 0
1 year ago
A section of a parallel-plate air waveguide with a plate separation of 7.11 mm is constructed to be used at 15 GHz as an evanesc
adell [148]

Answer:

the required minimum length of the attenuator is 3.71 cm

Explanation:

Given the data in the question;

we know that;

f_{c_1 = c / 2a

where f is frequency, c is the speed of light in air and a is the plate separation distance.

we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s

plate separation distance a = 7.11 mm = 0.0711 cm

so we substitute

f_{c_1 = 3 × 10¹⁰ / 2( 0.0711  )

f_{c_1 = 3 × 10¹⁰ cm/s / 0.1422  cm

f_{c_1 =  21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }

Now, we determine the minimum wavelength required

Each non propagating mode is attenuated by at least 100 dB at 15 GHz

so

Attenuation constant TE₁ and TM₁ expression is;

∝₁ = 2πf/c × √( (f_{c_1 / f)² - 1 )

so we substitute

∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )

∝₁ = 3.1079 × 10⁻⁷

∝₁ = 310.79 np/m

Now, To find the minimum wavelength, lets consider the design constraint;

20log₁₀e^{-\alpha _1l_{min = -100dB

we substitute

20log₁₀e^{-(310.7np/m)l_{min = -100dB

l_{min = 3.71 cm

Therefore, the required minimum length of the attenuator is 3.71 cm

6 0
2 years ago
Scientific investigations
snow_tiger [21]
I don’t know what you are trying to ask complete my but here are the steps to a scientific investigation i hope this helps you

6 0
2 years ago
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