The car's speed was zero at the beginning of the 12 seconds,
and 18 m/s at the end of it. Since the acceleration was 'uniform'
during that time, the car's average speed was (1/2)(0 + 18) = 9 m/s.
12 seconds at an average speed of 9 m/s ==> (12 x 9) = 108 meters .
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That's the way I like to brain it out. If you prefer to use the formula,
the first problem you run into is: You need to remember the formula !
The formula is D = 1/2 a T²
Distance = (1/2 acceleration) x (time in seconds)²
Acceleration = (change in speed) / (time for the change)
= (18 m/s) / (12 sec)
= 1.5 m/s² .
Distance = (1/2 x 1.5 m/s²) x (12 sec)²
= (0.75 m/s²) x (144 sec²) = 108 meters .
I think is A or B it depends on like what the trying to answer
It must gain an electron because if the proton number was to change it would no longer be the same element.
They have similar physical and chemical properties because of thier valence electrons
Answer:
The ball would have landed 3.31m farther if the downward angle were 6.0° instead.
Explanation:
In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).
We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.
So, first we need to determine the components of the velocity of the ball, like this:






we pick the positive one, so it takes 0.317s for the ball to hit on point A.
so now we can find the distance from the net to point A with this time. We can find it like this:



Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:







t= -0.9159s or t=0.468s
we pick the positive one, so it takes 0.468s for the ball to hit on point B.
so now we can find the distance from the net to point B with this time. We can find it like this:



So once we got the two distances we can now find the difference between them:

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.