The height of the tennis ball,relative to the ground is H=h max+h-->h max-the maximum height that the tennis ball reaches relative to the roof of the building; h-the height of the building;h max =v0^2/2g=24,2m(g=10m/s^2).H=gt^2/2=>24,2+h=gt^2/2=>h=gt^2/2-24,2=180,6m
Answer:
20 meters.
Explanation:
In the graph, the x-axis (the horizontal axis) represents the time, while the y-axis (the vertical axis) represents the distance.
If we want to find the distance covered in the first T seconds, you need to find the value T in the horizontal axis.
Once you find it, we draw a vertical line, in the point where this vertical line touches the graph, we now draw a horizontal line. This horizontal line will intersect the y-axis in a given value. That value is the total distance travelled by the time T.
In this case, we want to find the total distance that David ran in the first 4 seconds.
Then we need to find the value 4 seconds in the horizontal axis. Now we perform the above steps, and we will find that the correspondent y-value is 20.
This means that in the first 4 seconds, David ran a distance of 20 meters.
Answer:m1v1 + m2v2 = (m1f + m2f)vf. 3000kg(10.0m/s) + (15000kg)(0.0m/s) = (18000kg)(vf).
Explanation:
Answer:
twice
Explanation:
From magnification = height of image / height of object
Distance of image/ distance of object = magnification
If the distance and height of the object represents the initial light distance and the exposed surface respectively.
And similarly the distance and height of the image represents the final light distance and the exposed surface respectively.
Hence the new image exposure would be twice as large.
If we use the formula our point of investigation is Height of image,
H2= D2/D1× H1
H2 = 2D2/D1 × H1
H2 = 2H1
