Answer:
final velocity will be44.72m/s
Explanation:
HEIGHT=h=100m
vi=0m/s
vf=?
g=10m/s²
by using third equation of motion for bodies under gravity
2gh=(vf)²-(vi)²
evaluating the formula
2(10m/s²)(100m)=vf²-(0m/s)²
2000m²/s²=vf²
√2000m²/s²=√vf²
44.72m/s=vf
Answer:
Explanation:
Given
Power Supplied ![[tex]P_{input}=4\ kW](https://tex.z-dn.net/?f=%5Btex%5DP_%7Binput%7D%3D4%5C%20kW)
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Efficiency of the motor 
and 
So, vacuum cleaner delivers a power of
Answer:
2.5 x 10⁷ J
Explanation:
F = thrust of the engine = 2.3 x 10⁵ N
d = distance traveled = 87 m
Work done by the engine is given as
W = F d = (2.3 x 10⁵) (87) = 200.1 x 10⁵ J
W' = Net work done
W'' = work done by catapult
KE₀ = initial kinetic energy = 0 J
KE = final kinetic energy = 4.5 x 10⁷ J
Net work done is given as
W' = KE - KE₀
W' = 4.5 x 10⁷ J
We know that
W' = W + W''
4.5 x 10⁷ = 2.001 x 10⁷ + W''
W'' = 2.5 x 10⁷ J
Answer:
106 single crossover events
Explanation:
Total number of offspring = 1250
<u>Determine the number of offspring that would be expected </u>
( Rf between the genes vg and Pr ) * ( Total number of progeny )
= 0.125 * 1250 ≈ 156 ( value for single and double crossovers )
next :
( value of the Rf of double crossovers ) * ( Total number of progeny )
= (0.322 * 0.125) = 0.04025 * 1250 = 50
Hence the number of offspring expected to obtain/represent a single crossover event between pr and vg
= 156 - 50 = 106
