Answer:
C) 64lb
Explanation:
use the linearity method to find the weight of nadir on another planet, it is applied as follows.
Nadir Weight in earth ⇒ Nadir weight in another planet
Vince Weigh in eart ⇒ X
our goal is to find the weight of vince in another planet (X), for this we multiply the diagonal that continents the data and divide among the remaining
140pounds ⇒ 56lb
160pounds ⇒ X
X=
Vince weigh on the other planet is C) 64lb
Oil is less dense than water so it tends to float on the top of the water. Hope this Helps!
Answer:
<em>A.The rabbits in the new habitat will have lower genetic variation than the rabbits in the original habitat. </em>
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Explanation:
If two animals of opposite sex are isolated from a larger group of animal, and made to reproduce. They will produce offspring with similar genetic makeup. If this offspring still remain isolated, and continue to interbreed within themselves for a number of consecutive generations, their offspring will all be very closely related genetically. Situations like this just as with the two rabbits in the question leads to a lower genetic variation within the offspring of the two animals.
Animals need to reproduce within a larger group in order to increase genetic variation. Increasing genetic variation reduces the risk of been sucked into a gene pool. A lower genetic variation reduces the fitness of the animals involved. It is only an advantage in cases in which the the original pair are resistant to a deadly disease. In this case all the offspring also develop this immunity. Mostly the effects of a lower genetic variation leaves negative impacts, and animals try to avoid this by preferring to interbreed with unrelated partner
Answer:
a) The current density ,J = 2.05×10^-5
b) The drift velocity Vd= 1.51×10^-15
Explanation:
The equation for the current density and drift velocity is given by:
J = i/A = (ne)×Vd
Where i= current
A = Are
Vd = drift velocity
e = charge ,q= 1.602 ×10^-19C
n = volume
Given: i = 5.8×10^-10A
Raduis,r = 3mm= 3.0×10^-3m
n = 8.49×10^28m^3
a) Current density, J =( 5.8×10^-10)/[3.142(3.0×10^-3)^2]
J = (5.8×10^-10) /(2.83×10^-5)
J = 2.05 ×10^-5
b) Drift velocity, Vd = J/ (ne)
Vd = (2.05×10^-5)/ (8.49×10^28)(1.602×10^-19)
Vd = (2.05×10^-5)/(1.36 ×10^10)
Vd = 1.51× 10^-5
Answer:
Explanation:
Using the magnification formula.
Magnification = Image distance(v)/object distance(u) = Image Height(H1)/Object Height(H2)
M = v/u = H1/H2
v/u = H1/H2...1
3) Given the radius of curvature of the concave lens R = 20cm
Focal length F = R/2
f = 20/2
f = 10cm
Object distance u = 5cm
Object height H2= 5cm
To get the image distance v, we will use the mirror formula
1/f = 1/u+1/v
1/v = 1/10-1/5
1/v = (1-2)/10
1/v =-1/10
v = -10cm
Using the magnification formula
(10)/5 = H1/5
10 = H1
H1 = 10cm
Image height of the peg is 10cm
4) If u = 15cm
1/v = 1/f-1/u
1/v = 1/10-1/15
1/v = 3-2/30
1/v = 1/30
v = 30cm
30/15 = H1/5
15H1 = 150
H1/= 10cm
5) if u = 20cm
1/v = 1/f-1/u
1/v = 1/10-1/20
1/v = 2-1/20
1/v = 1/20
v = 20cm
20/20 = H1/5
20H1 = 100
H1 = 5cm
6) If u = 30cm
1/v = 1/f-1/u
1/v = 1/10-1/30
1/v = 3-1/30
1/v = 2/30
v = 30/2 cm
v =>15cm
15/30 = Hi/5
30H1 = 75
H1 = 75/30
H1 = 2.5cm
