V = 1/3 Bh v = 1/3 (13 ac)(43560ft^2/ac)(481ft) v = 90793560 ft^3 * 0.3048m/ft * 0.3048m/ft * 0.3048m/ft = 2570987m^3
Answer:
Explanation:
Given
Weight of Person 
Cave is
deep
Breaking stress 
Net Force on Person




The shortest time such that the person can be taken out of cave

where
h=distance moved
t=time
a=acceleration



The answer is Codominance, AB blood cells inherit both A and B blood types so they are codominance
Answer:
y <8 10⁻⁶ m
Explanation:
For this exercise, they indicate that we use the Raleigh criterion that establishes that two luminous objects are separated when the maximum diffraction of one of them coincides with the first minimum of the other.
Therefore the diffraction equation for slits with m = 1 remains
a sin θ = λ
in general these experiments occur for oblique angles so
sin θ = θ
θ = λ / a
in the case of circular openings we must use polar coordinates to solve the problem, the solution includes a numerical constant
θ = 1.22 λ / a
The angles in these measurements are taken in radians, therefore
θ = s / R
as the angle is small the arc approaches the distance s = y
y / R = 1.22 λ / s
y = 1.22 λ R / a
let's calculate
y = 1.22 500 10⁻⁹ 0.42 / 0.032
y = 8 10⁻⁶ m
with this separation the points are resolved according to the Raleigh criterion, so that it is not resolved (separated)
y <8 10⁻⁶ m
Answer:
94.1 m
Explanation:
From Coulombs law,
F = Gm1m2/r²................... Equation 1
where F = force, m1 = first mass, m2 = second mass, G = universal constant, r = distance of separation.
Make r the subject of the equation,
r = √(Gm1m2/F)................. Equation 2
Given: F = 7×10² N, m1 = 15×10⁷ kg, m2 = 62×10⁷ kg,
Constant: G = 6.67×10⁻¹¹ Nm²/kg²
Substitute into equation 2
r = √( 6.67×10⁻¹¹×15×10⁷×62×10⁷/7×10²)
r √(886.16×10)
r √(88.616×10²)
r = 9.41×10
r = 94.1 m.
Hence the distance of separation = 94.1 m