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1 year ago

10. Put the following in order of increasing size:

Physics

1 answer:

Blababa [14]1 year ago

3 0

In the increasing size order, they are Universe, Andromeda galaxy, solar system, Arcturus, Sun, Earth.

The universe, cosmos contains all things. It has countless billions of galaxies.
The Andromeda Galaxy, together with the Triangulum Galaxy and about 30 other smaller galaxies, is the largest galaxy in the Local Group, which also contains the Milky Way.
A solar system comprises a star and all its planets, asteroids, comets, and other celestial bodies. It is considerably bigger in the universe, cosmos than a star.
Arcturus is a huge red star in the Northern Hemisphere of the Earth's atmosphere. When the sun has evolved into a red giant, it will look like Arcturus. Arcturus has a diameter that is roughly 25 times greater than the sun and earth.

Learn more about the universe here:

brainly.com/question/8705671

#SPJ9

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What was created 300 years ago by scientist carolus linnaeus

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<span>the classification system</span>

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2 years ago

Which type of line is shown on the map

emmainna [20.7K]

Answer:

Explanation:

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3 years ago

A water discharge of 9 m3/s is to flow through this horizontal pipe, which is 0.98 m in diameter. If the head loss is given as 1

Mashutka [201]

Answer: The power required by the pump to produce a discharge of 9m³/s is 5990joules/secs.

Explanation: The given parameters from the questions are:

Flow rate Q = 9m³/s, Diameter D = 0.98m, acceleration a = 1.0, head loss(Pressure P) is given by the function 10v²/2g.

STEP 1. Find the velocity of water in the pipe from the equation:

Diameter D = (√4.Q/π.v), where v is the velocity, and Q is flow rate

Making v subject of the formula gives:

v = 4Q/π.√D =[ 4 × 9m³/s / 3.142 × (√0.98m)] = 11.69m/s.

STEP 2. Find the pressure from the relationship, P = 10v²/2g, NB. g = a

P = 10 × (11.69m/s)² / 2× 1.0m/s²

P = 683.25N/m² or Pascal.

STEP 3. Find force exerted by the pump;

Recall that Pressure P = Force/Area

But Area A = π.r², where r = D/2

Therefore, A = π.(D/2)²

A = 3.142 × [0.98m/2]² = 0.75m²

Therefore, Force = Pressure × Area

Force F = 683.25N/m² × 0.75m²

F = 512.44N.

STEP 4. Find work done

Work done W by the pump is = Force × distance d moved by the water

W = F . d

Also recall that flow rate Q = Velocity/time.

Q = v/t, we can write t = v/Q.

Time t = 11.69m/s / 9m³/s = 1.298s

Also recall that velocity v = distance d/time t, v = d/t, making d subject of formula gives v × t

Distance d = v × t = 11.69m/s × 1.298s = 15.17m.

Hence,

Work Done W = Force × distance

W = 512.44N × 15.17m = 7775.56Nm or joules.

Lastly, Power P = Work done/ time

P = 7775.56joules/1.298s

P = 5990.4joules/s.

8 0

2 years ago

A capacitor is not the most efficient device for storing energy. Batteries can store more energy in much less space. For example

photoshop1234 [79]

Answer:

$A=2.49\times 10^6\ m^2$

Explanation:

Given that

Stored energy E

$U=10^6\ J$

We know that stored energy in capacitor given as

$U=\dfrac{1}{2}CV^2$

Given that

$V=10^4\ V$

$U=\dfrac{1}{2}CV^2$

$10^6=\dfrac{1}{2}\times C\times (10^4)^2$

C= 0.02 F

Electric filed E = 9 x 10^6 V/m

We know that

V = E .d

$10^4=9\times 10^6\times d$

d=1.11 mm

We know that

$C=\dfrac{\varepsilon _oA}{d}$

$0.02=\dfrac{8.89\times 10^{-12}A}{1.11\times 10^{-3}}$

$A=2.49\times 10^6\ m^2$

This is area area of the plates.

5 0

3 years ago

If the specific surface energy for magnesium oxide is 1.0 J/m2 and its modulus of elasticity is (225 GPa), compute the critical

alexdok [17]

Answer:

The critical stress required for the propagation of an initial crack $\sigma_{c}$ = 21.84 M pa

Explanation:

Given data

Modulus of elasticity E = 225 × $10^{9}$ $\frac{N}{m^{2} }$

Specific surface energy for magnesium oxide is $\gamma_{s}$ = 1 $\frac{J}{m^{2} }$

Crack length (a) = 0.3 mm = 0.0003 m

Critical stress is given by $\sigma_{c}^{2} }$ = $\frac{2 E \gamma}{\pi a}$ -------- (1)

⇒ 2 E $\gamma_{s}$ = 2 × 225 × $10^{9}$ × 1 = 450 × $10^{9}$

⇒ $\pi$ a = 3.14 × 0.0003 = 0.000942

⇒ Put these values in equation 1 we get

⇒ $\sigma_{c}^{2} }$ = $\frac{450 }{0.000942} 10^{9}$

⇒ $\sigma_{c}^{2} }$ = 4.77 × $10^{14}$

⇒ $\sigma_{c}$ = 2.184 × $10^{7}$ $\frac{N}{m^{2} }$

⇒ $\sigma_{c}$ = 21.84 $\frac{N}{mm^{2} }$

⇒ $\sigma_{c}$ = 21.84 M pa

This is the critical stress required for the propagation of an initial crack.

4 0

3 years ago