Explanation:
The volume of the bread decreases, making the bread appear more compact, and smaller in size. The mass stays the same, it won't change unless part of the bread is removed. The density increases, the air bubbles inside of the bread get squished down, causing the bread to be smaller, and in turn, causing it to be more solid.
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During either one, the sun, moon, and Earth are lined up in the same straight line. The difference is whether the moon or the Earth is the one in the "middle".
Answer:
5080.86m
Explanation:
We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:
![y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}](https://tex.z-dn.net/?f=y_1%3Dy_%7B01%7D%2Bv_%7B01%7Dt%2B%5Cfrac%7Ba_1t%5E2%7D%7B2%7D)
![v_1=v_{01}+a_1t](https://tex.z-dn.net/?f=v_1%3Dv_%7B01%7D%2Ba_1t)
We must consider that it's launched from the ground (
) and from rest (
), with an upwards acceleration
that lasts a time t=9.7s.
We calculate then the height achieved in part 1:
![y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m](https://tex.z-dn.net/?f=y_1%3D%280m%29%2B%280m%2Fs%29t%2B%5Cfrac%7B%2828m%2Fs%5E2%29%289.7s%29%5E2%7D%7B2%7D%3D1317.26m)
And the velocity achieved in part 1:
![v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s](https://tex.z-dn.net/?f=v_1%3D%280m%2Fs%29%2B%2828m%2Fs%5E2%29%289.7s%29%3D271.6m%2Fs)
We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 (
) and its initial velocity is the one achieved in part 1 (
), now in free fall, which means with a downwards acceleration
. For the data we have it's faster to use the formula
, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:
![v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s](https://tex.z-dn.net/?f=v_%7B02%7D%5E2%2B2a_2%28y_2-y_%7B02%7D%29%3Dv_2%5E2%3D0m%2Fs)
Then, to get
, we do:
![2a_2(y_2-y_{02})=-v_{02}^2](https://tex.z-dn.net/?f=2a_2%28y_2-y_%7B02%7D%29%3D-v_%7B02%7D%5E2)
![y_2-y_{02}=-\frac{v_{02}^2}{2a_2}](https://tex.z-dn.net/?f=y_2-y_%7B02%7D%3D-%5Cfrac%7Bv_%7B02%7D%5E2%7D%7B2a_2%7D)
![y_2=y_{02}-\frac{v_{02}^2}{2a_2}](https://tex.z-dn.net/?f=y_2%3Dy_%7B02%7D-%5Cfrac%7Bv_%7B02%7D%5E2%7D%7B2a_2%7D)
And we substitute the values:
![y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m](https://tex.z-dn.net/?f=y_2%3Dy_%7B02%7D-%5Cfrac%7Bv_%7B02%7D%5E2%7D%7B2a_2%7D%3D%281317.26m%29-%5Cfrac%7B%28271.6m%2Fs%29%5E2%7D%7B2%28-9.8m%2Fs%5E2%29%7D%3D5080.86m)
Explanation: Electrostatic force is directly related to the charge of each object. So if the charge of one object is doubled, then the force will become two times greater.