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Marianna [84]
3 years ago
7

What are some uses of the electromagnetic

Physics
1 answer:
den301095 [7]3 years ago
8 0
If you mean electromagnetic waves
Radio waves can be used well in radio signals
Infrared in remote controls
X rays for x-rays
Gamma rays in a nuclear power plant
Uv rays can cause cancer but they transmit vitamin D
You might be interested in
What experiment should I make using Gravitational Force? <br><br>PLEASE HELP ME :)
Bogdan [553]

You could try the "Spinning Bucket" or the "Center Of Gravity" experiment. There are plenty more that you could research! Hope this helped :)

8 0
3 years ago
Uma partícula com carga “Q”, no vácuo, gera um potencial elétrico de 600 volts a uma distância de 6,0 m, determine o valor dessa
garik1379 [7]

Answer:

q =  400 nC

the correct answer is b

Explanation:

The expression for the electric potential of a point charge is

            V = k q / r

they ask us for the electrical charge

          q = V r / k

let's calculate

         Q = 600 6.0 / 9 10⁹

         Q = 4 10⁻⁷ C

let's reduce to nC

          Q = 4 10⁻⁷ C (10⁹ nC / 1C)

          q = 4 10² nC = 400 nC

the correct answer is b

Traslate

La expresión para el potencial eléctrico de una carga puntual es

            V = k q/r

nos piden la carga eléctrica

          q= V r /k

          calculemos

         Q= 600  6,0 / 9 10⁹

         Q=  4 10⁻⁷ C

reduzcamos a nC

          Q = 4 10⁻⁷ C(10⁹ nC/1C )  

          q = 4 10² nC = 400 nC

la respuesta correcta es b

8 0
3 years ago
Objects A and B are separated by 2m and the force of attraction between them is 8 x 10^-10 N . If the mass of A is 8kg what is t
Jobisdone [24]

Answer:

Only object a and b???????

5 0
2 years ago
What is the quantity of heat energy required to convert 10g cube of ice at -30oC to steam at 120oC. also draw a graph of tempera
Nadusha1986 [10]

Answer:

The amount of energy required is 31.1692 kJ .

Explanation:

The transitions of water are as follows,

(243K ice to 273K ice to 273K water to 373k water to 373K steam to 393K steam)

The following required data is,

specific heat capacity of ice= 2.1 kJ/kg

specific heat capacity of water= 4.2 kJ/kg

specific heat capacity of ice= 1.996 kJ/kg

specific latent heat of fusion=334 kJ/kg

specific latent heat of vapourisation = 2260 kJ/kg

FORMULAS:-

temperature change=mc(T2-T1)

phase transistion = mL,

where, m=mass , c=specific heat capacity ,L = latent heat of fusion or vapourisation ,(T2-T1)= temparature change

Thus the total amount of heat is,

=mc(T4-T3) + mL + mc(T3-T2) + mL + mc(T2-T1)

=10(10^{-3} )(2.1)(30)+10(10^{-3} )(334)+10(10^{-3} )(100)(4.2)+10(10^{-3} )(2260) +10(10^{-3} )(20)(1.996)

=(0.3)(2.1) + (3.34) + (4.2) +  (22.6) + (0.2)(1.996)

=31.1692 kJ.

3 0
3 years ago
A biker pedals hard to ride his bike to the top of a 44 m hill. He decides to let his bike coast down the hill, and is having so
Dafna11 [192]

Answer:

The bikers speed at the top of other hill is <u>25.82 m/s.</u>

Explanation:

Considering the biker is riding on a frictionless surface.

∴ There is no non-conservative or external force acting on the biker.

Hence we can conserve the energy of biker and bike as a system.

Let,

h_{1} = 44m

h_{2} = 10m

Since the biker starts from rest , his initial speed v_{1} = 0 m/s

Let final speed of the bike at the top of other hill be v_{2}.

∴ Initial Energy (at the top of 44m hill) = mgh_{1}

  Final Energy  (at the top of 10m hill) =  mgh_{2} + \frac{1}{2}mv_{2} ^{2}.

Conserving both the energies , we get

mgh_{1} = mgh_{2} + \frac{1}{2}mv_{2} ^{2}

∴ v_{2} = \sqrt{2g(h_{1}-h_{2} )}

Substituting the values for g , h_{1} , h_{2} , we get

v_{2} = 25.82 m/s

6 0
3 years ago
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