Answer:
a)906.5 Nm^2/C
b) 0
c) 742.56132 N•m^2/C
Explanation:
a) The plane is parallel to the yz-plane.
We know that
flux ∅= EAcosθ
3.7×1000×0.350×0.700=906.5 N•m^2/C
(b) The plane is parallel to the xy-plane.
here theta = 90 degree
therefore,
0 N•m^2/C
(c) The plane contains the y-axis, and its normal makes an angle of 35.0° with the x-axis.
therefore, applying the flux formula we get
3.7×1000×0.3500×0.700×cos35°= 742.56132 N•m^2/C
Answer:
(a) 2800 J
(b) 31.28 %
Explanation:
Heat intake, Q1 = 8950 J
Heat discard, Q2 = 6150 J
(a) Work = intake heat energy - discarded heat energy
W = Q1 - Q2 = 8950 - 6150 = 2800 J
(b) Efficiency = 1 - Q2 / Q1
Efficiency = 1 - (6150 / 8950) = 0.3128 = 31.28%
Hello!
In chemistry, atomic mass is normally written in amu. This stands for atomic mass units.
Within an atom, there are three main parts: protons, neutrons, and electrons. Between these three, electrons are extremely light. However, protons and neutrons both weigh 1 amu.
Thus, this means to get the atomic mass, add together the protons + neutrons.
Hope this helps!
Answer:
A
Explanation:
According to law of conservation of momentum, the total momentual in the system will be conserved
This question is incomplete, the complete question is;
Now we will examine the electric field of a dipole. The magnitude and direction of the electric field depends on the distance and the direction. We will investigate in detail just two directions. With charges available in the simulation (all the charges are either positive or negative 1 nC increments).
how do you create a dipole with dipole moment 1 x 10-9 Cm with a direction for the dipole moment pointing to the right. Make a table below that shows the amounts of charge and the distance between the charges. There are many correct answers
Answer:
Given the data in question;
Dipole moment P = 1 × 10⁻⁹ C.m
now dipole pointing to the right;
P→
(-) ---------------->(+) 
d
so let distance between the dipoles be d
∴ P = d
Let
= 1 nC
so
P = d
1 × 10⁻⁹ = 1 × 10⁻⁹ × d
d = (1 × 10⁻⁹) / (1 × 10⁻⁹)
d = 1 m
Also Let
= 2 nC
so
P = d
1 × 10⁻⁹ = 2 × 10⁻⁹ × d
d = (1 × 10⁻⁹) / (2 × 10⁻⁹)
d = 0.5 m
Also Let
= 3 nC
so
P = d
1 × 10⁻⁹ = 3 × 10⁻⁹ × d
d = (1 × 10⁻⁹) / (3 × 10⁻⁹)
d = 0.33 m
such that;
charge distance
1 nC 1.00 m
2 nC 0.50 m
3 nc 0.33 m
4 nC 0.25 m
5 nC 0.20 m