__5. The study of weather patterns can predict the trajectory and intensity of this

Calculate the weight of a new fast-food sandwich that has a mass of 0.1 kg (approximately the mass of a quarter pound). Think of

iragen [17]

Weight = (mass) x (gravity)

If you plan to sell these things on Earth, then the acceleration of gravity in the neighborhood of your drive-throughs will be 9.81 m/s².

Weight of each sandwich = (0.1 kg) x (9.81 m/s²).

Weight of each sandwich = 0.981 Newton.

This is only 1.9% less than 1 even Newton.

You should start by setting up one restaurant in New York, one in Chicago, one in LA, and maybe one in Miami or Tulsa. Sell it with a different name in each place, and see which name sells best.

You might want to try calling it

-- Isaac's burger

-- Gravity grub

-- Prism Patty

-- Mass 'o Meat

-- Unit-wich

and see if anything catches on.

I think I'd simply call it a "Newton Unit".

5 0

2 years ago

Two isolated, concentric, conducting spherical shells have radii R1 = 0.500 m and R2 = 1.00 m, uniform charges q1=+2.00 µC and q

scZoUnD [109]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

a E = $1.685*10^3 N/C$

b E = $36.69*10^3 N/C$

c E = 0 N/C

d $V = 6.7*10^3 V$

e $V = 26.79*10^3V$

f V = $34.67 *10^3 V$

g $V= 44.95*10^3 V$

h $V= 44.95*10^3 V$

i $V= 44.95*10^3 V$

Explanation:

From the question we are given that

The first charge $q_1 = 2.00 \mu C = 2.00*10^{-6} C$

The second charge $q_2 =1.00 \muC = 1.00*10^{-6}$

The first radius $R_1 = 0.500m$

The second radius $R_2 = 1.00m$

$Generally \ Electric \ field = \frac{1}{4\pi\epsilon_0}\frac{q_1+\ q_2}{r^2}$

And $Potential \ Difference = \frac{1}{4\pi \epsilon_0} [\frac{q_1 }{r}+\frac{q_2}{R_2} ]$

The objective is to obtain the the magnitude of electric for different cases

And the potential difference for other cases

Considering a

r = 4.00 m

$E = \frac{((2+1)*10^{-6})*8.99*10^9}{16}$

$= 1.685*10^3 N/C$

Considering b

$r = 0.700 m \ , R_2 > r > R_1$

This implies that the electric field would be

$E = \frac{1}{4\pi \epsilon_0}\frac{q_1}{r^2}$

This because it the electric filed of the charge which is below it in distance that it would feel

$E = 8*99*10^9 \frac{2*10^{-6}}{0.4900}$

= $36.69*10^3 N/C$

Considering c

r = 0.200 m

=> $r$

The electric field = 0

This is because the both charge are above it in terms of distance so it wont feel the effect of their electric field

Considering d

r = 4.00 m

=> $r > R_1 >r>R_2$

Now the potential difference is

$V =\frac{1}{4\pi \epsilon_0} \frac{q_1 + \ q_2}{r} = 8.99*10^9 * \frac{3*10^{-6}}{4} = 6.7*10^3 V$

This so because the distance between the charge we are considering is further than the two charges given

Considering e

r = 1.00 m $R_2 = r > R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{1.00} \frac{1.00*10^{-6}}{1.00} ] = 26.79 *10^3 V$

Considering f

$r = 0.700 m \ , R_2 > r > R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.700} \frac{1.0*10^{-6}}{1.00} ] = 34.67 *10^3 V$

Considering g

$r =0.500\m , R_1 >r =R_1$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

Considering h

$r =0.200\m , R_1 >R_1>r$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

Considering i

$r =0\ m \ , R_1 >R_1>r$

$V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2} ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V$

8 0

2 years ago

A radio station broadcast on a frequency of 3.7 mhz what is the energy of the radio wave

Ratling [72]

A radio station broadcast on a frequency of 3.7 mhz what is the energy of the radio wave A radio station broadcasts its programmes at a wavelength of 500 m. Find the frequency of the radiowaves transmitted by the radio station, if the speed of radiowaves in air is 3 x 108 m/s. Ans: 6 x 10 Hz

<h3>What is radio station ?</h3>

Radio broadcasting is the act of sending audio (sound), occasionally together with accompanying metadata, across radio waves to radio receivers used by the general public. Unlike satellite radio, which uses a satellite in Earth's orbit, terrestrial radio broadcasting uses a land-based radio station to transmit radio waves. The listener needs a broadcast radio receiver to hear the material (radio). A radio network with which stations frequently have affiliations provide content in a standard radio format, whether through broadcast syndication, simulcasting, or both. Radio stations use a variety of modulations to transmit their signals, including FM (frequency modulation), which is an older analog audio standard, and AM (amplitude modulation).

To learn more about radio station from the given link:

brainly.com/question/26439029

#SPJ4

3 0

1 year ago

HELP ASAP PLS ILL GIVE YOU BRAINLIEST

lys-0071 [83]

Answer:

chinese

Explanation:

8 0

2 years ago

Read 2 more answers

The si unit of measurement is a way for scientists to?

Vera_Pavlovna [14]

Have a universal record base. Everyone is able to understand the data compiled since the same measurement systems are being used around the world. This is just to simplify all of the information.

7 0

2 years ago