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3 years ago

The light of that star actually gives off

2 answers:

7 0

-I believe the star gives off energy-, With most stars, like our sun, hydrogen is being converted into Helium, a process which gives off energy that heats the star.

Minchanka [31]3 years ago

5 0

Absolute magnitude - The light that a star actually gives off

Apparent magnitude - The light seen from Earth

You might be interested in

The Statue of Liberty is made of copper that has turned green because it had undergone a change. What can be said about this cha

Sergeeva-Olga [200]

D. because the statue has a new chemical called patina. It is no longer copper.

3 0

2 years ago

Read 2 more answers

How much time would be saved if the delicious chicken breast were thawed on the kitchen counter instead, given that room tempera

Art [367]

The time that would be saved if the delicious chicken breast were thawed on the kitchen counter instead, given that room temperature is around 69 F could be calculated by :
tf - ti = 5.7

hope this helps

5 0

3 years ago

A hot air balloon is moving vertically upwards at a velocity of 3m/s. A sandbag is dropped when the balloon reaches 150m. How lo

gregori [183]

This is a perfect opportunity to stuff all that data into the general equation for the height of an object that has some initial height, and some initial velocity, when it is dropped into free fall.

   H(t) = (H₀) + (v₀ T) + (1/2 a T²)

Height at any time 'T' after the drop =

(initial height) +
      (initial velocity) x (T) +
   (1/2) x (acceleration) x (T²) .

For the balloon problem ...

-- We have both directions involved here, so we have to define them:

   Upward = the positive direction

   Initial height = +150 m
   Initial velocity = + 3 m/s

   Downward = the negative direction

   Acceleration (of gravity) = -9.8 m/s²

Height when the bag hits the ground = 0 .

   H(t) = (H₀) + (v₀ T) + (1/2 a T²)

   0 = (150m) + (3m/s T) + (1/2 x -9.8 m/s² x T²)

   -4.9 T² + 3T + 150 = 0

Use the quadratic equation:

   T = (-1/9.8) [ -3 plus or minus √(9 + 2940) ]

   = (-1/9.8) [ -3 plus or minus 54.305 ]

   = (-1/9.8) [ 51.305 or -57.305 ]

      T = -5.235 seconds or 5.847 seconds .

(The first solution means that the path of the sandbag is part of
the same path that it would have had if it were launched from the
ground 5.235 seconds before it was actually dropped from balloon
while ascending.)

Concerning the maximum height ... I don't know right now any other
easy way to do that part without differentiating the big equation.
So I hope you've been introduced to a little bit of calculus.

H(t) = (H₀) + (v₀ T) + (1/2 a T²)

   H'(t) = v₀ + a T

The extremes of 'H' (height) correspond to points where h'(t) = 0 .

Set    v₀ + a T = 0

+3 - 9.8 T = 0

Add 9.8 to each side: 3   = 9.8 T

Divide each side by 9.8 :   T = 0.306 second

That's the time after the drop when the bag reaches its max altitude.

Oh gosh ! I could have found that without differentiating.

- The bag is released while moving UP at 3 m/s .

- Gravity adds 9.8 m/s of downward speed to that every second.
So the bag reaches the top of its arc, runs out of gas, and starts
falling, after
   (3 / 9.8) = 0.306 second .

At the beginning of that time, it's moving up at 3 m/s.
At the end of that time, it's moving with zero vertical speed).
Average speed during that 0.306 second = (1/2) (3 + 0) = 1.5 m/s .

Distance climbed during that time = (average speed) x (time)

   = (1.5 m/s) x (0.306 sec)

   = 0.459 meter (hardly any at all)

   But it was already up there at 150 m when it was released.

It climbs an additional 0.459 meter, topping out at 150.459 m,
then turns and begins to plummet earthward, where it plummets
to its ultimate final 'plop' precisely 5.847 seconds after its release.

We can only hope and pray that there's nobody standing at
Ground Zero at the instant of the plop.

I would indeed be remiss if were to neglect, in conclusion,
to express my profound gratitude for the bounty of 5 points
that I shall reap from this work. The moldy crust and tepid
cloudy water have been delicious, and will not soon be forgotten.

6 0

3 years ago

"Determine the magnitude of the net force of gravity acting on the Moon during an eclipse when it is directly between Earth and

spayn [35]

Answer:

Net force = 2.3686 × 10^(20) N

Explanation:

To solve this, we have to find the force of the earth acting on the moon and the force of the sun acting on the moon and find the difference.

Now, from standards;

Mass of earth;M_e = 5.98 × 10^(24) kg

Mass of moon;M_m = 7.36 × 10^(22) kg

Mass of sun;M_s = 1.99 × 10^(30) kg

Distance between the sun and earth;d_se = 1.5 × 10^(11) m

Distance between moon and earth;d_em = 3.84 × 10^(8) m

Distance between sun and moon;d_sm = (1.5 × 10^(11)) - (3.84 × 10^(8)) = 1496.96 × 10^(8) m

Gravitational constant;G = 6.67 × 10^(-11) Nm²/kg²

Now formula for gravitational force between the earth and the moon is;

F_em = (G × M_e × M_m)/(d_em)²

Plugging in relevant values, we have;

F_em = (6.67 × 10^(-11) × 5.98 × 10^(24) × 7.36 × 10^(22))/(3.84 × 10^(8))²

F_em = 1.9909 × 10^(20) N

Similarly, formula for gravitational force between the sun and moon is;

F_sm = (G × M_s × M_m)/(d_sm)²

Plugging in relevant values, we have;

F_se = (6.67 × 10^(-11) × 1.99 × 10^(30) ×

7.36 × 10^(22))/(1496.96 × 10^(8))²

F_se = 4.3595 × 10^(20) N

Thus, net force = F_se - F_em

Net force = (4.3595 × 10^(20) N) - (1.9909 × 10^(20) N) = 2.3686 × 10^(20) N

8 0

2 years ago

Is -3 a solution to equation 3x -5 = 4+ 2x

Elodia [21]

Answer:

No!

Explanation:

The correct answer is x = 9.

5 0

3 years ago