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2 years ago

The peak magnetic field strength in a residential microwave oven is 9. 2 x 10-5 t. what is the intensity of the microwave?

Physics

1 answer:

Alexandra [31]2 years ago

7 0

The intensity of the microwave is 10.09 × 10⁵ W/m².

The unit of magnetic induction is the tesla (T). The magnetizing force, which induces the lines of force through a material, is called the field intensity, H (or H-field), and by convention has the units ampere per meter (A m−1) .

The portion of a material's magnetic field that results from an external current and is not intrinsic to the material itself is known as the magnetic field strength, also known as magnetic intensity or magnetic field intensity. It is measured in amperes per meter and represented as the vector H.

Magnetic Field, B = 9.2 × 10⁻⁵ T

$I_{avg} = \frac{CB_0^2}{2\mu _0}$

$= \frac{(2.998*10^8 m/s) (9.2 * 10^-^5 T)^2}{2 ( 4\pi * 10^-^7)}$

$= 10.09$ × $10^5 \frac{W}{m^2}$

Therefore, the intensity of the microwave is 10.09 × 10⁵ W/m².

Learn more about intensity here:

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3 years ago

From a window that is 20 m from the ground a stone with a speed of 10m / s is thrown vertically upwards. Calculate:

Oduvanchick [21]

consider the motion in upward direction as positive and down direction as negative

Y₀ = initial position of the stone = 20 m

v₀ = initial velocity of the stone = 10 m/s

a = acceleration = - 9.8 m/s²

Y = final position of the stone when it reach the maximum height

v = final velocity at the maximum height = 0 m/s

t = time taken to reach the maximum height

Using the equation

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0² = 10² + 2 (- 9.8) (Y - 20)

Y = 25.1 m

also using the equation

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inserting the values

0 = 10 + (- 9.8) t

t = 1.02 sec

consider the motion in upward direction as positive and down direction as negative

Y₀ = initial position of the stone = 20 m

v₀ = initial velocity of the stone = 10 m/s

a = acceleration = - 9.8 m/s²

Y = final position of the stone when it reach the ground = 0 m

t = time taken to reach the ground

Using the equation

Y = Y₀ + v₀ t + (0.5) a t²

0 = 20 + 10 t + (0.5) (- 9.8) t²

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3 years ago

Water enters the constant 130-mm inside-diameter tubes of a boiler at 7 MPa and 65°C and leaves the tubes at 6 MPa and 450°C wit

snow_lady [41]

The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

Explanation:

When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.

As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.

Since, the inlet volume flow is measured as the product of velocity with the area.

Inlet volume flow=Inlet velocity*Area*time

And the mass flow rate is

Mass flow rate in the inlet=density*area*inlet velocity*time

Mass flow rate in the outlet=density*area*outlet velocity*time

Since, the time and area is constant, the inlet and outlet will be same as

(Mass inlet)/(density*inlet velocity)=Area*Time

(Mass outlet)/(density*outlet velocity)=Area*Time

As the ratio of mass to density is termed as specific volume, then

(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

Inlet velocity= (Specific volume inlet)/(Specific volume outlet)*Outlet velocity

As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity

$Inlet velocity = \frac{0.001017}{0.05217}*72 =1.4035 m/s$

So, the inlet velocity is 1.4035 m/s.

Then the inlet volume will be

$Inlet volume = inlet velocity*area of circle=\pi r^{2}*inlet velocity$

As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s

$Inlet volume = 1.4*3.14*65*65*10^{-6} =0.019 \frac{m^{3} }{s}$

So, the inlet volume is 0.019 m³/s.

Thus, the inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

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