QUICK WILL MARK BRAINLIEST

With a thunderstorm brewing, an electric field of magnitude 2.0 × 102 newtons/coulomb exists at a certain point in the earth’s a

Aleks04 [339]

Answer:

The correct option is (b).

Explanation:

Given that,

Electric field, $E=2\times 10^2\ N/C$

We need to find the magnitude of the force on the electron as a result of the electric field.

We know that, the electric force is given by :

$F=qE\\\\=1.6\times 10^{-19}\times 2\times 10^2\\\\F=3.2\times 10^{-17}\ N$

So, the required force on the electron is equal to $3.2\times 10^{-17}\ N$ .

5 0

3 years ago

You can carry a 50 N weight up a flight of stairs that is 3 m high in 10 s. Then you carry 40 N up two flights of stairs in 20 s

d1i1m1o1n [39]

You can do 50 and 10 and carry the 30 and it would be in the same power so you’ll have the same energy

5 0

3 years ago

Light emitting diode (LEDs) light bulbs have become required in recent years, but do they make financial sense? Suppose a typica

kogti [31]

Answer:

The break even cost is $0.0063825

Explanation:

Break-even cost is the amount of money, or change in value, which equates to the amount at which an asset must be sold to equal the cost of acquiring it. For easier understanding it can be thought the amount of money for which a product or service must be sold to cover the costs of manufacturing or providing it.

Wattage = W

Cost per kilo watt hour = C

Number of hours per year = H

Price per bulb/CFL = P

Discount rate = 11%

Life of bulb = 2 years

Price of bulb = $0.39

Wattage consumption of bulb per hours = 60

Life of CFL = 24 years

Price of CFL = $3.10

Wattage consumption of CFL per hour = 15

Calculate the Equated Annual Cost (EAC) of bulb

EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 2years)}/ (PVIFA 11%, 2years)

PVIFA 11%, 2years = Annuity PV Factor = [1 – {(1 + r)^(-n)}]/r, where r is the rate per period and n is the number per periods

PVIFA 11%, 2 years = [1 – {(1 + 0.11)^(-2)}]/0.11 = 1.712523 (for 2 years)

PVIFA 11%, 24 years = [1 – {(1 + 0.11)^(-24)}]/0.11 = 8.348136 (for 2 years)

Calculate the EAC of bulb

EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 2 years)}/ (PVIFA 11%, 2 years)

EAC = {- 0.39 - (60/1000 x H x C) x (1.712523)}/ (1.712523)

EAC = {-0.39 – (51.37570 x C)}/ 1.712523, consider this equation 1

Calculate the EAC of CFL

EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 24 years)}/ (PVIFA 11%, 24 years)

EAC = {- 3.10 - (15/1000 x 500 x C) x (8.348136)}/ (8.348136)

EAC = {-3.10 – (62.61102 x C)}/8.348137, consider this equation 2

Equate 1 and 2 to find the amount of C

{-0.39 – (51.37570 x C)}/ 1.712523 = {-3.10 – (62.61102 x C)}/8.348137

{-0.39 – (51.37570 x C) x 8.348137} = {-3.10 – (62.61102 x C) x 1.712523}

C = $0.0063825

Thus, the break- even cost per kilo – watt is $0.0063825

3 0

3 years ago

What is matter made of ? Will mark as brainliest to answer

Masja [62]

Answer:

Explanation:

All matter is made up of atoms, which are turned up of protons, neutrons and electrons.

They bond together to make up matter

8 0

3 years ago

Read 2 more answers

A stereo speaker produces a pure \"E\" tone, with a frequency of 329.6 Hz. What is the period of the sound wave produced by the

olya-2409 [2.1K]

Answer:

0.003034 s

1.035 m

4.5 m

Explanation:

$f$ = frequency of the tone = 329.6 Hz

$T$ = Time period of the sound wave

we know that, Time period and frequency are related as

$T =\frac{1}{f}\\T =\frac{1}{329.6}\\T = 0.003034 s$

$v$ = speed of the sound in the air = 341 ms⁻¹

wavelength of the sound is given as

$\lambda =\frac{v}{f} \\\lambda =\frac{341}{329.6}\\\lambda = 1.035 m$

$v$ = speed of the sound in the water = 1480 ms⁻¹

wavelength of the sound in water is given as

$\lambda =\frac{v}{f} \\\lambda =\frac{1480}{329.6}\\\lambda = 4.5 m$

8 0

3 years ago