Fnety = (FT)(sin 32°) – Fg
Or the answer B, I checked it.
Vf= 60m/s
vs = 15m/s
t=20s
a = vf-vs/t
a = 45m/s / 20s
a = 2,25 m/s²
S = vs *t + 1/2 a*t²
S = 15 m/s*20s +1/2 * 2,25m/s² * 400 s²
S = 300m + 450m
S = 750m
v average = 750m/20s
v average = 37,5 m/s
Answer:
L' = 1.231L
Explanation:
The transmission coefficient, in a tunneling process in which an electron is involved, can be approximated to the following expression:

L: width of the barrier
C: constant that includes particle energy and barrier height
You have that the transmission coefficient for a specific value of L is T = 0.050. Furthermore, you have that for a new value of the width of the barrier, let's say, L', the value of the transmission coefficient is T'=0.025.
To find the new value of the L' you can write down both situation for T and T', as in the following:

Next, by properties of logarithms, you can apply Ln to both equations (1) and (2):

Next, you divide the equation (3) into (4), and finally, you solve for L':

hence, when the trnasmission coeeficient has changes to a values of 0.025, the new width of the barrier L' is 1.231 L
Answer:
Wet surfaces areaA=+25.3ft^2
Explanation:
Using F= K×A× S^2
Where F= drag force
A= surface area
S= speed
Given : F=996N S=20mph A= 83ft^2
K = F/AS^2=996/(83×20^2)
K= 996/33200 = 0.03
1215= (0.03)× A × 18^2
1215=9.7A
A=1215/9.7=125.3ft^2