To solve this problem, we know that:
1 Albert = 88 meters
1 A = 88 m
The first thing we have to do is to square both sides of
the equation:
(1 A)^2 = (88 m)^2
1 A^2 = 7,744 m^2
<span>Since it is given that 1 acre = 4,050 m^2, so to reach
that value, 1st let us divide both sides by 7,744:</span>
1 A^2 / 7,744 = 7,744 m^2 / 7,744
(1 / 7,744) A^2 = 1 m^2
Then we multiply both sides by 4,050.
(4050 / 7744) A^2 = 4050 m^2
0.523 A^2 = 4050 m^2
<span>Therefore 1 acre is equivalent to about 0.52 square
alberts.</span>
Answer:
Explanation:
We define the linear density of charge as:
Where L is the rod's length, in this case the semicircle's length L = πr
The potential created at the center by an differential element of charge is:
where k is the coulomb's constant
r is the distance from dq to center of the circle
Thus.
Potential at the center of the semicircle
Answer:
d = 105 m
Explanation:
Speed of a car, v = 21 m/s
We need to find the distance traveled by the dar during those 5 s before it stops. Let the distance is d. It can be calculated as :
d = v × t
d = 21 m/s × 5 s
d = 105 m
So, it will cover 105 m before it stops.
Answer:
cindi
Explanation:
cindi's work done is larger than all the other students combined
