All categories

3 years ago

What kind of electromagnetic waves do computers and microwave ovens produce?

1 answer:

5 0

Microwaves are a type of wave that are sandwiched between radio waves and infrared radiation on the electromagnetic spectrum. In the case of microwave ovens, the commonly used wave frequency is roughly 2,450 megahertz (2.45gigahertz).

You might be interested in

A slab of insulating material has thickness 2d and is oriented so that its faces are parallel to the yz-plane and given by the p

Anika [276]

Answer:

Electric field at (x, y, z): $E=\frac{\rho x}{\epsilon_o}$

Explanation:

(a) The slab is of an insulating material and has a uniform charge distribution. We can visualize this as infinite number of point charges, distributed throughout the slab, equally spaced apart. So if we (hypothetically) start to calculate the electric field due to each charge at x = 0, we shall always find a charge at a mirrored position about the x = 0 plane (within x = -d and x = d) and hence will cancel out the electric field.

A simpler example would be an infinitely long wire of uniform charge distribution. Any point on the wire will have zero electric field has there are essentially equal number of charges on either side (the length of the wire being infinitely long)

(b) Let us take a cylinder as a Gaussian surface with base area A. We shall take advantage of the symmetry about x = 0 and shall position the cylinder perpendicular to the y-z plane with x = 0 being the mid-point. Now the electric flux will only flow out through the 2 bases of the cylinder. This is because the slab has infinite dimensions along y and z-axes (think of an infinite sheet of charge) and the electric field always starts out perpendicular to any surface of charges. If the Electric field at some point on the base of the cylinder be E, then total outgoing flux = 2EA

$\rho$ is the charge density, hence, $Q_{enclosed}=\rho\times volume=2\rho Al$

where $2l$ is the length of the cylinder and $l$ is the x-coordinate.

Therefore, using Gauss's law,

$2EA=\frac{2\rho Al}{\epsilon_o}$

or, $E=\frac{\rho l}{\epsilon_o}$

or, $E=\frac{\rho x}{\epsilon_o}$

where, $\epsilon_o$ = permittivity of free space.

7 0

3 years ago

An unknown substance has a mass of 11.9 g . When the substance absorbs 1.071×102 J of heat, the temperature of the substance is

Mashcka [7]

Answer: The most likely identity of the substance is iron.

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

$Q=m\times c\times \Delta T$

Q = Heat absorbed= $1.071\times 10^2$ Joules

m= mass of substance = 11.9 g

c = specific heat capacity = ?

Initial temperature of the water = $T_i$ = 25.0°C

Final temperature of the water = $T_f$ = 45.0°CChange in temperature , $\Delta T=T_f-T_i=(45-25)^0C=20^0C$

Putting in the values, we get:

$1.071\times 10^2=11.9\times c\times 20^0C$

$c=0.45J/g^0C$

The specific heat of 0.45 is for iron and thus the substance is iron.

3 0

4 years ago

A car moving at 44 km/ h skids 16 m with locked brakes. How far will the car skid with locked brakes at 110km/ h

svp [43]

Answer:

100 m

Explanation:

44 km/h = 12.2 m/s

110 km/h = 30.6 m/s

Given:

Δx = 16 m

v₀ = 12.2 m/s

v = 0 m/s

Find: a

v² = v₀² + 2aΔx

(0 m/s)² = (12.2 m/s)² + 2a (16 m)

a = -4.67 m/s²

Given:

v₀ = 30.6 m/s

v = 0 m/s

a = -4.67 m/s²

Find: Δx

v² = v₀² + 2aΔx

(0 m/s)² = (30.6 m/s)² + 2 (-4.67 m/s²) Δx

Δx = 100 m

7 0

3 years ago

To change a tire, you need to use a jack to raise one corner of your car. While doing so, you happen to notice that pushing the

Alja [10]

Answer:

Explanation:

The energy lost by the handle in lowering is equal to the energy gained by the car in raising the height.

As the mass of car is much more as compared to the jack, so the height raised by the car is very small as compared to the handle of jack.

7 0

3 years ago

Please help!!!!!!!!!

aleksandr82 [10.1K]

1. This question asks about velocity, so A and B are not correct. The car's velocity after 15 s with acceleration 2.00 m/s² would be

(2.00 m/s²) (15 s) = 30 m/s

[D]

2. Because Ima is slowing down to a stop, the acceleration is negative. Let x be the displacement of her vehicle during this motion. Then

0² - (30.0 m/s)² = 2 (-8.00 m/s²) x

==> x = (30.0 m/s)²/(2 (8.00 m/s²)) = 56.25 m ≈ 65.3 m

[A]

3. Since acceleration is constant, the average velocity is exactly the average of the initial and final velocities:

(21.0 m/s + 0 m/s)/2 = 10.5 m/s

The average (and thus instantaneous) acceleration during this time is equal to the change in velocity divided by the change in time:

(0 m/s - 21.0 m/s)/(6.00 s) = -3.50 m/s²

If x is the distance traveled as the car comes to a stop, then

0² - (21.0 m/s)² = 2 (-3.50 m/s²) x

==> x = (21.0 m/s)² / (2 (3.50 m/s²)) = 63.0 m

[A]

4.a. Assuming the sprinter's acceleration is constant, the average acceleration would be a such that

(11.5 m/s)² - 0² = 2 a (15.0 m)

==> a = (11.5 m/s)² / (2 (15.0 m)) ≈ 4.41 m/s²

4.b. By definition of average acceleration,

4.41 m/s² = (11.5 m/s - 0 m/s)/t

==> t = (11.5 m/s)/(4.41 m/s²) ≈ 2.61 s

5. At maximum height, any thrown object has zero velocity, so if it was thrown with an initial speed v, at its highest point we have

0² - v ² = 2 (-g) (91.5 m)

==> v = √(2g (91.5 m)) ≈ 42.3 m/s

(where I use g = 9.80 m/s²)

6.a. The brick's velocity after 7.0 s is

-g (7.0 s) = -68.6 m/s ≈ -69 m/s

6.b. The brick is presumably dropped from rest, so it is displaced by x such that

(-68.6 m/s)² - 0² = 2 (-g) x

==> x = -240.1 m ≈ -240 m

which is to say it falls a distance of 240 m. (The displacement is negative because we take its initial position to be the origin, and I took the downward direction to be negative.)

3 0

3 years ago