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Answer:
Explanation:
The acceleration of gravity is 9.8m/s^2.
So to calculate the time it will take to make the ball stop(which btw means the ball now reach its greatest height), use the formula V1=V0+at. V1 is the final velocity(which is 0), V0 is the starting velocity(which is 30m/s), and the a(cceleration) is 9.8m/s^2.
(You can ignore the fact "at" is -30 instead 30, it's because the directions two velocity travel are opposite. )
We can now know the time it takes to make the ball stop just by the gravitational force is about 3 sec.
Use another formula S=1/2at^2, to find out the S(height) is 1/2*9.8*3^2=44.1, which is approximately D.45m .
21) Acceleration from D to E: 
22) The acceleration of the bus from D to E is
Explanation:
21)
The acceleration of an object is equal to the rate of change of velocity of the object. Mathematically:
where
u is the initial velocity
v is the final velocity
t is the time elapsed
In this problem, we want to measure the acceleration of the bus from point D to point E. We have:
- Initial velocity at point D: u = 0
- Final velocity at point E: v = 5 m/s
- Time elapsed from D to E: t = 21 - 16 = 5 s
Therefore, the acceleration between D and E is
22) This question is the same as 21), so the result is the same.
Learn more about acceleration:
brainly.com/question/9527152
brainly.com/question/11181826
brainly.com/question/2506873
brainly.com/question/2562700
#LearnwithBrainly
Answer:
∑F = 10.2 N
Explanation:
We have:
Initial velocity: 0.5 m/s
Final velocity: 3 m/s
Time: 1.5 s
We have all of the components needed to calculate acceleration. Let's do that, shall we?
a = vf-vo/t
a = 2.5/1.5
a = 1.7 
/
Now, look at the Net Force equation:
∑F = ma
Plug in the variables, to get:
∑F = (6)(1.7)
∑F = 10.2 N (You can round this according to significant digits)
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.
