Answer:
The group that is exposed to the variable to be tested is called as experimental group.
Explanation:
An experiment consists of two kinds of group. They are experimental group and control group. The working of control group and experimental group are mostly same except that the variables will be changed in experimental group, while it is kept constant in control group. So at the end of experiments, the results from control group and experimental group are compared and the final outcome has been derived. So the given statement where the group is exposed to the variable to be tested or the independent variable is called as experimental group.
The volume of the cylinder is:
V= pi* (r^2) * h.
So the volume is = pi * (5.7/2)^2 * 12 = 306.2 in^2.
Since it is scaled up by the factor of 1.5, so we have to multiply each dimension with 1.5.
That means the diameter will be 8.55 and the height will be 18. so the scaled up volume will be
=pi * ((5.7*1.5)/2)^2 * (12*1.5)
The answer then would have to be 1033.5
Answer:
D
Explanation:
Look at the two of them, if you put them together the image is identical to that of the unknown sample.
Answer:
A. Electric potential energy into kinetic energy
Explanation:
The electric potential energy of the charged particles is converted into kinetic energy as the electron is released.
Explanation:
Formula for steady flow energy equation for the flow of fluid is as follows.
![m[h_{1} + \frac{V^{2}_{1}}{2}] + z_{1}g] + q = m[h_{1} + \frac{V^{2}_{1}}{2} + z_{1}g] + w](https://tex.z-dn.net/?f=m%5Bh_%7B1%7D%20%2B%20%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%7D%7B2%7D%5D%20%2B%20z_%7B1%7Dg%5D%20%2B%20q%20%3D%20m%5Bh_%7B1%7D%20%2B%20%5Cfrac%7BV%5E%7B2%7D_%7B1%7D%7D%7B2%7D%20%2B%20z_%7B1%7Dg%5D%20%2B%20w)
Now, we will substitute 0 for both
and
, 0 for w, 334.9 kJ/kg for
, 2726.5 kJ/kg for
, 5 m/s for
and 220 m/s for
.
Putting the given values into the above formula as follows.
![1 \times [334.9 \times 10^{3} J/kg + \frac{(5 m/s)^{2}}{2} + 0] + q = 1 \times [2726.5 \times 10^{3} + \frac{(220 m/s)^{2}}{2} + 0] + 0](https://tex.z-dn.net/?f=1%20%5Ctimes%20%5B334.9%20%5Ctimes%2010%5E%7B3%7D%20J%2Fkg%20%2B%20%5Cfrac%7B%285%20m%2Fs%29%5E%7B2%7D%7D%7B2%7D%20%2B%200%5D%20%2B%20q%20%3D%201%20%5Ctimes%20%5B2726.5%20%5Ctimes%2010%5E%7B3%7D%20%2B%20%5Cfrac%7B%28220%20m%2Fs%29%5E%7B2%7D%7D%7B2%7D%20%2B%200%5D%20%2B%200)
q = 6597.711 kJ
Thus, we can conclude that heat transferred through the coil per unit mass of water is 6597.711 kJ.