The answer is A : 302 I really hope this helps!
Answer:
a) 
b) 
Explanation:
Given:
- weight of the stone,

- initial velocity of vertical projection,

- air drag acting opposite to the motion of the stone,

The mass of the stone:



Now the acceleration of the stone opposite of the motion:

where:
d = deceleration


<u>In course of going up the net acceleration on the stone will be:</u>



a)
Now using the equation of motion:

where:
final velocity when the stone reaches at the top of the projectile = 0
h = height attained by the stone before starting to fall down


b)
during the course of descend from the top height of the projectile:
initial velocity, 
The acceleration will be:



here the gravity still acts downwards but the drag acceleration acts in the direction opposite to the motion of the stone, now the stone is falling down hence the drag acts upwards.
Using equation of motion:
(+ve acceleration because it acts in the direction of motion)


Given what we know, we can confirm that this result from the goalie is a clear indicator of room for improvement in the reaction speed and visual coordination for this area of the net.
<h3>How can the goalie improve reaction speeds to this area?</h3>
The key for situations like this is simply repetition. The more the goalie is able to practice with shots in this area of the net, the more muscle memory they will build regarding reacting to these shots, and therefore less time will be needed to block them in the future.
Therefore, we can confirm that this result from the goalie is a clear indicator of room for improvement in the reaction speed and visual coordination for this area of the net.
To learn more about reaction speeds visit:
brainly.com/question/8186329?referrer=searchResults
V=IR
60-V
The current that passes through a 10-ohm resistor = I
I=60/10
6 amperes
Answer:
A.
ice → lemonade it is the correct answer of this question