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tresset_1 [31]
4 years ago
6

Two photographers are competing for business in town. Andrea uses only film photography and Keira uses only digital photography.

Keira claims that the digital technology is superior, and her advertisement claims that the images can be stored, shared, and copied more easily. They can even be sent long distances using a series of wave pulses to transmit the information. Andrea replies that digital images can be stolen and altered more easily.
Do the advantages outweigh the disadvantages, or do both sides have valid points? Support your answer with at least three reasons.
Physics
1 answer:
Alla [95]4 years ago
8 0

Answer:

The advantages outweigh the disadvantages in my opinion.

Explanation:

The advantages outweigh the disadvantages because when using digital photography, pictures can be shared, copied and stored with the click of a button. When using film photography, results are likely to be lower quality overall than when using digital photography. In addition to this, film photography is, in many ways, much more inconvenient to use. In conclusion, the advantages of using digital photography outweigh the minimal disadvantages that come with it all.

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An open container holds ice of mass 0.555 kg at a temperature of -16.6 ∘C . The mass of the container can be ignored. Heat is su
s2008m [1.1K]

Answer: A. 23.59 minutes.

              B. 249.65 minutes

Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.

<em>Latent heat </em><em>of fusion </em>is the total amount of heat rejected from the unit mass of water at 0 degree Celsius to convert completely into ice of 0 degree Celsius (and the heat required for vice-versa process).

<em>Specific heat capacity</em> of a substance is the amount of heat required by the unit mass of a substance to raise its temperature by 1 kelvin.

Here, <u>given that</u>:

  • mass of ice, m= 0.555 kg
  • temperature of ice, T= -16.6°C
  • rate of heat transfer, q=820 J.min^{-1}
  • specific heat of ice, c_{i}= 2100 J.kg^{-1}.K^{-1}
  • latent heat of fusion of ice, L_{i}=334\times10^{3}J.kg^{-1}

<u>Asked:</u>

1. Time require for the ice to start melting.

2. Time required to raise the temperature above freezing point.

Sol.: 1.

<u>We have the formula:</u>

Q=mc\Delta T

Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.

Q= 0.555\times2100\times16.6

Q= 19347.3 J

Now, we require 19347.3 joules of heat to bring the ice to 0°C  and then on further addition of heat it starts melting.

∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.

t=\frac{Q}{q}

=\frac{19347.3}{820}

= 23.59 minutes.

Sol.: 2.

Next we need to find the time it takes before the temperature rises above freezing from the time when heating begins.

<em>Now comes the concept of Latent  heat into the play, the temperature does not starts rising for the ice as soon as it reaches at 0°C it takes significant amount of time to raise the temperature because the heat energy is being used to convert the phase of the water molecules from solid to liquid.</em>

From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.

<u>We have the equation:</u> latent heat, Q_{L}= mL_{i}

Q_{L}= 0.555\times334\times10^{3}= 185370 J

<u>Now  the time required for supply of 185370 J:</u>

t=\frac{Q_{L}}{q}

t=\frac{185370}{820}

t= 226.06 minutes

∴ The time it takes before the temperature rises above freezing from the time when heating begins= 226.06 + 23.59

= 249.65 minutes

8 0
3 years ago
In 1650, an individual calculated that the earth was about 6,000 years old. since then what has happened to scientists of the ea
yuradex [85]
C is the correct answer to your question.
6 0
4 years ago
Read 2 more answers
Steve is planning his annual Spring Break road trip. He pulls out his map and draws out his route to visit the five locations th
Aleksandr [31]

Answer:

His average speed.

Explanation:

Average speed = Total distance travelled / total time .

In the given case odometer measures total distance travelled . Dividing it by time will give average speed .

Average velocity can not be the answer as it is measured by dividing displacement by time . Displacement here is zero because initial and final point are same . Instantaneous velocity is given by the speedometer . It keeps on changing all the time . It is actually speed at a particular point of time . It can not be average velocity or speed .

8 0
3 years ago
Jane does 5 joules of work when she closes her bedroom window. She applies a force of 6 newtons to do the job. How far does she
insens350 [35]

<u>Statement</u><u>:</u>

Jane does 5 joules of work when she closes her bedroom window. She applies a force of 6 newtons to do the job.

<u>To </u><u>find </u><u>out:</u>

The displacement of the window when she pulled it.

<u>Solution</u><u>:</u>

  • Work done (W) = 5 J
  • Force (F) = 6 N
  • Let the displacement of the window be s.
  • We know, the formula of work done, i.e., W = Fs
  • Putting the values in the above formula, we get
  • 5 J = 6 N × s
  • or, s = (5 ÷ 6) m
  • or, s = 0.83 m

<u>Answer</u><u>:</u>

She pulls the window by 0.83 m.

Hope you could understand.

If you have any query, feel free to ask.

4 0
3 years ago
A 1300 kg car starts at rest and rolls down a hill from a height of 10.0 m. It then moves across a
Makovka662 [10]

Answer:

0.51 m

Explanation:

Using the principle of conservation of energy, change in potential energy equals to the change in kinetic energy of the spring.

Kinetic energy, KE=½kx²

Where k is spring constant and x is the compression of spring

Potential energy, PE=mgh

Where g is acceleration due to gravity, h is height and m is mass

Equating KE=PE

mgh=½kx²

Making x the subject of formula

x=\sqrt {\frac {2mgh}{k}}

Substituting 9.81 m/s² for g, 1300 kg for m, 10m for h and 1000000 for k then

x=\sqrt \frac {2*1300*9.81*10}{1000000}=0.50503465227646m\\x\approx 0.51 m

5 0
3 years ago
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