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3 years ago

If the magnetic field in a traveling EM wave has a peak value of 17.9nT, what is the peak value of the electric field strength?

Physics

1 answer:

Doss [256]3 years ago

4 0

Answer:

5.37 N/C

Explanation:

Peak value of magnetic field, Bo = 17.9 nT = 17.9 x 10^-9 T

The electromagnetic wave is produced when an oscillating electric and magnetic field interacts each other perpendicularly.

The direction of propagation of electromagnetic wave is perpendicular to both electric and magnetic field.

the relation between the electric field and magnetic field amplitudes is given by

$c = \frac{E_{0}}{B_{0}}$

where, c be the velocity of light, Eo be the peak value of electric field strength, Bo is the peak value of magnetic field strength.

$3\times 10^{8} = \frac{E_{0}}{17.9\times 10^{-9}}}$

Eo = 5.37 N/C

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Answer:

Explanation:

The food chopping appliances in this scenario should adapt use of digital image processing systems to capture the images after the chopping begins. Certain variables can be defined like food particles size, area etc. This will help deciding a threshold for the variable and then the comparator used in the system will(run by real time operating system) compare the threshold size with the digital image size of particles. If the size is found small then the speed of motor should reduce meaning we have chopped the food as per our need. In case the chopping is not done aptly, the same can be enhanced by increasing the speed of motor till the final outcome is reached

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3 years ago

When a potential V is applied across a capacitor, a charge q is stored. What charge is stored if the potential across the capaci

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Answer:

q' = 0.5q

Hence, the charge stored is also halved

Explanation:

The capacitance of a capacitor is given by the following formula:

$C = \frac{q}{V}$ ------- eqn(1)

where,

C = Capacitance of the capacitor

q = charge stored in the capacitor

V = potential across the capacitor

Now, if we make potential half of its initial value:

$C = \frac{q'}{0.5V}\\\\CV = 2q'\\using\ eqn(1):\\\frac{q}{V}(V) = 2q'\\\\$

q' = 0.5q

Hence, the charge stored is also halved

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3 years ago

you went to move a 41 kg bookcase to a different place in the living room. if you push with a force of 65 N and the bookcase acc

ivann1987 [24]

Answer:

the force of friction = fk = (uk)n = 402(uk)

the net force acting to accelerate bookcase = ma = (41)(0.12) = 4.9 N

the {assumed horizontal} push force = 65 N

65 - fk = 4.9

fk = 65 - 4.9 = 60 N

uk = 60/402 = 0.15 <= ANS

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3 years ago

Responsibilities of a class are derived from ________________.

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Answer:

The correct option is a behavioural models of the the to-be system.

Explanation:

As the Use case analysis method generates the analysis classes list such that the classes are capable of performing the behavior needed to make the system function successfully. From these analysis classes list the responsibilities of each class are defined.

As the list is derived from the behavioural models of the system, thus option a is the correct option.

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3 years ago

A 0.600 kg block is attached to a spring with spring constant 15 N/m. While the block is sitting at rest, a student hits it with

gulaghasi [49]

Answer:

8.8 cm

31.422 cm/s

Explanation:

m = Mass of block = 0.6 kg

k = Spring constant = 15 N/m

x = Compression of spring

v = Velocity of block

A = Amplitude

As the energy of the system is conserved we have

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{0.6\times 0.44^2}{15}}\\\Rightarrow A=0.088\ m\\\Rightarrow A=8.8\ cm$

Amplitude of the oscillations is 8.8 cm

At x = 0.7 A

Again, as the energy of the system is conserved we have

$\dfrac{1}{2}kA^2=\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{k(A^2-x^2)}{m}}\\\Rightarrow v=\sqrt{\dfrac{15(0.088^2-(0.7\times 0.088)^2)}{0.6}}\\\Rightarrow v=0.31422\ m/s$

The block's speed is 31.422 cm/s

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3 years ago