C is the answer Because chemical energy Create heat And thermal power relates to heat energy
Answer:
two-slit interference model was proposed by Young d sin θ = m λ
Explanation:
The two-slit interference model was proposed by Young, it establishes that if a coherent source of light passes through two slits, the shape of the given pattern is a consequence of the relative phase difference between the two rays; mathematically it can be expressed by
d sin θ = m λ
m= 0, 1, 2, 3, ...
for constructive interference, that is, the two rays arrive with a number between wavelengths.
D is the distance between the slits, tea the angle between the two rays, m an integer and m the wavelength used.
In a simulation a pattern of slits of equal intensity and equally spaced is observed.
<span>Germanium
To determine which melts first, convert their melting temperatures so they're both expressed on same scale. It doesn't matter what scale you use, Kelvin, Celsius, of Fahrenheit. Just as long as it's the same scale for everything. Since we already have one substance expressed in Kelvin and since it's easy to convert from Celsius to Kelvin, I'll use Kelvin. So convert the melting point from Celsius to Kelvin for Gold by adding 273.15
1064 + 273.15 = 1337.15 K
So Germanium melts at 1210K and Gold melts at 1337.15K. Germanium has the lower melting point, so it melts first.</span>
Answer:
Explanation:
Given that,
Radius of solenoid R = 4cm = 0.04m
Turn per length is N/l = 800 turns/m
The rate at which current is increasing di/dt = 3 A/s
Induced electric field?
At r = 2.2cm=0.022m
µo = 4π × 10^-7 Wb/A•m
The magnetic field inside a solenoid is give as
B = µo•N•I
The value of electric field (E) can
only be a function of the distance r from the solenoid’s axis and it give as,
From gauss law
∮E•dA =qenc/εo
We can find the tangential component of the electric field from Faraday’s law
∮E•dl = −dΦB/dt
We choose the path to be a circle of radius r centered on the cylinder axis. Because all the requested radii are inside the solenoid, the flux-area is the entire πr² area within the loop.
E∮dl = −d/dt •(πr²B)
2πrE = −πr²dB/dt
2πrE = −πr² d/dt(µo•N•I)
2πrE = −πr² × µo•N•dI/dt
Divide both sides by 2πr
E =- ½ r•µo•N•dI/dt
Now, substituting the given data
E = -½ × 0.022 × 4π ×10^-7 × 800 × 3
E = —3.32 × 10^-5 V/m
E = —33.2 µV/m
The magnitude of the electric field at a point 2.2 cm from the solenoid axis is 33.2 µV/m
where the negative sign denotes counter-clockwise electric field when looking along the direction of the solenoid’s magnetic field.
