Answer:
The magnitude of the electric field is 8.47 N/C
Explanation:
Given;
uniform charge density, λ = 100 nC/m³
inner radii of the cylinder, r = 1.0 mm and 3.0 mm
distance from the symmetry axis, R = 2.0 mm
Area = 2πrl
Area =2π(2 x 10⁻)l
Volume = A x d
d = Volume / Area
the magnitude of the electric field
Therefore, the magnitude of the electric field is 8.47 N/C
To solve the problem we first need to define the nomenclature by which there is an X amount of water in our body. Let's call this a variable m, corresponding to the total mass in us and whose units are Kilograms.
We understand that one millionth of this mass, a small fraction of it, is the equivalent of that of a Water Molecule.
Assuming this to have the same body of water in our body that amount should be approximately
Density of water molecule
Answer:
T = 540 N (to two significant digits)
Explanation:
Let the crate dimension L be from strap attachment to floor contact
Let T be the strap tension
sum moments about the floor contact point to zero
mg[½Lcos25] - Tsin61[Lcos25] + Tcos61[Lsin25] = 0
L is common to all terms, so divides out.
½(71)(9.8)cos25 = T(sin61cos25 - cos61sin25)
T = (71)(9.8)cos25 / (2(sin61cos25 - cos61sin25))
T = 536.428020...
