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3 years ago

Someone be a kind soul and help me

Mathematics

1 answer:

8_murik_8 [283]3 years ago

7 0

Answer:

yes

Step-by-step explanation:

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Please help me thanks so much

Aleksandr [31]

I KILOGRAM IN GRAMS IS 1,000 grams, SO 7.4 TIMES 1000 IS 7,400

7 0

2 years ago

Read 2 more answers

Sketch the angle in standard position and draw an arrow representing the correct amount of rotation. Find the measure of two oth

ser-zykov [4K]

Answer:

The coterminal angles are : 270° , -450°

sketched angle is in third quadrant

Step-by-step explanation:

The sketch of the angle in standard position is attached below

The coterminal angles :

- 90 + 360 = 270°

-90 - 360 = -450°

Quadrant of the angle( -90° ) = Third quadrant

6 0

3 years ago

Suppose you have a light bulb that emits 50 watts of visible light. (Note: This is not the case for a standard 50-watt light bul

natali 33 [55]

Answer:

0.049168726 light-years

Step-by-step explanation:

The apparent brightness of a star is

$\bf B=\displaystyle\frac{L}{4\pi d^2}$

where

L = luminosity of the star (related to the Sun)

d = distance in ly (light-years)

The luminosity of Alpha Centauri A is 1.519 and its distance is 4.37 ly.

Hence the apparent brightness of Alpha Centauri A is

$\bf B=\displaystyle\frac{1.519}{4\pi (4.37)^2}=0.006329728$

According to the inverse square law for light intensity

$\bf \displaystyle\frac{I_1}{I_2}=\displaystyle\frac{d_2^2}{d_1^2}$

where

$\bf I_1=$ light intensity at distance $\bf d_1$

$\bf I_2=$ light intensity at distance $\bf d_2$

Let $\bf d_2$ be the distance we would have to place the 50-watt bulb, then replacing in the formula

$\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}$