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A 3.00 μF capacitor is charged to 480 V and a 4.00 μF capacitor is charged to 500 V . Part A These capacitors are then disconnec

pogonyaev

Q before connected = Q after connected C1V1+C2V2 = (C1+C2) V

C1= 3×10^-6 F

V1= 480v

C2= 4×10^-6 F

V2= 500v

(3×10^-6)×(480) + (4×10^-6)×(500) = (3×10^-6 + 4×10^-6) × V

Simplifying the above, we get:

( 1440× 10^-6) + (2000 ×10^-6) = (7 × 10^-6) × V.

Further simplified as:

3440 × 10^-6 = 7 × 10^-6 × V

Making V the subject

V = 491.43volts

Therefore the potential difference across each capacitor is 491.43v

4 0

3 years ago

A 100 kg box as showsn above is being pulled along the x axis by a student. the box slides across a rough surface, and its posit

Anit [1.1K]

Answer:

a) 2 m/s

b) i) $K.E = 50 (1.5t^2 + 2) ^2\\$

ii) $F = 3tm$

Explanation:

The function for distance is $x = 0.5t ^3 + 2t$

We know that:

Velocity = $v= \frac{d}{dt} x$

Acceleration = $a= \frac{d}{dt}v$

To find speed at time t = 0, we derivate the distance function:

$x = 0.5 t^3 + 2t\\v= x' = 1.5t^2 + 2$

Substitute t = 0 in velocity function:

$v = 1.5t^2 + 2\\v(0) = 1.5 (0) + 2\\v(0) = 2$

Velocity at t = 0 will be 2 m/s.

To find the function for Kinetic Energy of the box at any time, t.

$Kinetic \ Energy = \frac{1}{2} mv^2\\\\K.E = \frac{1}{2} \times 100 \times (1.5t^2 + 2) ^2\\\\K.E = 50 (1.5t^2 + 2) ^2\\$

We know that $Force = mass \times acceleration$

$a = v'(t) = 1.5t^2 + 2\\a = 3t$

$F = m \times a\\F= m \times 3t\\F = 3tm$

6 0

4 years ago

07. How do scientists use spectroscopy (the study of light frequencies released

Vera_Pavlovna [14]

Every element is able to be recognized individually in many different ways. A very easy and common way is using light absorption also known as spectroscopy. Every atom has electrons, and these electrons like to stay in their lowest-energy configuration. However, when photons collide with an electron it can increase it to a higher energy level.. This is absorption, and each element’s electrons absorb light at specific wavelengths related to the difference between energy levels in that atom. But the electrons want to return to their original levels, so they don’t hold onto the energy for long. When they emit the energy, they release photons with exactly the same wavelengths of light that were absorbed in the first place. An electron can release this light in any direction, so most of the light is emitted in directions away from our line of sight. Therefore, a dark line appears in the spectrum at that particular wavelength.

Because the wavelengths at which absorption lines occur are unique for each element, astronomers can measure the position of the lines to determine which elements are present in a target. The amount of light that is absorbed can also provide information about how much of each element is present.

8 0

3 years ago

If, as is typical, each of them breathes about 500 cm3 of air with each breath, approximately what volume of air (in cubic meter

deff fn [24]

Answer:

a) 12614.4 m^3

b) 28.8 m

Explanation:

The complete question is

Four astronauts are in a spherical space station. (a) If, as is typical, each of them breathes about 500 cm^3 of air with each breath, approximately what volume of air (in cubic meters) do these astronauts breathe in a year? (b) What would the diameter (in meters) of the space station have to be to contain all this air?

The average breathing rate is 12 breaths per minute

there are 60 minutes x 24 hours x 365 days in a year = 525600 minutes in a year

if an average human takes 12 breath per minute, then in a year an average human take 12 x 525600 = 6307200 breath

For the four astronauts, the amount of breath = 4 x 6307200 = 25228800 breath in total.

The volume of air per breath = 500 cm^3

1 cm^3 = 10^-6 m^3

500 cm^3 = $x$ m^3