Answer:
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Answer:
(L: Length, T: Time)
p: Dimension: L; unit: m
q: Dimension: L/T or (L)*(T)^-1; unit: m/s
r: Dimension: L/T^2 or (L)*(T)^-2; unit: m/s^2
Explanation:
since y is distance (Length), make all terms L distance.
p is same as y dimension ==> dimension: L; unit: m (meter)
qt dimension is L ==> q dimension :L/T; unit: m/s
rt^2 dimension is L ==> r dimension : L/T^2; unit: m/s^2
Answer: E/4 ( one - fourth of it electric field)
Explanation:
The electric field of a point charge is given below as
E =kq/r²
E = electric field,
K = electric constant
q = magnitude of electric charge
r = distance between point charge and electric field.
It can be seen that only E and r are the only variable here and also, E is inversely proportional to r²
Which implies that
E = k/r² , k = E × r²
E1 ×(r1)² = E2 × (r2)²
Let E1 = E, r1 =1, r2 = 2 and E2 =?
Let us substitute the parameters
E × 1 = E2 × 2²
E × 1 = E2 × 4
E = E2 × 4
E2 = E/4
Which implies that the electric field at the second distance (r =4) is one fourth of the initial electric field.
Answer:
2 m/s^2
Explanation:
from the question
v=15 m/s
t=7.5
a=?
from the first equation of motion
v=u+at
where,
v=final velocity
u=initial velocity
a=acceleration
t=time
from the question (u) will be zero because the body started at rest
v=u+at
15=(0)+a×7.5
15=7.5a
a=15/7.5
a=2 m/s^2