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
Hello, I wanted an answer from a mathematician. The number 1.04 is closer to the number 1, 2, 1.25 or 1.5.
2 answers:
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Answer:
The distance the block will slide before it stops is 3.3343 m
Explanation:
Given;
mass of bullet, m₁ = 20-g = 0.02 kg
speed of the bullet, u₁ = 400 m/s
mass of block, m₂ = 2-kg
coefficient of kinetic friction, μk = 0.24
Step 1:
Determine the speed of the bullet-block system:
From the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
v is the speed of the bullet-block system after collision
(0.02 x 400) + (2 x 0) = v (0.02 + 2)
8 = v (2.02)
v = 8/2.02
v = 3.9604 m/s
Step 2:
Determine the time required for the bullet-block system to stop
Apply the principle of conservation momentum of the system
when the system stops, vf = 0
3.9604 -2.352t = 0
2.352t = 3.9604
t = 3.9604/2.352
t = 1.684 s
Thus, time required for the system to stop is 1.684 s
Finally, determine the distance the block will slide before it stops
From kinematic, distance is the product of speed and time
Now, recall that t = 1.684 s
S = 3.9604(1.684) - 1.176(1.684)²
S = 6.6693 - 3.3350
S = 3.3343 m
Thus, the distance the block will slide before it stops is 3.3343 m
Answer:
a.Attractive
Explanation:
When it comes to charges, the charges which are alike repel each other and the charges which are different will attract each other.
Here, there is a proton and electron which are different particles hence, they will attract each other.
= Charge of electron and proton =
r = Distance between them = 997 nm
k = Coulomb constant =
Force is given by
The force of attraction between the particles will be