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Thepotemich [5.8K]
3 years ago
5

Which parts of the spectrum show the presence of elements in the stars atmosphere

Physics
2 answers:
olga_2 [115]3 years ago
5 0

Answer:

the one right behind 600 and the one in between 600 and 700

Explanation:

Yuliya22 [10]3 years ago
3 0

the one right behing 600 and the one in between 600 and 700

"please give me brainliest"

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A force of 100 newtons is used to move an object a distance of 15 meters with a power of 25 watts. Find the
valentina_108 [34]


work is distance * force so 15*100=1500

and to find time you know power = diastance * force / time

so 25=15*100/t

25=1500/t

25/1500=t

.016=time


5 0
3 years ago
How do I do this problem?
lesya692 [45]
Add then divide the hint clearly backs it up two so yeahh
3 0
3 years ago
Un avión de rescate en Alaska deja caer un paquete de provisiones a un grupo de exploradores extraviados. Si el avión viaja hori
posledela

Answer:

180.4 m

Explanation:

The package in relation to the point where it was released falls a certain distance that is calculated by applying the horizontal motion formulas , as the horizontal speed of the plane and the height above the ground are known, the time that It takes the package to reach its destination and then the horizontal distance (x) is calculated from where it was dropped, as follows:    

$V_{ox}=v_x = 40 \ m/s$

   h = 100 m  

    x =?

     Height formula h:

     $h=g \times \frac{t^2}{2}$

      Time t is cleared:

     $t = \sqrt{\frac{2h}{g}}$

      $t = \sqrt{\frac{2 \times 100}{9.8}}$

      t = 4.51 sec

 Horizontal distance formula x:

       $x=V_x \times t$

        x = 40 m / sec x 4.51 sec

        x = 180.4 m

4 0
3 years ago
An electron enters a region with a speed of 5×10^6m/s and is slowed down at the rate of 1.25×10^-4m/s². How far does the electro
Mashutka [201]

1) The distance travelled by the electron is 1\cdot 10^{17} m

2) The time taken is 4.0\cdot 10^{10}s

Explanation:

1)

The electron in this problem is moving by uniformly accelerated motion (constant acceleration), so we can use the following suvat equation

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled

For the electron in this problem,

u=5\cdot 10^6 m/s is the initial velocity

v = 0 is the final velocity (it comes to a stop)

a=-1.25\cdot 10^{-4} m/s^2 is the acceleration

Solving for s, we find the distance travelled:

s=\frac{v^2-u^2}{2a}=\frac{0-(5\cdot 10^6)^2}{2(-1.25\cdot 10^{-4})}=1\cdot 10^{17} m

2)

The total time taken for the electron in its motion can also be found by using another suvat equation:

v=u+at

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time taken

Here we have

u=5\cdot 10^6 m/s

v = 0

a=-1.25\cdot 10^{-4} m/s^2

And solving for t, we find the time taken:

t=\frac{v-u}{a}=\frac{0-5\cdot 10^6}{-1.25\cdot 10^{-4}}=4.0\cdot 10^{10}s

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

7 0
3 years ago
A kite suspended in the sky is flowing back and forth. Which type of friction is being described?
Mice21 [21]

The type of friction of a kite suspended in the sky that is flowing back and forth is fluid friction. The fluid here is the air that helps the kite move back and forth. The kite feels a drag force due to air which acts in the upward direction.

5 0
3 years ago
Read 2 more answers
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