Can u please help mee?

Explain, briefly, what the electrical field inside a microwave oven does to heat food.

yarga [219]

Answer:

Inside the strong metal box, there is a microwave generator called a magnetron. When you start cooking, the magnetron takes electricity from the power outlet and converts it into high-powered, 12cm (4.7 inch) radio waves. Thus the microwaves pass their energy onto the molecules in the food, rapidly heating it up.May 3, 2018

5 0

3 years ago

a 550 kg car accelerates from 15 m/s to 25 m/s in 15 s by applying a constant force. how large of a force is exerted? Please sho

Novay_Z [31]

You'll be using the equation f = m a, or force = mass x acceleration

First, you have to find the acceleration. The acceleration needed is the average acceleration over the 15 seconds is accelerated. So, you take the change in speed (25m/s - 15m/s) to get a change of 10m/s.
The average acceleration (acceleration per second) is found by dividing total acceleration by the time it took. So, it's 10 / 15, which equals .6. This is a, your acceleration
Now just plug it into the equation F = m a, because it already gives you the mass of the car
F = 550 x .6
Solve that to get F = 366.6. F is measured in Newtons (N), so your answer is 366.6N

3 0

3 years ago

Why is raw sewage a major pollutant in some countries but not in developed countries?

Damm [24]

Answer:because developed cities are more equipped to deal with raw sewage than less developed cities

Explanation:

3 0

3 years ago

A jet plane traveling 1890 km/h (525 m/s) pulls out of a dive by moving in an arc of radius 5.20 km. What is the plane's acceler

Tema [17]

Answer:

Acceleration of the plane, a = 5.4 g

Explanation:

It is given that,

Speed of the jet plane, v = 1890 km/h = 525 m/s

Radius of the arc, r = 5.20 km = 5200 m

The plane is moving in the circular path, the centripetal acceleration will act on it. It is given by :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(525\ m/s)^2}{5200\ m}$

$a=53.004\ m/s^2$

We know that, the value of g is, $g=9.8\ m/s^2$

$\dfrac{a}{g}=\dfrac{53.004\ m/s^2}{9.8\ m/s^2}=5.4$

$a=5.4\times g$

So, the acceleration of the plane is 5.4 g. Hence, this is the required solution.

6 0

3 years ago

Read 2 more answers

A tank whose bottom is a mirror is filled with water to a depth of 20.0 cm. A small fish floats motionless 7.0 cm under the surf

liberstina [14]

Answer:

The apparent depth of (a) the fish is 5.3 cm and (b) the image of the fish is 24.8 cm.

Explanation:

According to the following equation:

$\frac{n_{w} }{s} +\frac{n_{a} }{s'} = \frac{n_{a}- n_{w}}{R_{c} } \\$

where nw and na is the refractive indices of water (1.33) and air (1.00) respectively; s is the depth of the fish below the surface of the water; s' is the apparent depth of the fish from normal incidence and Rc is the radius of curvature of the mirror at the bottom of the tank.

Note that the bottom of the tank is assumed to be a flat mirror, therefore the radius of curvature is very large (R⇒∞).

Therefore, the above equation can be expressed as:

$\frac{n_{w}}{s} +\frac{n_{a}}{s'}=0$

Now we can solve for the apparent depth of the fish.

(a) $s'=-(\frac{n_{a}}{n_{w}})x s$ (Make s' subject of the formula from the above equation)

$s'=(\frac{1.00}{1.33} )x7cm$

∴ $s'=5.3 cm.$

(b) The motionless fish floats 13 cm above the mirror, therefore the image of the fish will be situated at 13 + 20 =33 cm away from the real fish.

Therefore, s = 33 cm

$s'=-(\frac{n_{a}}{n_{w}})x s$

$s'=(\frac{1.00}{1.33} )x 33 cm$

$s'=24.8 cm.$

NB: Here, it is assumed that the water is pure, as impurities may alter the refractive index of water.

8 0

4 years ago