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ra1l [238]
3 years ago
7

Problem (2): Two point charged particles are 4.41cm apart. They are moved and placed in a new position. The force between them i

s found to have been tripled. How far apart are they now?​
Physics
1 answer:
dlinn [17]3 years ago
3 0

Answer:

HI

Explanation:

You might be interested in
The number 14 is the 'mass number'. What does it tell us about this isotope?​
Svetradugi [14.3K]

Answer:

This show the most stable of atom of that element

Explanation:

The mass number of a element on the periodic table show the most stable atoms of that element.

4 0
3 years ago
A sample of copper with a mass of 1.80 kg, initially at a temperature of 150.0°C, is in a well-insulated container. Water at a t
user100 [1]

Answer:

the mass of water is 0.3 Kg

Explanation:

since the container is well-insulated, the heat released by the copper is absorbed by the water , therefore:

Q water + Q copper = Q surroundings =0 (insulated)

Q water = - Q copper

since Q = m * c * ( T eq - Ti ) , where m = mass, c = specific heat, T eq = equilibrium temperature and Ti = initial temperature

and denoting w as water and co as copper :

m w * c w * (T eq - Tiw) = - m co * c co * (T eq - Ti co) =  m co * c co * (T co - Ti eq)

m w = m co * c co * (T co - Ti eq) / [ c w * (T eq - Tiw) ]

We take the specific heat of water as c= 1 cal/g °C = 4.186 J/g °C . Also the specific heat of copper can be found in tables → at 25°C c co = 0.385 J/g°C

if we assume that both specific heats do not change during the process (or the change is insignificant)

m w = m co * c co * (T eq - Ti co) / [ c w * (T eq - Tiw) ]

m w= 1.80 kg *  0.385 J/g°C ( 150°C - 70°C) /( 4.186 J/g°C ( 70°C- 27°C))

m w= 0.3 kg

7 0
4 years ago
A closed cylindrical tank of radius 3.5 m and height 2m is made from
Dahasolnce [82]

Explanation:

Total surface area of cylinder:

2\pi \times r(h + r) \\ 2 \times \pi \times 3.5(3.5 + 2) \\ 2\pi \times 19.25 = 120.951317 \\ 120.95  \: {cm}^{2}

sheet of metal required = 120.95 cm^2

3 0
3 years ago
Read 2 more answers
Without using a micrometer screw gauge, how do I find the average diameter of a long piece of thin wire using a metre rule and a
Mice21 [21]

Answer:

Wind the long piece of thin wire around the uniform glass rod multiple times, find the length of the total diameters using the metre ruler, and divide by the number of times you wound it around the rod.

Explanation:

Since the diameter of one long piece of thin wire is too thin to be measured by a metre ruler, you can wind it multiple times and push it side by side to get a length you can measure.

For example, if you wound it around 20 times and the total length of 20 diameters of the wire side-by-side is 2.0 cm, one winding, which is the diameter would be 2.0cm ÷ 20 = 0.10cm or 1mm.

5 0
3 years ago
After data are collected, they are often arranged in a
statuscvo [17]
Any sort of graph depending on the data it could be a bar graph , line graph any graph really
3 0
3 years ago
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