Answer:
<em>0.228 nm</em>
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Explanation:
Atomic mass number of copper = 64
but an atomic mass unit = 1.66 x 10^-27 kg
therefore, the mass of the copper atom m = 64 x 1.66 x 10^-27 kg = 1.06 x 10^-25 kg
The number of atoms in this mass n = ρ/m
where ρ is the density of copper = 8.96 x 10^3 kg/m^3
==> n = (8.96 x 10^3)/(1.06 x 10^-25) = 8.45 x 10^28 atoms/m^3
We know that the volume occupied by this amount of atoms n =
where a is the lattice constant
equating, we have
8.45 x 10^28 =
a = 4.389 x 10^9
we also know that
d = 1/a
where d is the smallest distance between the two copper atom.
d = 1/(4.389 x 10^9) = 2.28 x 10^-10 m
==> <em>0.228 nm</em>
Specific Gravity of the fluid = 1.25
Height h = 28 in
Atmospheric Pressure = 12.7 psia
Density of water = 62.4 lbm/ft^3 at 32F
Density of the Fluid = Specific Gravity of the fluid x Density of water = 1.25 x 62.4
Density of the Fluid p = 78 lbm/ft^3
Difference in pressure as we got the differential height, dP = p x g x h dP = (78 lbm/ft^3) x (32.174 ft/s^2) x (28/12 ft) [ 1 lbf / 32.174 ft/s^2] [1 ft^2 /
144in^2]
Difference in pressure = 1.26 psia
(a) Pressure in the arm that is at Higher
P = Atmospheric Pressure - Pressure difference = 12.7 - 1.26 = 11.44 psia
(b) Pressure in the tank that is at Lower
P = Atmospheric Pressure + Pressure difference = 12.7 + 1.26 = 13.96psia
Each time you add a bulb it would probably get more dim. I think. Hope this helps XD
Power in a wire where current is flowing can be calculated from the product of the square of the current and the resistance. Resistance is equal to the product of resistivity and length divided by the area of the wire. We do as follows:
Resistance = 2.44 × 10-8 ( 0.11) / (π)(0.0009)^2 = 1.055x10^-3 <span>Ω
P = I^2R = .170^2 (</span>1.055x10^-3 ) = 3.048x10^-5 W
Answer:
A)0.00966 N/C
B) counterclockwise direction
Explanation:
We are given;
Diameter of the metal ring; d = 4.3 cm
Radius;r = 2.15 cm = 0.021- m
Initial magnetic field, B = 1.12 T
Rate of decrease of the magnetic field;dB/dt = 0.23 T/s
Now, as a result of change in magnetic field, an emf will be induced in it. Thus, , electric field is induced and given by the formula :
∫E•dr = d/dt∫B.A •dA
This gives;
E(2πr) = dB/dt(πr²)
Gives;. 2E = dB/dt(r)
E = dB/dt × 2r
We are given;
E = 0.23 × 2(0.021)
E = 0.00966 N/C
The magnitude of the electric field induced in the ring has a magnitude of 0.00966 N/C
B) The direction of electric field will be in a counterclock wise direction when viewed by someone on the south pole of the magnet