Answer:
W = 0.012 J
Explanation:
For this exercise let's use Hooke's law to find the spring constant
F = K Δx
K = F / Δx
K = 3 / (0.16 - 0.11)
K = 60 N / m
Work is defined by
W = F. x = F x cos θ
in this case the force and the displacement go in the same direction therefore the angle is zero and the cosine is equal to 1
W = ∫ F dx
W = k ∫ x dx
we integrate
W = k x² / 2
W = ½ k x²
let's calculate
W = ½ 60 (0.19 -0.17)²
W = 0.012 J
Answer:
E = 7.334 KeV
Explanation:
given,
initial energy = 60 keV
Δ x = 30 cm
μ(for soft tissue) = 0.7 cm⁻¹ (taken from table)
E = 60 × 0.1224
E = 7.334 KeV
energy of x-ray photon after travelling through 30 cm soft tissue in body is E = 7.334 KeV
Answer:
I dunno for sure but i think its C " reduced wear and tear".
Explanation:
I dont really have an explanation for it, it just kinda feels right. Ill update this with the actual answer when i submit the test.
A. Decrease the rate of electrical flow
more resistance = lkess current