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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:

brainly.com/question/24319848

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Answer:

<em>Explanation below</em>

Explanation:

<u>Speed vs Velocity </u>

These are two similar physical concepts. They only differ in the fact that the velocity is vectorial, i.e. having magnitude and direction, and the speed is scalar, just the magnitude regardless of the direction. They are strongly related to the concepts of displacement and distance, which are the vectorial and scalar versions of the space traveled by a moving object. The velocity can be computed as

$\displaystyle \vec v=\frac{\vec r}{t}$

Where $\vec r$ is the position vector and t is the time. The speed is

$\displaystyle v=\frac{d}{t}$

To compute $\vec r$ , we only need to know the initial and final positions and subtract them. To compute d, we need to add all the distances traveled by the object, regardless of their directions.

Maggie walks to a friend's house, located 1500 meters from her place. The initial position is 0 and the final position is 1500 m. The displacement is

$\vec r=1500\ m \text{ to the south}$

and the velocity is

$\displaystyle \vec v=\frac{1500}{45}=33.33\ m/s\text{ to the south}$

Now, we know Maggie had to make three different turns of direction to finally get there. This means her distance is more than 1500 m. Let's say she walked 500 m in all the turns, then the distance is

$d=1500+500=2000\ m$

If she took the same time to reach her destiny, she would have to run faster, because her average speed is

$\displaystyle v=\frac{2000}{45}=44.44\ m/s$

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3 years ago

I will make brainliest! PLS HELPPP!!

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Answer:

1.A compound is when two substances are put together and there is a reaction creating a new substance Ex. H20. A mixture is when two substances are put together, there is no reaction and they can be separated easier than compounds. Ex. salad dressing.

2.Atomic bonds

3.Compound

4.Element

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2 years ago

A uniform disk with mass 35.2 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stati

Sergio [31]

Answer:

a) v = 1.01 m/s

b) a = 5.6 m/s²

Explanation:

If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:

$\tau = I * \alpha (1)$

Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.
Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:
τ = F*r (2)
For a solid uniform disk, the rotational inertia regarding an axle passing through its center is just I = m*r²/2 (3).
Replacing (2) and (3) in (1), we can solve for α, as follows:

$\alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)$

Since the angular acceleration is constant, we can use the following kinematic equation:

$\omega_{f}^{2} - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)$

Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:

$0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)$

Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:

$\omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)$

Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:

$v = \omega * r (8)$

where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of the disk, r becomes the radius of the disk, 0.200 m.
Replacing this value and (7) in (8), we get:

$v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)$

There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:

$a_{t} = \alpha * r (9)$

where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.
Replacing this value and (4), in (9), we get:

$a_{t} = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)$

Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.
The radial acceleration is just the centripetal acceleration, that can be expressed as follows:

$a_{c} = \omega^{2} * r (11)$

Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).
Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.
Replacing this value and (7) in (11) we get:

$a_{c} = \omega^{2} * r = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)$

The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.
Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows:

$a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)$

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3 years ago

Energy that flows from somthing with a higher temperature to something with a lower

spayn [35]

Answer:

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Explanation:

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2 years ago

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A. through a relatively short distance.

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