Answer:
The electric force on q₁ is .
Explanation:
Given that,
Charge on
Distance
Charge on at origin
Distance
Charge on
We need to calculate the electric force on q₁
Using formula of electric force
Put the value into the formula
Negative sign shows the attraction force.
We need to calculate the electric force F₁₃
Positive sign shows the repulsive force.
We need to calculate the net electric force
Hence, The electric force on q₁ is .
Answer:
Exposure to very high levels of radiation, such as being close to an atomic blast, can cause acute health effects such as skin burns and acute radiation syndrome (“radiation sickness"). It can also result in long-term health effects such as cancer and cardiovascular disease
Answer:
Induced emf of the wire is 6.36 Volts.
Explanation:
It is given that,
Length of the wire, l = 75 cm = 0.75 m
Magnetic field, B = 0.53 T
Velocity, v = 16 m/s
The wire is moving straight up in the magnetic field. So, an emf is induced in the wire. It is given by :
So, the induced emf of the wire is 6.36 V. Hence, the correct option is (b) "6.36 V".
Answer: Category 6 cable
Explanation:
Category 6 cable also called Cat 6 is a standard form of twisted cable pair that is used in Ethernet as well as in other physical network layers that is compatible with the category 5/5e and Cat 3 standard backwards.
Cat 6 has a 500MHZ with better alien crosstalk feature which permits 10 GBASE - 1 to be run at 100 meters maximum which is the same distance at which other Ethernet variants are run.
Cat 6 cable is used in cabling infrastructure such as 10BASE-T (Ethernet) 100BASE-TX (fast ethernet) etc.
Ans:
12500 N/C
Explanation:
Side of square, a = 2.42 m
q = 4.25 x 10^-6 C
The formula for the electric field is given by
where, K be the constant = 9 x 10^9 Nm^2/c^2 and r be the distance between the two charges
According to the diagram
BD =
where, a be the side of the square
So, Electric field at B due to charge at A
EA = 6531.32 N/C
Electric field at B due to charge at C
Ec = 6531.32 N/C
Electric field at B due to charge at D
ED = 3265.66 N/C
Now resolve the components along X axis and Y axis
Ex = EA + ED Cos 45 = 6531.32 + 3265.66 x 0.707 = 8840.5 N/C
Ey = Ec + ED Sin 45 = 6531.32 + 3265.66 x 0.707 = 8840.5 N/C
The resultant electric field at B is given by
E = 12500 N/C
Explanation: