Ox:vₓ=v₀
x=v₀t
Oy:y=h-gt²/2
|vy|=gt
tgα=|vy|/vₓ=gt/v₀=>t=v₀tgα/g
y=0=>h=gt²/2=v₀²tg²α/2g=>tgα=√(2gh/v₀²)=√(2*10*20/24²)=√(400/576)=0.83=>α=tg⁻¹0.83=39°
cosα=vₓ/v=v₀/v=>v=v₀/cosα=24/cos39°=24/0,77=31.16 m/s
Ec=mv²/2=2*31.16²/2=971.47 J=>Ec≈0.97 kJ
Frequency = 1/time period = 1/0.05 = 20s^-1.
Answer:
Hello, I believe it would have a negative charge considering protons have a positive charge while elctrons have a negative charge
Explanation:
Answer: option D. the ratio of the population of male deer is not constant.
Explanation:
The bar graph permits to compare the results for two different populations: male and female deer in a very easy visual way.
These features are remarkable:
- The polulation of male deer (blue bars) decrease from 1961 to 1971, then increase in the next 10 year, decrease in the next decade, and increase for the next two decades. So, its trend is erratic, with ups and downs.
This discards the option A, which states that the population of male deer increases each decade from 1961 to 2011.
- The population of female deer (purple or brown bars) decreases every decade.
This discards the option B. which states that when the polulation of male deer increases, the poluplation of female deer also increases.
- The populations never are equal, hence this discards the option C.
- Since, one popultion increases and decreases, while the other population only decreases, you conclude that the ratio of the population of male deer to female deer is not constant, which is the option D.
Explanation:
We need to apply the conservation law of linear momentum to two dimensions:
Let 
= momentum of the 1st ball
= momentum of the 2nd ball
In the x-axis, the conservation law can be written as
or
Since we are dealing with identical balls, all the m terms cancel out so we are left with
Putting in the numbers, we get
In the y-axis, there is no initial y-component of the momentum before the collision so we can write
or
Taking the ratio of the sine equation to the cosine equation, we get
or
Solving now for 
,
