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3 years ago

11

Write an essay for me pls about Proxima Centaurithis will help a lot

1 answer:

lawyer [7]3 years ago

4 0

Answer:

Proxima Centauri is a small, low-mass star located 4.244 light-years (1.301 pc) away from the Sun in the southern constellation of Centaurus. Its Latin name means the "nearest [star] of Centaurus". This object was discovered in 1915 by Robert Innes and is the nearest-known star to the Sun.

Star system: Alpha Centauri

Category: Star, Flare star

Constellation: Centaurus

Discoverer: Robert T. A. Innes

Explanation:

u should make it into an essay

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A 0.85 N force exists between a 7.1 * 10 ^ - 6 * C charge 5.4 m away. What is the magnitude of the second charge ? Please show w

katovenus [111]

Answer:

Explanation:

Force between charge is given by the following expression

F = k Q₁ Q₂ / R² , k = 9 x 10⁹ , Q₁ and Q₂ are charges , R is distance between charges .

Putting the given values ,

.85 = 9 x 10⁹ x 7.1 x 10⁻⁶ x Q₂ / 5.4²

Q₂ = .85 x 5.4² / (9 x 10⁹ x 7.1 x 10⁻⁶ )

= .38788 x 10⁻³ C .

= 387.88 x 10⁻⁶ C .

4 0

3 years ago

A 34.0 %-efficient electric power plant produces 800 MW of electric power and discharges waste heat into 20∘C ocean water. Suppo

mojhsa [17]

Answer:

77647

Explanation:

$\eta$ = Efficiency = 34%

Power used in 1 home = 0.02 MW

Total power is

$P=\dfrac{800}{\eta}\\\Rightarrow P=\dfrac{800}{0.34}\\\Rightarrow P=2352.94117\ MW$

Waste of power

$2352.94117-800=1552.94117\ W$

Number of homes would be given by

$n=\dfrac{1552.94117}{0.02}=77647.0585\ homes$

The number of homes that could be heated with the waste heat of this one power plant is 77647

8 0

3 years ago

Look at the variables for this lab. Which variable would we have to change in order to change the amount of current flowing thro

____ [38]

Where are the variables from the lab?

3 0

2 years ago

A horizontal sheet of negative charge has a uniform electric field E = 3000N/C. Calculate the electric potential at a point 0.7m

vesna_86 [32]

Answer:

Electric potential, E = 2100 volts

Explanation:

Given that,

Electric field, E = 3000 N/C

We need to find the electric potential at a point 0.7 m above the surface, d = 0.7 m

The electric potential is given by :

$V=E\times d$

$V=3000\ N/C\times 0.7\ m$

V = 2100 volts

So, the electric potential at a point 0.7 m above the surface is 2100 volts. Hence, this is the required solution.

6 0

3 years ago

A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.

ivann1987 [24]

Answer:

v₂ = 176.24 m/s

Explanation:

given,

angle of projectile = 45°

speed = v₁ = 150 m/s

for second trail

speed = v₂ = ?

angle of projectile = 37°

maximum height attained formula,

$H_{max}= \dfrac{v^2 sin^2(\theta)}{g}$

now,

$H_{max}= \dfrac{v_1^2 sin^2(\theta_1)}{g}$

$H_{max}= \dfrac{v_2^2 sin^2(\theta_2)}{g}$

now, equating both the equations

$\dfrac{v_2^2}{v_1^2}=\dfrac{sin^2(\theta_1)}{sin^2(\theta_2)}$

$\dfrac{v_2^2}{150^2}=\dfrac{sin^2(45^0)}{sin^2(37^0)}$

v₂² = 31061.79

v₂ = 176.24 m/s

velocity of projectile would be equal to v₂ = 176.24 m/s

8 0

3 years ago