Answer:
let us use the expression v = ./2gS in both the cases.
a) In the first case g = 9.8 m/s^2 and S = 10.3 m.
So v = 14.21 m/s
b) in the second case, g = 1.64 m/s^2 and S = 4.2 m = 3.71 m/s
00
Explanation:
a) The car’s speed just after leaving the icy portion of the road is the first part
We have s = ut +
Here s = 130 m, a = -1.2
, u = 76 mile/hr = 33.975 m/s

After 14.72 seconds cars speed is given by v =u +at
u = 33.975m/s, a = -1.2 
v = 33.975-1.2*14.72 = 16.31 m/s
So velocity of car on leaving icy surface = 16.31 m/s
b) Time taken after icy surface
0 = 16.31 - 7.10*t
t = 2.23 seconds
Distance traveled during this time

Total distance traveled = 130+18.74 = 148.74 m
c) Total time taken = 14.72+2.23=16.95 seconds
ΔU = Q + W
- heat in, Q +
- heat out, Q -
- does work , W -
- work in, W +
93-58 = 63 + W
35 = 63 + W
W = - 28 J (does work/being done by system)
I think it is force of acceleration but I'm not sure.. hope this helps though!
Answer:2m
Explanation:
Wavelength=velocity/frequency
Wavelength=4/2=2m