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Naily [24]
3 years ago
6

When a rubber band is pulled back on your finger but not yet let go how is that potential energy?

Physics
1 answer:
loris [4]3 years ago
4 0
It is potential energy because the band is not in movement, th band has the potential to move.
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What is the work of a 520n student walking a distance of 3 m
Wewaii [24]

Answer:

W=1560 Joules

Explanation:

Use W=fd

so, W=520N(3m)= 1560J

8 0
3 years ago
An atom emits a light wave with a wavelength of 449 nm. what type or color of light does this represent?
Vinvika [58]
The visible spectrum ranges from 390 nm to 700 nm. the visible spectrum consist of the red ( 620 - 750 nm ) , orange ( 590 - 620 nm ) , yellow ( 570 - 590 nm ) , green ( 495 - 570 nm ) , blue ( 450 - 495 nm ) and violet ( 380 - 450 nm ) so the wave length 449 nm will produce a violet color
4 0
3 years ago
What is the half-reaction that occurs at the cathode during the electrolysis of molten potassium bromide?
polet [3.4K]

The complete ionization of KBr into its constituents is:<span>
<span>KBr (s)  --->  K+ (aq)  +  Br- (aq)</span></span>

<span>
During electrolysis, oxidation takes place at the anode electrode. This means that an ion is stripped off its electron hence becoming more positive:
<span>2 Br- (aq)  --->  Br2 (g) + 2e- </span></span>

We can see that Bromine gas Br2 is evolved at the anode. 

<span>
<span>Meanwhile at the cathode, the reduction reaction occurs. Which means that the electron from the anode electrode is used to make an ion more negative:
<span>2K+ (aq)  +  2e-  --->  2K (s) </span></span>
Hence, through reduction, solid potassium is deposited on the plate.</span>

 

 

Half reactions:

<span>Anode: 2 Br- (aq)  --->  Br2 (g) + 2e- </span>                       

<span>Cathode: 2K+ (aq)  +  2e-  --->  2K (s) </span>

7 0
3 years ago
How do clouds become stars ?
kenny6666 [7]
Stars form inside relatively dense concenstrations of interstellar gas and dust known as molecular clouds.





hope it helps
3 0
3 years ago
Read 2 more answers
The greatest height reported for a jump into an airbag is 99.4 m by stuntman Dan Koko. In 1948 he jumped from rest from the top
vaieri [72.5K]

Answer:

v = 44,16 m/s

Explanation:

We will fixate our reference in the starting point from where Dan jumped of, at the top of the Casino. Therefore, the displacement made when dan reached the airbag would be of y= -99,4 m viewed from our reference. We describe the motion of dan with the equation:

v_y^2 =v_0^2 +2ay

Dan jumped from the rest, that means that the initial velocity v_0=0, therefore:

 v_y^2 =2ay \rightarrow v_y = \sqrt{2ay}

Since Dan is moving in the negative axis regarding our reference point, we take the negative root of the equation.

v_y=-√(2*(-9,81 m/s^2 )*(-99,4 m) )=44,1613 m/s  v_y =- \sqrt{2*(-9,81 m/s^2)*(-99,4 m)} = 44,1613 m

So, if we don’t take the air resistance into account, Dan would have achieved an velocity of 44,16 m/s when he reached the airbag.

I hope everything was clear with my explanation. If you need anything else, let me know. Have a great day :D

7 0
3 years ago
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