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Rudik [331]
2 years ago
6

What is the angular momentum of a 0.25 kg mass rotating on the end of a piece of

Physics
1 answer:
finlep [7]2 years ago
4 0

L = r x p = rmv = mr²ω

L = 0.25 x 0.75² x 12.5 = 1.758

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How much potential energy does a 1kg mess have 10m off the ground?
777dan777 [17]
Potential Energy = mass x gravitational acceleration x height
potential Energy = 1 x 9.8 x 10 = 98 joules
3 0
3 years ago
What is the power of 10 when 0.00503 is written in scientific notation?
svetoff [14.1K]

Answer:

Negative 3

Explanation:

Bc scientific notation is the zeros either ahead or behind the actual numbers

4 0
3 years ago
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A car moves with a speed of 30 metres per second calculate the distance travelled in 30 seconds
Leni [432]

30x30=900

The answer is 900 meters after 30 seconds

7 0
2 years ago
Suppose light from a 632.8 nm helium-neon laser shines through a diffraction grating ruled at 520 lines/mm. How many bright line
Leya [2.2K]

Answer:

1 bright fringe every 33 cm.

Explanation:

The formula to calculate the position of the m-th order brigh line (constructive interference) produced by diffraction of light through a diffraction grating is:

y=\frac{m\lambda D}{d}

where

m is the order of the maximum

\lambda is the wavelength of the light

D is the distance of the screen

d is the separation between two adjacent slit

Here we have:

\lambda=632.8 nm = 632.8\cdot 10^{-9} m is the wavelength of the light

D = 1 m is the distance of the screen (not given in the problem, so we assume it to be 1 meter)

n=520 lines/mm is the number of lines per mm, so the spacing between two lines is

d=\frac{1}{n}=\frac{1}{520}=1.92\cdot 10^{-3} mm = 1.92\cdot 10^{-6} m

Therefore, substituting m = 1, we find:

y=\frac{(632.8\cdot 10^{-9})(1)}{1.92\cdot 10^{-6}}=0.330 m

So, on the distant screen, there is 1 bright fringe every 33 cm.

6 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
2 years ago
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