Answer:
The mass flow rate is 2.37*10^-4kg/s
The exit velocity is 34.3m/s
The total flow of energy is 0.583 KJ/KgThe rate at which energy leave the cooker is 0.638KW
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The directions arrow<span> is </span>always<span> going the wrong </span>way<span>.</span>
For this problem, we use the Coulomb's law written in equation as:
F = kQ₁Q₂/d²
where
F is the electrical force
k is a constant equal to 9×10⁹
Q₁ and Q₂ are the charge of the two objects
d is the distance between the two objects
Substituting the values:
F = (9×10⁹)(-22×10⁻⁹ C)(-22×10⁻⁹ C)/(0.10 m)²
F = 0.0004356 N
Answer:
Yeah ice floats on water.
Observation
Example in those areas were ice is found like Antarctica ice is found on top of water.
Answer:
4.75 m/s
Explanation:
The computation of the velocity of the existing water is shown below:
Data provided in the question
Tall = 2 m
Inside diameter tank = 2m
Hole opened = 10 cm
Bottom of the tank = 0.75 m
Based on the above information, first we have to determine the height which is
= 2 - 0.75 - 0.10
= 2 - 0.85
= 1.15 m
We assume the following things
1. Compressible flow
2. Stream line followed
Now applied the Bernoulli equation to section 1 and 2
So we get
![\frac{P_1}{p_g} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{p_g} + \frac{v_2^2}{2g} + z_2](https://tex.z-dn.net/?f=%5Cfrac%7BP_1%7D%7Bp_g%7D%20%2B%20%5Cfrac%7Bv_1%5E2%7D%7B2g%7D%20%2B%20z_1%20%3D%20%5Cfrac%7BP_2%7D%7Bp_g%7D%20%2B%20%5Cfrac%7Bv_2%5E2%7D%7B2g%7D%20%2B%20z_2)
where,
P_1 = P_2 = hydrostatic
z_1 = 0
z_2 = h
Now
![\frac{v_1^2}{2g} + 0 = \frac{v_2^2}{2g} + h\\\\V_2 < < V_1 or V_2 = 0\\\\Therefore\ \frac{v_1^2}{2g} = h\\\\v_1^2 = 2gh\\\\ v_1 = \sqrt{2gh} \\\\v_1 = \sqrt{2\times 9.8\times 1.15}](https://tex.z-dn.net/?f=%5Cfrac%7Bv_1%5E2%7D%7B2g%7D%20%2B%200%20%3D%20%5Cfrac%7Bv_2%5E2%7D%7B2g%7D%20%2B%20h%5C%5C%5C%5CV_2%20%3C%20%3C%20V_1%20or%20V_2%20%3D%200%5C%5C%5C%5CTherefore%5C%20%20%5Cfrac%7Bv_1%5E2%7D%7B2g%7D%20%3D%20h%5C%5C%5C%5Cv_1%5E2%20%3D%202gh%5C%5C%5C%5C%20v_1%20%3D%20%5Csqrt%7B2gh%7D%20%5C%5C%5C%5Cv_1%20%3D%20%5Csqrt%7B2%5Ctimes%209.8%5Ctimes%201.15%7D)
= 4.7476 m/sec
= 4.75 m/s