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BaLLatris [955]

3 years ago

9

Which component of a galaxy is found in the space between the stars and gives the appearance of bright, colorful clouds?(2 point

s)
1. Clouds
2. Gas
3. Object
4. Star

2 answers:

DerKrebs [107]3 years ago

4 0

Answer:

stars

Explanation:

fgiga [73]3 years ago

3 0

Star

. ....................

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Identify the basic structure of a chemical equation

guajiro [1.7K]

The basic structure would be:

Reactants → Products

6 0

3 years ago

Three resistors, 21 Ω, 58 Ω, and 64 Ω, are connected in series, and a 0.50-A current passes through them. What are (a) the equiv

adoni [48]

<h2>a) Equivalent resistance is 143 Ω</h2><h2>b) Potential difference is 71.5 V</h2>

Explanation:

When resistors are connected in series, effective resistance is given by

$R_{eff}=R_1+R_2+R_3+......$

Here

R₁ = 21Ω

R₂ = 58Ω

R₃ = 64Ω

a) $R_{eff}=R_1+R_2+R_3\\\\R_{eff}=21+58+64\\\\R_{eff}=143\Omega$

Equivalent resistance is 143 Ω

b) We know

Potential difference = Current x Resistance

V = IR

I = 0.5 A

R = 143Ω

Substituting

V = 0.5 x 143 = 71.5 V

Potential difference is 71.5 V

6 0

3 years ago

A set of statements that belong together as a group and contribute to the function definition is known as a(n) ________.

dsp73

Block. hope this helps.

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3 years ago

How much time does it take a rhino running at 40 m/s to run 250 m?

lions [1.4K]

Answer:

Rhinoceroses are odd-toed ungulates native to sub-Saharan Africa and southern Asia, though all five living species have hugely contracted in range and number due to the influence of humans. Despite their titanic, tank-like bulk, rhinos can be amazingly swift: The fastest may reach at least 50 kilometers per hour (31 mph).

Explanation:

sana makatulong

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3 years ago

You’ve made the finals of the science Olympics. As one of your tasks you’re given 1.0 g of copper and asked to make a cylindrica

Pani-rosa [81]

Answer:

Length = 2.92 m

Diameter = 0.11 mm

Explanation:

We have $m = dl D \ \ \& \ \ \ R = \frac{\rho l}{A}$ , where:

$l$ is the length

$m = 1.0 g = 1 \times 10^{-3} \ kg\\R = 1.3 \ \Omega\\\rho = 1.7 \times 10^{-8} \Omega m\\d = 8.96 \ g/cm^3 = 8960 kg/m^3$

We divide the first equation by the second equation to get:

$\frac{m}{R} = \frac{d A^2}{\rho}$

$A^2 = \frac{m \rho}{dR} \\\\A^2 = \frac { 1 \times 10^{-3} \times 1.7 \times 10^{-8}}{8960 \times 1.3}\\\\A^2 = 1.5 \times 10^{-15}\\\\ A= 3.8 \times 10^{-8} \ m^2$

Using this Area, we find the diameter of the wire:

$D = \sqrt{\frac{4A}{\pi}}$

$D = \sqrt{\frac{4 \times 3.8 \times 10^{-8} }{\pi}}$

$D = 0.00011 \ m = 1.1 \times 10^ {-4} = 0.11 \ mm$

To find the length, we multiply the two equations stated initially:

$mR = d\rho l^2\\\\l^2 = \frac{mR}{d\rho} \\\l^2 = \frac {1.0 \times 10^{-3} \times 1.3}{8960 \times 1.7\times 10^{-8}}$

$l^2 = 8.534\\l = 2.92 \ m$

8 0

3 years ago

Read 2 more answers