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2 years ago

If the star Alpha Centauri were moved to a distance 10 times farther than it is now, its parallax angle would

Physics

1 answer:

steposvetlana [31]2 years ago

3 0

Answer:

B. get smaller

Explanation:

The parallax angle of a star measured with respect to the Earth is inversely proportional to the distance of the star from the Earth:

$\theta = \frac{1}{d}$

where

$\theta$ is the parallax angle, measured in arcsec

d is the distance between the star and the Earth, measured in parsec

Therefore, if the star Alpha Centauri is moved farther from Earth, then d increases, and therefore the parallax angle will get smaller.

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Electricity from the local power company costs 7.22 cents/kWh. A homeowner wishes to pay a maximum of 25 cents per month for the

mestny [16]

You haven't said how much power the stereo uses. It matters !

Whatever that number is, the maximum hours per month is

(3460) divided by (the # of watts the stereo uses when it's playing) .

7 0

3 years ago

Read 2 more answers

The graph illustrates the activity level of three common digestive enzymes, across a range of pH values. Which enzyme is likely

vaieri [72.5K]

Answer:

(A) Pepsin

Explanation:

From the graph it is clear that pepsin is the only enzyme which works in highly acidic condintion in the digestive system.

less than 7 the liquid is acidic
above 7 the liquid is basic
at 7 the liquid is neutral

It has an optimum pH of about 1.5 at which its activity level is 8.5 as shown in graph.

7 0

3 years ago

. Consider the equation =0+0+02/2+03/6+04/24+5/120, where s is a length and t is a time. What are the dimensions and SI units of

Olegator [25]

Answer:

See Explanation

Explanation:

Given

$s=s_0+v_0t+\frac{a_0t^2}{2}+ \frac{j_0t^3}{6}+\frac{S_0t^4}{24}+\frac{ct^5}{120}$

Solving (a): Units and dimension of $s_0$

From the question, we understand that:

$s \to L$ --- length

$t \to T$ --- time

Remove the other terms of the equation, we have:

$s=s_0$

Rewrite as:

$s_0=s$

This implies that $s_0$ has the same unit and dimension as $s$

Hence:

$s_0 \to L$ --- dimension

$s_o \to$ Length (meters, kilometers, etc.)

Solving (b): Units and dimension of $v_0$

Remove the other terms of the equation, we have:

$s=v_0t$

Rewrite as:

$v_0t = s$

Make $v_0$ the subject

$v_0 = \frac{s}{t}$

Replace s and t with their units

$v_0 = \frac{L}{T}$

$v_0 = LT^{-1}$

Hence:

$v_0 \to LT^{-1}$ --- dimension

$v_0 \to$ $m/s$ --- unit

Solving (c): Units and dimension of $a_0$

Remove the other terms of the equation, we have:

$s=\frac{a_0t^2}{2}$

Rewrite as:

$\frac{a_0t^2}{2} = s_0$

Make $a_0$ the subject

$a_0 = \frac{2s_0}{t^2}$

Replace s and t with their units [ignore all constants]

$a_0 = \frac{L}{T^2}\\$

$a_0 = LT^{-2$

Hence:

$a_0 = LT^{-2$ --- dimension

$a_0 \to$ $m/s^2$ --- acceleration

Solving (d): Units and dimension of $j_0$

Remove the other terms of the equation, we have:

$s=\frac{j_0t^3}{6}$

Rewrite as:

$\frac{j_0t^3}{6} = s$

Make $j_0$ the subject

$j_0 = \frac{6s}{t^3}$

Replace s and t with their units [Ignore all constants]

$j_0 = \frac{L}{T^3}$

$j_0 = LT^{-3}$

Hence:

$j_0 = LT^{-3}$ --- dimension

$j_0 \to$ $m/s^3$ --- unit

Solving (e): Units and dimension of $s_0$

Remove the other terms of the equation, we have:

$s=\frac{S_0t^4}{24}$

Rewrite as:

$\frac{S_0t^4}{24} = s$

Make $S_0$ the subject

$S_0 = \frac{24s}{t^4}$

Replace s and t with their units [ignore all constants]

$S_0 = \frac{L}{T^4}$

$S_0 = LT^{-4$

Hence:

$S_0 = LT^{-4$ --- dimension

$S_0 \to$ $m/s^4$ --- unit

Solving (e): Units and dimension of $c$

Ignore other terms of the equation, we have:

$s=\frac{ct^5}{120}$

Rewrite as:

$\frac{ct^5}{120} = s$

Make $c$ the subject

$c = \frac{120s}{t^5}$

Replace s and t with their units [Ignore all constants]

$c = \frac{L}{T^5}$

$c = LT^{-5}$

Hence:

$c \to LT^{-5}$ --- dimension

$c \to m/s^5$ --- units

4 0

3 years ago

Who was the first to hypothesize that electron orbit a positively charged nucleus

Eva8 [605]

I think the answer is ruthorford

6 0

3 years ago

21. A 60 kg student on a scooter is pushed with a constant force of 40 N across a horizontal concrete driveway at a constant spe

DedPeter [7]

The magnitude of the force of friction is 40 N

Explanation:

To solve the problem, we just have to analyze the forces acting on the student and the scooter along the horizontal direction. We have:

- The constant pushing force forward, of magnitude F = 40 N

- The frictional force, acting backward, $F_f$

Since the two forces are in opposite direction, the equation of motion is

$F-F_f = ma$

where

m is the mass of the student+scooter

a is the acceleration

However, here the scooter is moving at constant speed: this means that its acceleration is zero, so

a = 0

And therefore,

$F - F_f = 0\\F_f = F$

which means that the magnitude of the force of friction is also equal to 40 N.

Learn more about force of friction here:

brainly.com/question/6217246

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#LearnwithBrainly

3 0

3 years ago