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soldier1979 [14.2K]
3 years ago
9

A light-year is the distance light travels in one year (at speed = 2.998 × 108 m/s). (a) how many meters are there in 11.0 light

-years? m (b) an astronomical unit (au) is the average distance from the sun to earth, 1.50 × 108 km. how many au are there in 11.0 light-years? au (c) what is the speed of light in au/h? au/h
Physics
1 answer:
larisa [96]3 years ago
8 0
<span>The answers are as follows:

(a) how many meters are there in 11.0 light-years?

11.0 light years ( 365 days / 1 year ) ( 24 h / 1 day ) ( 60 min / 1 h ) ( 60 s / 1 min ) ( 2.998x10^8 m/s ) = 1.04x10^17 m

(b) an astronomical unit (au) is the average distance from the sun to earth, 1.50 × 108 km. how many au are there in 11.0 light-years?

1.04x10^17 m ( 1 au / </span>1.50 × 10^8 km <span>) ( 1 km / 1000 m) = 693329.472 au

(c) what is the speed of light in au/h? au/h

</span>2.998 × 10^8 m/s ( 1 au / 1.50 × 10^8 km ) ( 1 km / 1000 m) ( 3600 s / 1 h ) = 7.1952 au/h

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a

Explanation:

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Answer:

This question is asking to identify the following variables:

Independent variable (IV): TYPE OF SOIL

Dependent variable (DV): HEIGHT AND NUMBER OF LEAVES

Control group: None in this experiment

Constant: SAME ROSE PLANT, SAME TIME INTERVAL (1 WEEK)

Explanation:

Independent variable in an experiment is the variable that is manipulated or changed by the experimenter in order to effect a measurable outcome. In this case, the independent variable is the TYPE OF SOIL used.

Dependent variable is the measurable variable that responds to changes made to the independent variable. In this experiment, the dependent variable is the HEIGHT AND NUMBER OF LEAVES of each rose.

Constants or control variable is the variable that is kept unchanged or constant for all groups throughout the experiment. In this experiment, the constants are SAME ROSE PLANT, SAME TIME INTERVAL (1 WEEK).

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3 years ago
Calculate the electric field at the center of a square
pantera1 [17]

Answer:

E_y=1175510.2\ N.C^{-1}

The Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

Explanation:

Given:

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  • charge on one corner of the square, q_1=+45\times 10^{-6}\ C
  • charge on the remaining 3 corners of the square,q_2=q_3=q_4=-27\times 10^{-6}\ C

<u>Distance of the center from each corners</u>=\frac{1}{2} \times diagonals

diagonal=\sqrt{52.5^2+52.5^2}

diagonal=74.25\cm=0.7425\ m

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E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}

<u>For charge q_1 we have the field lines emerging out of the charge since it is positively charged:</u>

E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}

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<u>Force by each of the charges at the remaining corners:</u>

E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}

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E_y=E_1-E_4

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