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1 year ago

Which is brighter in our sky, a star with apparent magnitude 2 or a star with apparent magnitude 7?

Physics

1 answer:

Lina20 [59]1 year ago

7 0

The star with apparent magnitude 2 is more brighter than 7.

To find the answer, we have to know about apparent magnitude.

<h3>What is apparent magnitude?</h3>

100 times as luminous as a star with an apparent brightness of 7 is a star with a magnitude of 2.
The apparent magnitude of bigger stars is always smaller.
The brightest star in the night sky is Sirius.
The brightness of a star or other celestial object perceived from Earth is measured in apparent magnitude (m).
The apparent magnitude of an object is determined by its inherent luminosity, its distance from Earth, and any light extinction brought on by interstellar dust in the path of the observer's line of sight.

Thus, we can conclude that, the star with apparent magnitude 2 is more brighter than 7.

Learn more about the apparent magnitude here:

brainly.com/question/350008

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I need a science fair project that doesn't need anything on the pic and it can be about any subject thanks.

tigry1 [53]

you could make a self propelled car all you need is cardboard, wheels, and a balloons or rubber bands

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3 years ago

James threw a ball vertically upward with a velocity of 41.67ms-1 and after 2 second David threw a ball vertically upward with a

Reptile [31]

Answer:

When have passed 3.9[s], since James threw the ball.

Explanation:

First, we analyze the ball thrown by James and we will find the final height and velocity by the time two seconds have passed.

We'll use the kinematics equations to find these two unknowns.

$y=y_{0} +v_{0} *t+\frac{1}{2} *g*t^{2} \\where:\\y= elevation [m]\\y_{0}=initial height [m]\\v_{0}= initial velocity [m/s] =41.67[m/s]\\t = time passed [s]\\g= gravity [m/s^2]=9.81[m/s^2]\\Now replacing:\\y=0+41.67 *(2)-\frac{1}{2} *(9.81)*(2)^{2} \\\\y=63.72[m]\\$

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can find the velocity after 2 seconds.

$v_{f} =v_{o} +g*t\\replacing:\\v_{f} =41.67-(9.81)*(2)\\\\v_{f}=22.05[m/s]$

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can take these values calculated as initial values, taking into account that two seconds have already passed. In this way, we can find the time, through the equations of kinematics.

$y=y_{o} +v_{o} *t-\frac{1}{2} *g*t^{2} \\y=63.72 +22.05 *t-\frac{1}{2} *(9.81)*t^{2} \\\\y=63.72 +22.05 *t-4.905*t^{2} \\$

As we can see the equation is based on Time (t).

Now we can establish with the conditions of the ball launched by David a new equation for y (elevation) in function of t, then we match these equations and find time t

$y=y_{o} +v_{o} *t+\frac{1}{2} *g*t^{2} \\where:\\v_{o} =55.56[m/s] = initial velocity\\y_{o} =0[m]\\now replacing\\63.72 +22.05 *t-(4.905)*t^{2} =0 +55.56 *t-(4.905)*t^{2} \\63.72 +22.05 *t =0 +55.56 *t\\63.72 = 33.51*t\\t=1.9[s]$

Then the time when both balls are going to be the same height will be when 2 [s] plus 1.9 [s] have passed after David throws the ball.

Time = 2 + 1.9 = 3.9[s]

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2 years ago

1. What are three appliances in your house that use a motor that turns electrical energy into mechanical energy?

AleksAgata [21]

A vacuum is an electrical motor and<span> which it converts electrical energy into mechanical energy.

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3 years ago

a boy throws a ball straight up into the air. it reaches its highest point after 4 seconds.how fast was the ball going when it l

baherus [9]

Answer:

Explanation:

The most important thing to remember about parabolic motion in physics is that when an object reaches its max height, the velocity right there at the highest point is 0. Use this one-dimensional motion equation to solve this problem:

v = v₀ + at and filling in:

0 = v₀ + (-9.8)(4.0) **I put in 4.0 for time so we have more than just 1 sig fig here**

0 = v₀ - 39 and

-v₀ = -39 so

v₀ = 39 m/s

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2 years ago

2. The substances that undergo change in a chemical reaction are called

AVprozaik [17]

Reactants

In a chemical reaction, the substances that undergo change are called reactants. The new substances formed as a result of that change are called products.

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3 years ago