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Alla [95]
2 years ago
5

If this charged soot particle is now isolated (that is, removed from the electric field described in the previous part), what wi

ll be the magnitude E of the electric field due to the particle at distance 1.00 meter from the particle?
Physics
1 answer:
kicyunya [14]2 years ago
7 0

The magnitude of the electric field due to the particle at the given distance is 9\times 10^9 q.

<h3>Electric field strength</h3>

The electric field strength of a charged particle is the force per unit charge in the given field.

The electric field strength of a charge is given as;

E = \frac{F}{q} \\\\E = \frac{kq}{r^2}

where;

  • k is Coulomb's constant
  • q is the charge
  • r is the distance

E = \frac{9\times 10^9 \times q}{1^2} \\\\E = 9\times 10^9 q

Thus, the magnitude of the electric field due to the particle at the given distance is 9\times 10^9 q.

Learn more about electric field here: brainly.com/question/14372859

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When a fixed amount of ideal gas goes through an isobaric expansion A) its internal (thermal) energy does not change.B) the gas
Bingel [31]
<h2>Answer: its temperature must increase.</h2>

Explanation:

In an isobaric process the pressure remains constant, which means the initial pressure and the final pressure will be the same.

In addition, during this thermodynamic process, the volume of the ideal gas expands or contracts in such a way that the variation of pressure \Delta P is neutralized.

Now, according to the First law of Thermodynamics that establishes the conservation of energy:

\Delta U=\Delta Q-\Delta W   (1)

Where:

\Delta U is the internal energy

\Delta Q is the heat transferred

\Delta W is the work

Now, for an isobaric process:

\Delta W=P\Delta V    (2)

Where:

P is the pressure (<u>always positive</u>)

\Delta V is the volume variation of the gas

<u />

<u>Here we have two possible results:</u>

-If the gas expands (positive \Delta V), the work is positive.

-If the gas compresses (negative \Delta V), the work is negative.

In this case we are talking about the first result (work is positive).

Then, according to the above, equation (1) can be written as follows:

\Delta U=\Delta Q - P\Delta V   (3)

Clearing \Delta Q:

\Delta Q=\Delta U+P \Delta V    (4)

Then, for an ideal gas in an isobaric process, part of the heat (Q) added to the system will be used to do work (positive in this case) and the other part <u>will increase the internal energy</u>, hence <u>the temperature will increase as well.</u>

7 0
3 years ago
2. Find the time taken by the bus to reach the stop. need only group B, 2 answer
Molodets [167]

Answer:

t = 2 seconds

Explanation:

In 2nd question, the question is given the attached figure.

Initial speed of the bus, u = 0

Acceleration of the bus, a = 8 m/s²

Final speed, v = 16 m/s

We need to find the time taken by the car to reach the stop. Acceleration of an object is given by :

a=\dfrac{v-u}{t}

t is time taken

t=\dfrac{v-u}{a}\\\\t=\dfrac{16-0}{8}\\\\t=2\ s

The bus will take 2 seconds to reach the stop.

3 0
3 years ago
WHAT IS THE DENSTY OF ALL THE LAYER IN THE ATMOSPHERE
Verdich [7]

Answer:

The troposphere starts at the Earth's surface and extends 8 to 14.5 kilometers high (5 to 9 miles). This part of the atmosphere is the most dense. Almost all weather is in this region.

Explanation:

8 0
2 years ago
Read 2 more answers
In which of the following would the particles move most rapidly? a. ice at -20 °C b. water at 20 °C c. steam at 110 °C d. boilin
Artemon [7]

Answer:

C. steam at 110

Explanation:

3 0
3 years ago
A piano string having a mass per unit length equal to 4.70 10-3 kg/m is under a tension of 1 400 N. Find the speed with which a
sweet [91]

Answer:

The speed of the sound wave on the string is 545.78 m/s.

Explanation:

Given;

mass per unit length of the string, μ = 4.7 x 10⁻³ kg/m

tension of the string, T = 1400 N

The speed of the sound wave on the string is given by;

v = \sqrt{\frac{T}{\mu} }

where;

v is the speed of the sound wave on the string

Substitute the given values and solve for speed,v,

v = \sqrt{\frac{T}{\mu} }\\\\v = \sqrt{\frac{1400}{4.7*10^{-3}} }\\\\v = \sqrt{297872.34}\\\\v = 545.78 \ m/s

Therefore, the speed of the sound wave on the string is 545.78 m/s.

3 0
2 years ago
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