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3 years ago

“What is the stance we take when we first get possession of the Basketball?”

Physics

1 answer:

vesna_86 [32]3 years ago

4 0

Answer:

Photosynthesis is a process used by plants and other organisms to convert light energy into chemical energy that, through cellular respiration, can later be released to fuel the organism's

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A car battery with a 12 V emf and an internal resistance of 0.11 Ω is being charged with a current of 56 A. What are (a) the pot

denpristay [2]

Answer:

Part a)

V = 18.16 V

Part b)

$P_r = 345 Watt$

Part c)

P = 672 Watt

Part d)

V = 5.84 V

Part e)

$P_r = 345 Watt$

Explanation:

Part a)

When battery is in charging mode

then the potential difference at the terminal of the cell is more than its EMF and it is given as

$\Delta V = E + i r$

here we have

$E = 12 V$

$i = 56 A$

$r = 0.11$

now we have

$\Delta V = 12 + (0.11)(56) = 18.16 V$

Part b)

Rate of energy dissipation inside the battery is the energy across internal resistance

so it is given as

$P_r = i^2 r$

$P_r = 56^2 (0.11)$

$P_r = 345 W$

Part c)

Rate of energy conversion into EMF is given as

$P_{emf} = i E$

$P_{emf} = (56)(12)$

$P_{emf} = 672 Watt$

Now battery is giving current to other circuit so now it is discharging

now we have

Part d)

$V = E - i r$

$V = 12 - (56)(0.11)$

$V = 12 - 6.16 = 5.84 V$

Part e)

now the rate of energy dissipation is given as

$P_r = i^2 r$

$P_r = 56^2 (0.11)$

$P_r = 345 W$

7 0

3 years ago

A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet

lesya692 [45]

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

Current flowing in the wire, I = 4.00 mA
Initial diameter of the wire, d₁ = 4 mm = 0.004 m
Final diameter of the wire, d₂ = 1 mm = 0.001 m
Length of wire, L = 2.00 m
Density of electron in the copper, n = 8.5 x 10²⁸ /m³

The initial area of the copper wire;

$A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2$

The final area of the copper wire;

$A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2$

The initial drift velocity of the electrons is calculated as;

$v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s$

The final drift velocity of the electrons is calculated as;

$v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7} \ m/s$

The change in the mean drift velocity is calculated as;

$\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s$

The time of motion of electrons for the initial wire diameter is calculated as;

$t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s$

The time of motion of electrons for the final wire diameter is calculated as;

$t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s$

The average acceleration of the electrons is calculated as;

$a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2$

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

7 0

3 years ago

An uniform electric field of magnitude E = 100 N/C is oriented along the positive y-axis. What is the magnitude of the flux of t

Ede4ka [16]

Answer:

The magnitude of the flux of electric field through a square of surface area is zero.

Explanation:

$E=100 NC^{-1}\\\\A=2 m^2\\\\Electic\,\,flux\,\,flux\,\,is\,\,given\,\,as:\\\\\phi_E=E.A\,cos\,\theta$

It is given that square box is parallel to yz-plane which has normal vector perpendicular to plane in x-direction. Angle between normal vector of area and electric field is 90°. Substituting in (1)

$\phi_E=E.A\,cos\,(90^o)\\\\\phi_E=0$

4 0

4 years ago

If a layer was deposited but does not appear in the rock record, what has occured

nalin [4]

I think you forgot to give the choices along with the question. I am answering the question based on my research and knowledge. If a layer was deposited but does not appear in the rock record, the thing that happened is erosion. I hope that this is the ans wer that has actually come to your desired help.

5 0

3 years ago

Read 2 more answers

Need help with this question

Alika [10]

Answer:

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Digital art

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Digital art is an artistic work or practice that uses digital technology as part of the creative or presentation process. Since the 1960s, various names have been used to describe the process, including computer art and multimedia art.[1] Digital art is itself placed under the larger umbrella term new media art.[2][3]

Maurizio Bolognini, Programmed Machines (Nice, France, 1992–97). An installation at the intersection of digital art and conceptual art (computers are programmed to generate flows of random images which nobody would see).

The image of the computer virus Chernobyl, created by Ukrainian new media artist Stepan Ryabchenko in 2011.

Irrational Geometrics digital art installation 2008 by Pascal Dombis

Joseph Nechvatal birth Of the viractual 2001 computer-robotic assisted acrylic on canvas

The Cave Automatic Virtual Environment at the University of Illinois, Chicago

After some initial resistance,[4] the impact of digital technology has transformed activities such as painting, drawing, sculpture and music/sound art, while new forms, such as net art, digital installation art, and virtual reality, have become recognized artistic practices.[5] More generally the term digital artist is used to describe an artist who makes use of digital technologies in the production of art. In an expanded sense, "digital art" is contemporary art that uses the methods of mass production or digital media.[6]

Lillian Schwartz's Comparison of Leonardo's self portrait and the Mona Lisa based on Schwartz's Mona Leo. An example of a collage of digitally manipulated photographs

The techniques of digital art are used extensively by the mainstream media in advertisements, and by film-makers to produce visual effects. Desktop publishing has had a huge impact on the publishing world, although that is more related to graphic design. Both digital and traditional artists use many sources of electronic information and programs to create their work.[7] Given the parallels between visual and musical arts, it is possible that general acceptance of the value of digital visual art will progress in much the same way as the increased acceptance of electronically produced music over the last three decades.[8]

Digital art can be purely computer-generated (such as fractals and algorithmic art) or taken from other sources, such as a scanned photograph or an image drawn using vector graphics software using a mouse or graphics tablet.[9] Though technically the term may be applied to art done using other media or processes and merely scanned in, it is usually reserved for art that has been non-trivially modified by a computing process (such as a computer program, microcontroller or any electronic system capable of interpreting an input to create an output); digitized text data and raw audio and video recordings are not usually considered digital art in themselves, but can be part of the larger project of computer art and information art.[10] Artworks are considered digital painting when created in similar fashion to non-digital paintings but using software on a computer platform and digitally outputting the

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3 years ago