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3 years ago

Complete the following analogy.

Physics

2 answers:

soldi70 [24.7K]3 years ago

4 0

Barred spiral; the middle of this galaxy resembles a bar, hence the name :)

Olenka [21]3 years ago

3 0

The answer to your question is D. barred spiral

Hope I could help! :)

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Name this from taco bell and get brainlest answer :)

lys-0071 [83]

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5 0

3 years ago

Read 2 more answers

The Heat required to raise the temp. of 20 g water from 25 C to 36 C

Dennis_Churaev [7]

What is the question

5 0

3 years ago

A bowling ball (mass = 7.2 kg, radius = 0.11 m) and a billiard ball (mass = 0.38 kg, radius = 0.028 m) may each be treated as un

Semenov [28]

Answer:

Explanation:

Given that

Mass of bowling ball M1=7.2kg

The radius of bowling ball r1=0.11m

Mass of billiard ball M2=0.38kg

The radius of the Billiard ball r2=0.028m

Gravitational constant

G=6.67×10^-11Nm²/kg²

The magnitude of their distance apart is given as

r=r1+r2

r=0.028+0.11

r=0.138m

Then, gravitational force is given as

F=GM1M2/r²

F=6.67×10^-11×7.2×0.38/0.138²

F=9.58×10^-9N

The force of attraction between the two balls is

F=9.58×10^-9N

3 0

3 years ago

How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 23.0 cm in diameter to p

lyudmila [28]

Answer:

10573375000

$216.57162\ N/C$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

r = Distance = $\dfrac{d}{2}=\dfrac{23}{2}=11.5\ cm$

E = Electric field = 1150 N/C

Electric field is given by

$E=\dfrac{kq}{r^2}\\\Rightarrow q=\dfrac{Er^2}{k}\\\Rightarrow q=\dfrac{1150\times 0.115^2}{8.99\times 10^9}\\\Rightarrow q=1.69174\times 10^{-9}\ C$

Number of electrons is given by

$n=\dfrac{1.69174\times 10^{-9}}{1.6\times 10^{-19}}\\\Rightarrow n=10573375000$

Number of excess electrons is 10573375000

r = 0.115+0.15 = 0.265 m

$E=\dfrac{kq}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 1.69174\times 10^{-9}}{0.265^2}\\\Rightarrow E=216.57162\ N/C$

The electric field is $216.57162\ N/C$

4 0

3 years ago

What is the current through an electric heater with a resistance of 38  when the potential difference is 240 V?

Zinaida [17]

$\bf{ \underline{Given:- }}$

$\sf• \: Resistance \: (R) \: = 38 \: Ω$

$• \sf \: Potential \: difference \: (V) \: = 240 \: v$

$\bf{ \underline{To \: Find:- }}$

$• \sf \: The \: current \: through \: an \: electric \: heater.$

<h2> $\bf{ \underline{ Solution :-}}$ </h2>

$\bf \red{•\: The \: formula \: of \: Current \: (I) = \frac{V}{R}}$

$\sf \rightarrow I = \frac{240}{38}$

$\sf \rightarrow I = 6.32$

$\sf \pink{Answer :- \: The \: current \: through \: an \: electric \: heater \: is \: 6.32 \: A.}$

6 0

3 years ago