Constant acceleration of plane = 3m/s²
a) Speed of the plane after 4s
Acceleration = speed/time
3m/s² = speed/4s
S = 12m/s
The speed of the plane after 4s is 12m/s.
b) Flight point will be termed as the point the plane got initial speed, u, 20m/s
Find speed after 8s, v
a = 3m/s²
from,
a = <u>v</u><u> </u><u>-</u><u> </u><u>u</u>
t
3 = <u>v</u><u> </u><u>-</u><u> </u><u>2</u><u>0</u>
8
24 = v - 20
v = 44m/s
After 8s the plane would've 44m/s speed.
Answer:
10 feet.
Explanation:
Conductors are considered outside the building when they installed. The requirement of NEC is above 10 feet for the clearance after the final grade for service conductors and cables where they are connected with the building, above the sidewalks and the to other areas which are accessible to the pedestrians.
Answer:
divide the mass value by 1e+8
Answer:
The P site
Explanation:
After the formation of peptide linkage, the trna of P site becomes free, more to the E site and then slip away. The A site trna has dipeptide instead of a single aminoacid.
In the prescence of translocase and energy from GTP, the ribosome move in such a way the peptidyl bearing trna of A site come to lie with P sit. This exposes next codon to A site. A new aminoacid trna complex reaches the fresh codon exposed at A site.