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3 years ago

12

A racecar drives at a constant speed down a straight track. The car is in _?_

1 answer:

vagabundo [1.1K]3 years ago

7 0

Answer:

the answer is

the car is in motion

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3 years ago

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;

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

so

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

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

Three resistors connected in series have potential differences across them labeled /\V1 , /\V2 , and /\V3. What expresses the po

Brrunno [24]

Answer:

$\Delta V=\Delta V_1+\Delta V_2+\Delta V_3$

Explanation:

We are given that three resistors R1, R2 and R3 are connected in series.

Let

Potential difference across $R_1=\Delta V_1$

Potential difference across $R_2=\Delta V_2$

Potential difference across $R_3=\Delta V_3$

We know that in series combination

Potential difference , $V=V_1+V_2+V_3$

Using the formula

$\Delta V=\Delta V_1+\Delta V_2+\Delta V_3$

Hence, this is required expression for potential difference.

3 0

3 years ago