In a stationary situation, the weight of person is

This is the weight "felt" by the scale, which is basically the normal reaction applied by the scale on the person, and which uses the value of g (9.81) as reference to convert the weight (602.8 N) into a mass (62 kg).
When the person is in the elevator, the scale says 77 kg. The scale is still using the same value of conversion (9.81), so the apparent weight "felt" by the scale is

This is the normal reaction applied by the scale on the person, and which is directed upward. Besides this force, there is still the weight W of the person, acting downward. So, if we use Newton's second law:


where a is the acceleration of the elevator. If we solve for a, we find

The negative sign means the acceleration is in the opposite direction of g (which we take positive), so it means the elevator is going upward.
Answer:
weight at height = 100 N .
Explanation:
The problem relates to variation of weight due to change in height .
Let g₀ and g₁ be acceleration due to gravity , m is mass of the object .
At the surface :
Applying Newton's law of gravitation
mg₀ = G Mm / R²
At height h from centre
mg₁ = G Mm /h²
Given mg₀ = 400 N
400 = G Mm / R²
400 = G Mm / (6400 x 10³ )²
G Mm = 400 x (6400 x 10³ )²
At height h from centre
mg₁ = 400 x (6400 x 10³ )²/ ( 2 x 6400 x 10³)²
= 400 / 4
= 100 N .
weight at height = 100 N
Answer:
c) 2.02 x 10^16 nuclei
Explanation:
The isotope decay of an atom follows the equation:
ln[A] = -kt + ln[A]₀
<em>Where [A] is the amount of the isotope after time t, k is decay constant, [A]₀ is the initial amount of the isotope</em>
[A] = Our incognite
k is constant decay:
k = ln 2 / Half-life
k = ln 2 / 4.96 x 10^3 s
k = 1.40x10⁻⁴s⁻¹
t is time = 1.98 x 10^4 s
[A]₀ = 3.21 x 10^17 nuclei
ln[A] = -1.40x10⁻⁴s⁻¹*1.98 x 10^4 s + ln[3.21 x 10^17 nuclei]
ln[A] = 37.538
[A] = 2.01x10¹⁶ nuclei remain ≈
<h3>c) 2.02 x 10^16 nuclei</h3>
Answer choice d is correct