Answer:
Magnitude of the net force on q₁-
Fn₁=1403 N
Magnitude of the net force on q₂+
Fn₂= 810 N
Magnitude of the net force on q₃+
Fn₃= 810 N
Explanation:
Look at the attached graphic:
The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.
Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:
F= (k*q*q)/(d)²
F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N
Magnitude of the net force on q₁-
Fn₁x= 0
Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N
Fn₁=1403 N
Magnitude of the net force on q₃+
Fn₃x= 810- 810 cos 60° = 405 N
Fn₃y= 810*sin 60° = 701.5 N
Fn₃ = 810 N
Magnitude of the net force on q₂+
Fn₂ = Fn₃ = 810 N
The force of attraction between the opposite charges of the ions in an ionic compound is an ionic bond.
<u>Explanation:</u>
The transfer process of valence electron between atoms referred as ionic bond. This is a kind of chemical bonds which can create two oppositely charged ions. In the presence of ionic bonds, the metal loses electrons and becomes a positive charge cation, while non-metal accepts these electrons and becomes a negative charge anion.
Here, more than 1 electron can be emitted or received to meet the octet principle and the net charge of the compound should be zero. For example: Table salt. In this compound, sodium loses the electron to become 
, while the chlorine loses the electron to become 
.
Answer: Because the fuse can disconnect the circuit only when the excess current flows completely through the neutral. ... Since, neutral is not a live conductor coming from the source, disconnecting a neutral line can only open the current path through neutral. But, the live phase still carries the charge.
Explanation:
C. 340
Frequency is the number of wavelengths per second and since the length is 0.5 you multiply 0.5*686 and get 343.
the question not allowing for one position thus the answer is c
Given Information:
Power of bulb = w = 25 W
atts
distance = d = 9.5 cm = 0.095 m
Required Information:
Radiation Pressure = ?
Answer:
Radiation Pressure =7.34x10⁻⁷ N/m²
Explanation:
We know that radiation pressure is given by
P = I/c
Where I is the intensity of radiation and is given by
I = w/4πd²
Where w is the power of the bulb in watts and d is the distance from the center of the bulb.
So the radiation pressure becomes
P = w/c4πd²
Where c = 3x10⁸ m/s is the speed of light
P = 25/(3x10⁸*4*π*0.095²)
P = 7.34x10⁻⁷ N/m²
Therefore, the radiation pressure due to a 25 W bulb at a distance of 9.5 cm from the center of the bulb is 7.34x10⁻⁷ N/m²
