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3 years ago

What are three ways in which people use microwaves

Physics

2 answers:

lora16 [44]3 years ago

7 0

Answer:

Explanation:

Microwaves are the electromagnetic waves which are produced by the specially designed vacuum tubes such as magnetrons, gunn diodes, etc.

They are used in the following manner:

1. In Radar systems for the navigation.

2. in microwave ovens for the cooking purposes.

3. In observing the movements of trains on rails.

elena55 [62]3 years ago

3 0

Answer:

1.) Radar

2.) Cancer Treatment

4.) Cooking

Explanation:

Microwaves are a type of electromagnetic radiations, like radio waves, X-rays, Gamma Rays, ultraviolet radiation. Microwaves has a wide range of applications. It has frequency range 1GHz to 300GHz and has smaller wavelengths. As microwaves do not bend by any obstacles coming in their path, So, it can be used in radar. In cancer treatment, it is used to increase the temperature of tumor cells by high-power microwaves, results in turns cell membrane damage, which in results destruction of cancer cells. In cooking, microwave heat food like the sun rays your face i.e by means of radiation. It increases the energy of molecules inside food, which results in cooking or heating the food.

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Does potential energy increase,kinetic energy decrease when a book is placed on a shelf

DerKrebs [107]

Yes potential increases while kinetic decreases

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3 years ago

In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s. Ignore fr

spin [16.1K]

Answer:

a) $v_p=9.35m/s$

Explanation:

From the question we are told that:

Open cart of mass $M_o=50.0 kg$

Speed of cart $V=5.00m/s$

Mass of package $M_p=15.0kg$

Speed of package at end of chute $V_c=3.00m/s$

Angle of inclination $\angle =37$

Distance of chute from bottom of cart $d_x=4.00m$

Generally the equation for work energy theory is mathematically given by

$\frac{1}{2}mu^2+mgh=\frac{1}{2}mv_p^2$

Therefore

$\frac{1}{2}u^2+gh=\frac{1}{2}v_p^2$

$v_p=\sqrt{2(\frac{1}{2}u^2+gh)}$

$v_p=\sqrt{2(\frac{1}{2}v_c^2+gd_x)}$

$v_p=\sqrt{2(\frac{1}{2}(3)^2+(9.8)(4))}$

$v_p=9.35m/s$

4 0

3 years ago

The period (T) of an oscillating wave is 1/5s. What happens to the frequency (f) of the wave if T increases to 1/2s

Anastasy [175]

Frequency = 1/T
as the 5 is reduced, frequency is increase.
as 1 whole wave travels through a point in a lesser time now

6 0

3 years ago

R S ( M ) = 2 G M c 2 , where G is the gravitational constant and c is the speed of light. It is okay if you do not follow the d

padilas [110]

The provided question's answer is "Schwarzschild radius".

The conversion factor between mass and energy is the speed of light squared.

GM/r stands for gravitational potential energy, also known as energy per unit mass.

GM/rc² then has "mass per unit mass" units. In other words, as mass/mass splits out in a dimensional analysis, "dimensionless per unit."

The derivation yields a formula for time or space coordinate ratios requiring sqrt(1 - 2GM/rc²). This number becomes 0 when r=2GM/c2, or the formula becomes infinite if in the denominator. However, there is no justification for using c² as a conversion factor there. Consider the initial expression sqrt(1 - 2GM/rc²).

Assume that m is used as the test particle's mass instead of 1. Then you have sqrt(m - 2GMm/rc² and mass units. This expression denotes that the rest energy of the test mass m you introduced into the gravitational field is "gone" at that radius.

The 2 would be absent if the gravitational field were Newtonian. However, at the event horizon, Einstein gravity is slightly stronger than Newton gravity, resulting in the factor 2 in qualitative terms.

So, the given equation is of Schwarzschild radius.

Learn more about Schwarzschild radius here:

brainly.com/question/12647190

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2 years ago

Which of the following is necessary for the greenhouse effect?

Mice21 [21]

It is the atmosphere

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3 years ago