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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

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.

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3 years ago

Read 2 more answers

~~~NEED HELP ASAP~~~ Please solve each section and show all work for each section.

anastassius [24]

Explanation:

Forces on Block A:

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass $m_A$ as

$x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)$

$y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)$

Substituting (2) into (1), we get

$\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)$

where $f_N= \mu_kN$ , the frictional force on $m_A.$ Set this aside for now and let's look at the forces on $m_B$

Forces on Block B:

Let the x-axis be (+) up along the inclined plane. We can write the forces on $m_B$ as

$x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)$

$y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)$

From (5), we can solve for N as

$N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)$

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension T is given by

$T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)$

Substituting (7) into (4) we get

$m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba$

Collecting similar terms together, we get

$(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag$

$a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)$

Putting in the numbers, we find that $a = 1.4\:\text{m/s}$ . To find the tension T, put the value for the acceleration into (7) and we'll get $T = 21.3\:\text{N}$ . To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get $N = 50.9\:\text{N}$