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3 years ago

The distance from Earth to the star Epsilon Eridani is about 10.5 light years. Which of the following statements is true?

Physics

2 answers:

MaRussiya [10]3 years ago

5 0

Answer:

B. It takes a ray of light 10.5 years to travel from Earth to Epsilon Eridani.

Explanation:

Light year is the unit of the distance between two points. It is defined as the distance traveled by light from one point to other for given interval of time.

So here if the distance is given as 10.5 light years then it means the distance between Earth and the given star is same as that the distance traveled by the light in 10.5 years.

So here we can say that distance is

$d = 3\times 10^8 (10.5 \times 365\times 24 \times 3600)$

$d = 9.93 \times 10^{16} m$

so here the correct answer will be

B. It takes a ray of light 10.5 years to travel from Earth to Epsilon Eridani.

Simora [160]3 years ago

4 0

<span>The correct answer is B. - It would take a ray of light 10.5 light years to travel from Earth to Epsilon Eridani, or vice-versa. Using our current technology it would take far longer than 21.0 years for a space ship from Earth to travel that far - I would have to guess many hundreds of years.</span>

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What is the main determining factor in defining boundaries between layers of earth's atmosphere?

konstantin123 [22]

The main determining factor in defining boundaries between layers of earth's atmosphere would be temperature changes in these layers. Temperature is one essential property that varies in the atmosphere. Based from this variation, the atmosphere is divided into four major layers and further to three smaller layers - troposphere, tropopause, the stratosphere, stratopause, the mesosphere, mesopause, and the thermosphere.The troposphere is the layer that is nearest to the surface of the Earth. It is the part where humans, plants and animals survive. Also, it is the warmest layer of the atmosphere. And as we go higher the atmosphere, the temperature would drop making it much cooler.

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2 years ago

Read 2 more answers

Which of these is true about kinetic energy but not necessarily true about potential energy

cestrela7 [59]

Kinetic energy is never negative, but potential energy can be.

Potential energy depends on height above some reference level,
and you can pick any level you want as the reference. So, if the
object is below the reference level you pick, then its potential
energy relative to your reference level is negative.

What that means is: You have to lift it / do work on it / give it more
energy than it has now ... in order to move it to the reference level.

(That's exactly the situation with electrons bound to an atom. Their
energy is considered negative, because we have to do work and
give them more energy to rip them away from the atom.)
_____________________________________

Regarding the other choices:

-- Kinetic energy is scalar ... Yes. So is potential energy.

-- Kinetic energy increases with height ...
No. It doesn't, but potential energy does.

-- Kinetic energy depends on position ...
No. It doesn't, but potential energy does.

3 0

2 years ago

A 3200-lb car is moving at 64 ft/s down a 30-degree grade when it runs out of fuel. Find its velocity after that if friction exe

Karolina [17]

Answer:

The velocity is 40 ft/sec.

Explanation:

Given that,

Force = 3200 lb

Angle = 30°

Speed = 64 ft/s

The resistive force with magnitude proportional to the square of the speed,

$F_{r}=kv^2$

Where, k = 1 lb s²/ft²

We need to calculate the velocity

Using balance equation

$F\sin\theta-F_{r}=m\dfrac{d^2v}{dt^2}$

Put the value into the formula

$3200\sin 30-kv^2=m\dfrac{d^2v}{dt^2}$

Put the value of k

$3200\times\dfrac{1}{2}-v^2=m\dfrac{d^2v}{dt^2}$

$1600-v^2=m\dfrac{d^2v}{dt^2}$

At terminal velocity $\dfrac{d^2v}{dt^2}=0$

So, $1600-v^2=0$

$v=\sqrt{1600}$

$v=40\ ft/sec$

Hence, The velocity is 40 ft/sec.

4 0

3 years ago

The diagram below shows the velocity vectors for two cars that are moving relative to each other.

scoundrel [369]

Answer:

The answer is " $5 \ \frac{m}{s} \ west$ "

Explanation:

$\to \vec{V_1} = (25 \frac{m}{s}) (\hat{-i})\\\\\to \vec{V_2} = (20 \frac{m}{s}) (\hat{-i})\\\\$

velocity of car | respect to car :

$\to \vec{V_{12}} = \vec{V_1} - \vec{V_2}\\\\$

$=\vec{-25} \hat{i}+ \vec{20} \hat{i}\\\\= 5 \ \frac{m}{s} \ west$

7 0

2 years ago

three girls were pushing the same car with a net force of 450 N [N48°E]. Two of the girls were pushing with forces of 310 N [N25

ElenaW [278]

The net force is the vector

∑ F = (450 N) (cos(42°) i + sin(42°) j)

and two of the forces provided by the girls are

F₁ = (310 N) (cos(115°) i + sin(115°) j)

F₂ = (250 N) (cos(285°) i + sin(285°) j)

Then the force provided by the third girl is the vector

F₃ = ∑ F - F₁ - F₂

F₃ = ((450 N) cos(42°) - (310 N) cos(115°) - (250 N) cos(285°)) i

… … … + ((450 N) sin(42°) - (310 N) sin(115°) - (250 N) sin(285°)) j

F₃ ≈ (400.722 N) i + (261.635 N) j

So, the third girl provided a force of magnitude

||F₃|| = √((400.722 N)² + (261.635 N)²) ≈ 478.572 N ≈ 480 N

pointing in a direction

arctan((261.635 N)/(400.722 N)) ≈ 33.1409° ≈ 33°

relative to East which refers to 0°; that is, 33° N of E or E33°N. Since the other forces are given relative to North or South, we can write this direction as N57°E.

So, the third girl pushed with force 480 N [N57°E].

5 0

2 years ago