Answer:
13 km/h
Explanation:
Average speed = distance/time
Let the total distance and total time taken for the whole trip be d km and t hours respectively
Average speed for the whole trip = 82 km/h
d = 82t
The distance covered in the first half = d1/2
Time taken = t/2
Average speed = 69 km/h
69 = d1/2 ÷ t/2
d1 = 69t
The distance covered in the second half = d2/2
Time taken = t/2
Let the average sly for the see half be A
A = d2/2 ÷ t/2
d2 = At
d = d1 + d2
82t = 69t + At
At = 82t - 69t
At = 13t
A = 13t/t = 13 km/h
A. Equal and opposite
For every action, there is an equal and opposite reaction
Displacement is a vector quantity. So, you incorporate the vector calculations when you try to determine the resultant vector. This is the shortest path from the starting point to the endpoint. If they are moving on one axis only, you use sign conventions. For motions moving to the left, use the negative sign. If it's moving to the right, then use the positive sign. Now, it the object moves 2 km to the left, and 2 km also to the right, the displacement is zero.
Displacement = 2 km - 2km = 0
Generally, the equation is:
<span>Displacement = Distance of motion to the right - Distance of motion to the left</span>
Answer:
q_poly = 14.55 KJ/kg
Explanation:
Given:
Initial State:
P_i = 550 KPa
T_i = 400 K
Final State:
T_f = 350 K
Constants:
R = 0.189 KJ/kgK
k = 1.289 = c_p / c_v
n = 1.2 (poly-tropic index)
Find:
Determine the heat transfer per kg in the process.
Solution:
-The heat transfer per kg of poly-tropic process is given by the expression:
q_poly = w_poly*(k - n)/(k-1)
- Evaluate w_poly:
w_poly = R*(T_f - T_i)/(1-n)
w_poly = 0.189*(350 - 400)/(1-1.2)
w_poly = 47.25 KJ/kg
-Hence,
q_poly = 47.25*(1.289 - 1.2)/(1.289-1)
q_poly = 14.55 KJ/kg
Answer:
TATTCATTCATTA—TGATTT—ATTCG
Explanation:
A mutation is a permanent change in the nucleotide sequence of DNA. A mutation occurs during replication or recombination. It may be due to base substitutions, deletions and insertions. As per the question, DNA segment forms encodes for the enzyme pepsin is CATTGTTA.
Option TATTCATTCATTA—TGATTT—ATTCG is the correct answer. In the DNA segment which encodes pepsin, a purine base (G) guanine is replaced by another purine (A) adenine. This type of mutation is called a transition type point mutation.
Due to base substitution, the mutated segment CATTCATTA will nor encode pepsin.
