All categories

4 years ago

Wooden ball lies in a vessel with water so that half of it is immersed in water.

Physics

1 answer:

Ugo [173]4 years ago

5 0

Downward force acting on the ball is 19.6N

Net force acting on the ball is 1960V N

Explanation:

Given:

Mass of the ball, m = 2kg

Density of ball, σ = 800 kg/m³

Density of water, ρ = 1000 kg/m³

Downward force acting by the ball in the vessel = mg

where, g = 9.8m/s²

F = 2 X 9.8

F = 19.6N

Net force acting on the ball:

Fnet = (ρ - σ) Vg

where,

V is the volume of water

Fnet = (1000 - 800) V X 9.8

Fnet = 1960V N

If the volume is known, then substitute the value of V to find the net force.

Thus, Downward force acting on the ball is 19.6N

Net force acting on the ball is 1960V N

You might be interested in

38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobari

kolbaska11 [484]

Answer:

Final volumen first process $V_{2} = 98,44 cm^{3}$

Final Pressure second process $P_{3} = 1,317 * 10^{10} Pa$

Explanation:

Using the Ideal Gases Law yoy have for pressure:

$P_{1} = \frac{n_{1} R T_{1} }{V_{1} }$

where:

P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

$n_{1} = n_{2} = n$

In an isobaric process the pressure is constant so:

$P_{1} = P_{2}$

$\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }$

$\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }$

$V_{2} = \frac{T_{2} V_{1} }{T_{1} }$

Replacing : $T_{1} =786 K, T_{2} =1209 K, V_{1} = 64 cm^{3}$

$V_{2} = 98,44 cm^{3}$

Replacing on the ideal gases formula the pressure at this piont is:

$P_{2} = 3,92 * 10^{9} Pa$

For Temperature the ideal gases formula is:

$T = \frac{P V }{n R }$

For the second process you have that $T_{2} = T_{3}$ So:

$\frac{P_{2} V_{2} }{n R } = \frac{P_{3} V_{3} }{n R }$

$P_{2} V_{2} = P_{3} V_{3}$

$P_{3} = \frac{P_{2} V_{2}}{V_{3}}$

$P_{3} = 1,317 * 10^{10} Pa$

7 0

3 years ago

What kind of frequency do radio waves have? A. High frequency B. Low frequency

inn [45]

B low frequency it is the lowest frequency

7 0

3 years ago

How to find new pressure of gas if bottle explodes

KIM [24]

If the container explodes there is no pressure, becuase all your gas has escaped its container, there for, you ain’t got no gas

6 0

3 years ago

Need help with this question

Alika [10]

Answer:

Open main menu

Digital art

Language

Download PDF

Watch

Edit

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

6 0

3 years ago

A stone is thrown horizontally at 60.0 m/sm/s from the top of a very tall cliff. Calculate its horizontal position and vertical

svp [43]

Answer:

X-Positions: Y-Positions

x(0) = 0 y(0) = 0

x(2) = 120 m y(2) = 19.6 m

x(4) = 240 m y(4) = 78.4 m

x(6) = 360 m y(6) = 176.4 m

x(8) = 480 m y(8) = 313 m

x(10) = 600m y (10) = 490 m

Explanation:

X-Positions

First, we choose to take the horizontal direction as our x-axis, and the positive x-axis as positive.
After being thrown, in the horizontal direction, no external influence acts on the stone, so it will continue in the same direction at the same initial speed of 60. 0 m/s
So, in order to know the horizontal position at any time t, we can apply the definition of average velocity, rearranging terms, as follows:

$x = v_{ox} * t = 60.0 m/s * t(s)$

It can be seen that after 2 s, the displacement will be 120 m, and each 2 seconds, as the speed is constant, the displacement will increase in the same 120 m each time.

Y-Positions

We choose to take the vertical direction as our y-axis, taking the downward direction as our positive axis.
As both axes are perpendicular each other, both movements are independent each other also, so, in the vertical direction, the stone starts from rest.
At any moment, it is subject to the acceleration of gravity, g.
As the acceleration is constant, we can find the vertical displacement (taking the height of the cliff as the initial reference level), using the following kinematic equation:

$y = \frac{1}{2} * g* t^{2} = \frac{1}{2} * 9.8 m/s2 * t(s)^{2}$

Replacing by the values of t, we get the following vertical positions, from the height of the cliff as y = 0:
y(2) = 2* 9.8 m/s2 = 19.6 m
y(4) = 8* 9.8 m/s2 = 78.4 m
y(6) = 18*9.8 m/s2 = 176.4 m
y(8) = 32*9.8 m/s2 = 313.6 m
y(10)= 50 * 9.8 m/s2 = 490.0 m

5 0

3 years ago