Answer:
Δd = 
Explanation:
As 
, when the car is making full stop, 
. 
. Therefore,
Apply the same formula above, with 
and 
, and the car is starting from 0 speed, we have
As 
. After 
, the car would have traveled a distance of
Hence 
As 
we can simplify 
After t time, the train would have traveled a distance of 
Therefore, Δd would be 
Answer:
38 cm from q1(right)
Explanation:
Given, q1 = 3q2 , r = 60cm = 0.6 m
Let that point be situated at a distance of 'x' m from q1.
Electric field must be same from both sides to be in equilibrium(where EF is 0).
=> k q1/x² = k q2/(0.6 - x)²
=> q1(0.6 - x)² = q2(x)²
=> 3q2(0.6 - x)² = q2(x)²
=> 3(0.6 - x)² = x²
=> √3(0.6 - x) = ± x
=> 0.6√3 = x(1 + √3)
=> 1.03/2.73 = x
≈ 0.38 m = 38 cm = x
Force = mass × accelaration
Force = 0.25Kg × 196 m/s²
Force = 49 Newtons
Answer: d₂ = 170 mGya
Explanation:
the relationship between absonbed 'd' and exposure 'E' is given as;
D(Gv) = F . x (AS/xB)
F is a conversion coefficient depending on medium
so we can simply write
d₁/d₂ = x₁/x₂
Given that;
our x₁ = 60 mAs, x₂ = 120 mAs, d₁ = 85 mGya, d₂ = ?
from the given formula,
d₂ = (x₂d₁ / x₁)
now we substitute
d₂ = (120 × 85) / 60
d₂ = 170 mGya
∴ if 120 mAa is used, the new exposure will be 170 mGya
Sorry most of speak english on this website so i cannot help you
