Answer:
Explanation:
Given
mass of body 
radius of circle 
Time Period 
suppose 
is the angular velocity of revolution
therefore 
Centripetal acceleration 
therefore centripetal 
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There are three type of them nucleus,proton and lastly neutron
hope it helps
Answer:
512.5 mJ
Explanation:
Let the two identical charges be q = +35 µC and distance between them be r₁ = 46 cm. A charge q' = +0.50 µC located mid-point between them is at r₂ = 46 cm/2 = 23 cm = 0.23 m.
The electric potential at this point due to the two charges q is thus
V = kq/r₂ + kq/r₂
= 2kq/r₂
= 2 × 9 × 10⁹ Nm²/C² × 35 × 10⁻⁶ C/0.23 m
= 630/0.23 × 10³ V
= 2739.13 × 10³ V
= 2.739 MV
When the charge q' is moved 12 cm closer to either of the two charges, its distance from each charge is now r₃ = r₂ + 12 cm = 23 cm + 12 = 35 cm = 0.35 m and r₄ = r₂ - 12 cm = 23 cm - 12 cm = 11 cm = 0.11 cm.
So, the new electric potential at this point is
V' = kq/r₃ + kq/r₄
= kq(1/r₃ + 1/r₄)
= 9 × 10⁹ Nm²/C² × 35 × 10⁻⁶ C(1/0.35 m + 1/0.11 m)
= 315 × 10³(2.857 + 9.091) V
= 315 × 10³ (11.948) V
= 3763.62 × 10³ V
= 3.764 MV
Now, the work done in moving the charge q' to the point 12 cm from either charge is
W = q'(V' - V)
= 0.5 × 10⁻⁶ C(3.764 MV - 2.739 MV)
= 0.5 × 10⁻⁶ C(1.025 × 10⁶) V
= 0.5125 J
= 512.5 mJ
Answer: 3.75 mg
Explanation:
We need a dose of 2.50 mg per square meter of BSA, and we also know that the BSA of the child is 1.50 m^2, then if the dosage should be 2.50mg for one m^2, for 1,5 m^2 we should administrate an amount of:
D = 1.5 m^2*(2.50mg/m^2) = 1.5*2.5 mg = 3.75 mg
