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3 years ago

Organisms that live on or burrow into the ocean floor are

Physics

2 answers:

creativ13 [48]3 years ago

8 0

Benthos

Option b is the answer

Varvara68 [4.7K]3 years ago

3 0

The answer is B: benthos

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Suppose an electron and a proton move at the same speed. which particle has a longer de broglie wavelength? suppose an electron

BARSIC [14]

When both particles, the electron and the proton move at the same speed, they may have differences with their de Broglie wavelength, the particle that would have a longer wavelength would be the proton since the wavelength is in direct proportionality with the mass of the particle.

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3 years ago

Which of the following is most like an indicator of a chemical change

qaws [65]

Answer:

methyl orange, methyl red,phenoptalin, merhy red

Explanation:

all this following are indicators use to check the end point of a reaction

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3 years ago

What’s up world or people

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Hi how are you?

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3 years ago

Round your answers to one decimal place.this parallel circuit has two resistors at 15 and 40 ohms. what is the total resistance?

lubasha [3.4K]

1) The equivalent resistance of two resistors in parallel is given by:
$\frac{1}{R_{eq}}= \frac{1}{R_1}+ \frac{1}{R_2}$
so in our problem we have
$\frac{1}{R_{eq}} = \frac{1}{15 \Omega}+ \frac{1}{40 \Omega}=0.092 \Omega^{-1}$
and the equivalent resistance is
$R_{eq} = \frac{1}{0.092 \Omega^{-1}}=10.9 \Omega$

2) If we have a battery of 12 V connected to the circuit, the current in the circuit will be given by Ohm's law, therefore:
$I= \frac{V}{R_{eq}}= \frac{12 V}{10.9 \Omega}=1.1 A$

4 0

3 years ago

Read 2 more answers

Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ

bija089 [108]

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; $H_0 = 51km/s/Mly$

Age of the universe; $t = \ ?$

We know that, the reciprocal of the Hubble's constant ( $H_0$ ) gives an estimate of the age of the universe ( $t$ ). It is expressed as:

$Age\ of\ Universe; t = \frac{1}{H_0}$

Now,

Hubble's constant; $H_0 = 51km/s/Mly$

We know that;

$1\ light\ years = 9.46*10^{15}m$

$1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m$

Therefore;

$H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly} \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 = 5.39 *10^{-18}s^{-1}\\$

Now, we input this Hubble's constant value into our equation;

$Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t = \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years$

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Learn more: brainly.com/question/14019680

6 0

3 years ago