Data:
u=0 m/s is the initial velocity of the plane
v=62 m/s is the final velocity of the plane (at which the plane takes off)
a=1.7 m/s^2 is the acceleration of the plane
To find the minimum distance S the plane needs to take off, we can use the following equation:
Re-arranging it and substituting the numbers, we find
Cell are small and us human can't see them with our own eyes. it is in possible to see cell without a microscope
This is just testing your ability to recall that kinetic energy is given by:
<span>k.e. = ½mv² </span>
<span>where m is the mass and v is the velocity of the particle. </span>
<span>The frequency of the light is redundant information. </span>
<span>Here, you are given m = 9.1 * 10^-31 kg and v = 7.00 * 10^5 m/s. </span>
<span>Just plug in the values: </span>
<span>k.e. = ½ * 9.1 * 10^-31 * (7.00 * 10^5)² </span>
<span>k.e. = 2.23 * 10^-19 J
so it will be d:2.2*10^-19 J</span>
W = 1/2k*x^2.
k = spring constant = 2500 n/m.
x = distance = 4 cm = 0.04m (convert to same units).
W = 1/2(2500)(0.04)^2 = 2J.
Answer:
a). 
kJ/kg
b). 
kJ/kg-K
Explanation:
a). The energy rate balance equation in the control volume is given by
kJ/kg
b). Entropy produced from the entropy balance equation in a control volume is given by
kJ/kg-K
