All categories

3 years ago

Differences Between light year and astronomical unit in two points .

1 answer:

5 0

One astronomical unit (AU) is the average distance from the Earth to the Sun, which is about 93-million miles or about 8.5 light-minutes.

A light-year is the distance light travels in one year in a vacuum, which is about 6-trillion miles. One light-year is about 64,500 times longer than one AU.

You might be interested in

Mrs. Buckley asked her physical science class to look at the graph and tell which of the following explanations are most likely

iren [92.7K]

I think it's D. Options C and D are true. The graph shows an increase in speed from points 6 to 12, which means Option C is true. It shows an even bigger increase in speed from points 24 to 30, and since bikes go faster when traveling downhill, I would think that Option D was correct as well.

3 0

3 years ago

Read 2 more answers

A student squeezes several oranges to make a glass of orange juice. The juice contains pieces of orange pulp mixed with the juic

Mars2501 [29]

In chemistry, a suspension is a heterogeneous mixture containing solid particles that are sufficiently large for sedimentation. Usually they must be larger than one micrometer. A Suspension is a heterogeneous mixture in which the solute particles do not dissolve but get suspended throughout the bulk of the medium. a solution is a homogeneous mixture composed of two or more substances. In such a mixture, a solute is a substance dissolved in another substance, known as a solvent.. it is considered a solution and suspension because the orange juice mix with water while the pulp did not and retained as solid

7 0

3 years ago

Read 2 more answers

A train at a constant 79.0 km/h moves east for 27.0 min, then in a direction 50.0° east of due north for 29.0 min, and then west

ivolga24 [154]

Answer:

Magnitude of avg velocity, $|v_{avg}| = 18.9 km/h$

$\theta' = 56.85^{\circ}$

Given:

Constant speed of train, v = 79 km/h

Time taken in East direction, t = 27 min = $\frac{27}{60} h$

Angle, $\theta = 50^{\circ}$

Time taken in $50^{\circ}$ east of due North direction, t' = 29 min = $\frac{29}{60} h$

Time taken in west direction, t'' = 37 min = $\frac{27}{60} h$

Solution:

Now, the displacement, 's' in east direction is given by:

$\vec{s} = vt = 79\times \frac{27}{60} = 35.5\hat{i} km$

Displacement in $50^{\circ}$ east of due North for 29.0 min is given by:

$\vec{s'} = vt'sin50^{\circ}\hat{i} + vt'cos50^{\circ}\hat{j}$

$\vec{s'} = 79(\frac{29}{60})sin50^{\circ}\hat{i} + 79(\frac{29}{60})cos50^{\circ}\hat{j}$

$\vec{s'} = 29.25\hat{i} + 24.54\hat{j} km$

Now, displacement in the west direction for 37 min:

$\vec{s''} = - vt''hat{i} = - 79\frac{37}{60} = - 48.72\hat{i} km$

Now, the overall displacement,

$\vec{s_{net}} = \vec{s} + \vec{s'} + \vec{s''}$

$\vec{s_{net}} = 35.5\hat{i} + 29.25\hat{i} + 24.54\hat{j} - 48.72\hat{i}$

$\vec{s_{net}} = 16.03\hat{i} + 24.54\hat{j} km$

(a) Now, average velocity, $v_{avg}$ is given:

$v_{avg} = \frac{total displacement, \vec{s_{net}}}{total time, t}$

$v_{avg} = \frac{16.03\hat{i} + 24.54\hat{j}}{\frac{27 + 29 + 37}{60}}$

$v_{avg} = 10.34\hat{i} + 15.83\hat{j}) km/h$

Magnitude of avg velocity is given by:

$|v_{avg}| = \sqrt{(10.34)^{2} + (15.83)^{2}} = 18.9 km/h$

(b) angle can be calculated as:

$tan\theta' = \frac{15.83}{10.34}$

$\theta' = tan^{- 1}\frac{15.83}{10.34} = 56.85^{\circ}$

6 0

3 years ago

HELP! how did michael jordan influence future basketball players to make prodcuts?

nataly862011 [7]

Answer:

By encouraging them.

Explanation:

#CARRYONLEARNING

5 0

3 years ago

Read 2 more answers

The pulley system below uses a gasoline engine to raise a drill head up through a smooth drill pipe. The engine provides a const

polet [3.4K]

NB: The diagram of the pulley system is not shown but the information provided is sufficient to answer the question

Answer:

Power = 2702.56 W

Explanation:

Let the power consumed be P

Energy expended = E = mgh

height, h = 5 m

E = 80 * 9.8 * 5

E = 3920 J

$Power = \frac{Energy}{time}$

To calculate the time, t

From F = ma

F = 900 N

900 = 80 a

a = 900/80

a = 11.25 m/s²

From the equation of motion, $s = ut + 0.5at^{2}$

The drill head starts from rest, u = 0 m/s

$5 = 0 * t + (0.5*11.25*t^{2} )\\5 = 5.625t^{2}\\t^{2} = 5/5.625\\t^{2} = 0.889\\t = 0.943 s$

Power, P = E/t

P = 3920/0.0.943

P = 4157.79 W

But Efficiency, E = 0.65

P = 0.65 * 4157.79

Power = 2702.56 W

6 0

3 years ago