Answer:
Just consider the resistance as water flowing through a pipe. If the pipe is too small in radius (consider thin wire as small pipe) water can’t flow easly. If the pipe is big in radius (consider thick wire as big pipe) water can flow easly. So flowing water through a small pipe for a long distance will inversly affect the normal flow where as flowing water through a big pipe for a small distance will not affect too much the normal flow.
So long & thin wire has high resistance. Short & thick wire has low resistance.
B. A wire that is 2 m long and has a cross-sectional area of 0.066
By using a tangent line.
I need a bit more info to be sure but that's not wrong.
Define
g = 9.8 m/s², acceleration due to gravity, positive downward.
Assume that wind resistance may be neglected.
Frog A:
u = 0.551 m/s, launch velocity, upward.
When the frog lands back on the pad, its vertical position is zero, and its vertical velocity will be 0.551 m/s downward.
If the time of flight is t, then
(0.551 m/s)*(t s) - 0.5*(9.8 m/s²)*(t s)² = 0
0.551t - 4.9t² = 0
t = 0, or t = 0.1124 s
t = 0 corresponds to launch, and t = 0.1124 s corresponds to landing.
Frog B:
Launch velocity is 1.75 m/s
When t = 0.1124 s, the position of the frog is
s = (1.75 m/s)(0.1124 s) - 0.5*(9.8 m/s²)*(0.1124 s)²
= 0.135 m
The velocity of frog B is
v = (1.75 m/s) - (9.8 m/s²)*(0.1124 s)
= 0.6485 m/s
Answer:
When frog A lands on the ground,
Frog B is 0.135 m above ground and its velocity is 0.649 m/s upward.
Skin is specific
Macrophages specific
All are specific except fever
Answer:
Explanation:
Constant pressure molar heat capacity Cp = 29.125 J /K.mol
If Cv be constant volume molar heat capacity
Cp - Cv = R
Cv = Cp - R
= 29.125 - 8.314 J
= 20.811 J
change in internal energy = n x Cv x Δ T
n is number of moles , Cv is molar heat capacity at constant volume , Δ T is change in temperature
Putting the values
= 20 x 20.811 x 15
= 6243.3 J.
