An isotonic solution is <span>a solution in which concentration or solute is equal to that of a cell placed in it. Thus, the system is in dynamic equilibrium, and so water molecules flow in both directions.
The correct answer is <u>C. w</u></span><span><u>ater molecules flow in both directions at the same rate.</u></span>
oxygen's atomic number is 8
1s22s22p4 is the electron configuration
2+2=4 4+4=8
Answer:
(a). The initial velocity is 28.58m/s
(b). The speed when touching the ground is 33.3m/s.
Explanation:
The equations governing the position of the projectile are
![(1).\: x =v_0t](https://tex.z-dn.net/?f=%281%29.%5C%3A%20x%20%3Dv_0t)
![(2).\: y= 15m-\dfrac{1}{2}gt^2](https://tex.z-dn.net/?f=%282%29.%5C%3A%20y%3D%2015m-%5Cdfrac%7B1%7D%7B2%7Dgt%5E2)
where
is the initial velocity.
(a).
When the projectile hits the 50m mark,
; therefore,
![0=15-\dfrac{1}{2}gt^2](https://tex.z-dn.net/?f=0%3D15-%5Cdfrac%7B1%7D%7B2%7Dgt%5E2)
solving for
we get:
![t= 1.75s.](https://tex.z-dn.net/?f=t%3D%201.75s.)
Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that
![50m = v_0(1.75s)](https://tex.z-dn.net/?f=50m%20%3D%20v_0%281.75s%29)
which gives
![\boxed{v_0 = 28.58m/s.}](https://tex.z-dn.net/?f=%5Cboxed%7Bv_0%20%3D%2028.58m%2Fs.%7D)
(b).
The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,
![v_x = 28.58m/s.](https://tex.z-dn.net/?f=v_x%20%3D%2028.58m%2Fs.)
the vertical component of the velocity is
![v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.](https://tex.z-dn.net/?f=v_y%20%3D%20gt%20%5C%5Cv_y%20%3D%20%289.8m%2Fs%5E2%29%281.75s%29%5C%5C%5C%5C%7Bv_y%20%3D%2017.15m%2Fs.)
which gives a speed
of
![v = \sqrt{v_x^2+v_y^2}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7Bv_x%5E2%2Bv_y%5E2%7D)
![\boxed{v =33.3m/s.}](https://tex.z-dn.net/?f=%5Cboxed%7Bv%20%3D33.3m%2Fs.%7D)
Answer:
It's D
Explanation:
Sorry if im wrong, correct me in comments and tell them actual answer if I'm wrong.
Time taken to reach water :
![t = \dfrac{D}{v}\\\\t=\dfrac{6.78}{8.06}\ s\\\\t=0.84\ s](https://tex.z-dn.net/?f=t%20%3D%20%5Cdfrac%7BD%7D%7Bv%7D%5C%5C%5C%5Ct%3D%5Cdfrac%7B6.78%7D%7B8.06%7D%5C%20s%5C%5C%5C%5Ct%3D0.84%5C%20s)
Now, initial vertical speed , u = 0 m/s.
By equation of motion :
![h = ut +\dfrac{at^2}{2}](https://tex.z-dn.net/?f=h%20%3D%20ut%20%2B%5Cdfrac%7Bat%5E2%7D%7B2%7D)
Here, a = g = acceleration due to gravity = 9.8 m/s².
So,
![h = 0(t) +\dfrac{9.8\times 0.84^2}{2}\\\\h=3.46\ m](https://tex.z-dn.net/?f=h%20%3D%200%28t%29%20%2B%5Cdfrac%7B9.8%5Ctimes%200.84%5E2%7D%7B2%7D%5C%5C%5C%5Ch%3D3.46%5C%20m)
Therefore, the height of the bridge is 3.46 m.
Hence, this is the required solution.