The wind speed increased but in the same direction.
It is A/1 because the wind speed did increase and it is still going the same way clockwise because it also did not move that much.
I'm pretty sure the answer is B: <span>establish public doubt
Hope this helps!</span>
Repeat the experiment
Explanation:
To increase the validity of the results obtained from the single experiment, the students should be encouraged to repeat the experiments more number of times as much as possible.
In an experiment, scientist always try to limit errors by making precise and accurate observation. A single observation does not really represent a precise and accurate finding. When an experiment is repeated as often as possible, the reliability of the conclusion drawn from the hypothesis testing will improve and the results can be accepted to be valid.
A single observation/experiment is not valid enough.
Learn more:
Experiments brainly.com/question/5096428
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(a) 154.5 N
Let's divide the motion of the sprinter in two parts:
- In the first part, he starts with velocity u = 0 and accelerates with constant acceleration
for a total time
During this part of the motion, he covers a distance equal to
, until he finally reaches a velocity of
. We can use the following suvat equation:
![s_1 = u t_1 + \frac{1}{2}a_1t_1^2](https://tex.z-dn.net/?f=s_1%20%3D%20u%20t_1%20%2B%20%5Cfrac%7B1%7D%7B2%7Da_1t_1%5E2)
which reduces to
(1)
since u = 0.
- In the second part, he continues with constant speed
, covering a distance of
in a time
. This part of the motion is a uniform motion, so we can use the equation
(2)
We also know that the total time is 10.0 s, so
![t_1 + t_2 = 10.0 s\\t_2 = (10.0-t_1)](https://tex.z-dn.net/?f=t_1%20%2B%20t_2%20%3D%2010.0%20s%5C%5Ct_2%20%3D%20%2810.0-t_1%29)
Therefore substituting into the 2nd equation
![s_2 = a_1 t_1 (10-t_1)](https://tex.z-dn.net/?f=s_2%20%3D%20a_1%20t_1%20%2810-t_1%29)
From eq.(1) we find
(3)
And substituting into (2)
![s_2 = \frac{2s_1}{t_1^2}t_1 (10-t_1)=\frac{2s_1}{t_1}(10-t_1)=\frac{20 s_1}{t_1}-2s_1](https://tex.z-dn.net/?f=s_2%20%3D%20%5Cfrac%7B2s_1%7D%7Bt_1%5E2%7Dt_1%20%2810-t_1%29%3D%5Cfrac%7B2s_1%7D%7Bt_1%7D%2810-t_1%29%3D%5Cfrac%7B20%20s_1%7D%7Bt_1%7D-2s_1)
Solving for t,
![s_2+2s_1=\frac{20 s_1}{t_1}\\t_1 = \frac{20s_1}{s_2+2s_1}=\frac{20(45)}{55+2(45)}=6.2 s](https://tex.z-dn.net/?f=s_2%2B2s_1%3D%5Cfrac%7B20%20s_1%7D%7Bt_1%7D%5C%5Ct_1%20%3D%20%5Cfrac%7B20s_1%7D%7Bs_2%2B2s_1%7D%3D%5Cfrac%7B20%2845%29%7D%7B55%2B2%2845%29%7D%3D6.2%20s)
So from (3) we find the acceleration in the first phase:
And so the average force exerted on the sprinter is
![F=ma=(66 kg)(2.34 m/s^2)=154.5 N](https://tex.z-dn.net/?f=F%3Dma%3D%2866%20kg%29%282.34%20m%2Fs%5E2%29%3D154.5%20N)
b) 14.5 m/s
The speed of the sprinter remains constant during the last 55 m of motion, so we can just use the suvat equation
![v_1 = u +a_1 t_1](https://tex.z-dn.net/?f=v_1%20%3D%20u%20%2Ba_1%20t_1)
where we have
u = 0
is the acceleration
is the time of the first part
Solving the equation,
![v_1 = 0 +(2.34)(6.2)=14.5 m/s](https://tex.z-dn.net/?f=v_1%20%3D%200%20%2B%282.34%29%286.2%29%3D14.5%20m%2Fs)