<u>Answer:</u>
The acceleration of the car is 
<u>Explanation:</u>
In the question it is given that car initially heads north with a velocity
. It then accelerates for
and in the end its velocity is
.
initial velocity 
time 
final velocity 
The equation of acceleration is


The value of acceleration is positive, here since the car is speeding up. If it was slowing down the value of acceleration would be negative.
In a parallel circuit, the voltage across each of the components is the same, and the total current is the sum of the currents through each component. If two or more components are connected in parallel they have the same potential difference ( voltage) across their ends.
Relationship between alleles of a single gene, the effect on a phenotype of one allele masks the contribution to the second one at the same locus.
I hope this helps! :)
F= ma
f=200*5
force = 1000 Newtons
Answer:
a) 1955.6 J
b) 1955.6 J
c) 2607 N
d) 110373 J
Explanation:
a) 
b) Work done on bullet should be increase in KE = 1955.6 J
c) Work done = Force \times displacement = KE of bullet

d) Work done to stop 
