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denis23 [38]
3 years ago
9

What are the change in internal energy if 250 j of hear is added to a system of 80 j of work is done by the system

Physics
1 answer:
Sphinxa [80]3 years ago
5 0

Answer:

Explanation:

dU= dq+w

dU is change in internal energy of the system

dq is the amount heat added or released by the system which be positive or negative respectivelý

And w is the amount of work done by the system or on the system which will be positive or negative respectively.

Hence,

dU= 250+80= 330 J

The change will be positive

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Noble gasses ( insert gases)
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Two electric charges, held a distance, dd, apart experience an electric force of magnitude, FF, between them. If one of the char
lorasvet [3.4K]

Answer:

F'=2F

Explanation:

The Coulomb's law states that the magnitude of the electrostatic force between two charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them:

F=\frac{kq_1q_2}{d^2}

In this case, we have q_1'=2q_1:

F'=\frac{kq'_1q_2}{d^2}\\F'=\frac{k(2q_1)q_2}{d^2}\\F'=2\frac{kq_1q_2}{d^2}\\F'=2F

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3 years ago
In a thundercloud there may be an electric charge of +40 C near the top of the cloud and -40 C near the bottom of the cloud. The
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Answer:

Electric force, F=-3.59\times 10^6\ N

Explanation:

Given that,

Electric charge 1, q_1=+40\ C

Electric charge 2, q_2=-40\ C

Distance, d=2\ km=2\times 10^3\ m

To find,

The electric force between these two sets of charges.

Solution,

There exists a force between two electric charges and this force is called electrostatic force. It is equal to the product of electric charged divided by square of distance between them.

F=k\dfrac{q_1q_2}{d^2}

k is the electrostatic constant

F=8.99\times 10^9\times \dfrac{40\times (-40)}{(2\times 10^3)^2}

F=-3.59\times 10^6\ N

So, the electric force between these two sets of charges is -3.59\times 10^6\ N.

5 0
3 years ago
Which is a compound<br> A Kr<br> B O<br> 2<br><br> c cl2<br> d CuF2
Volgvan
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Review. For a certain type of steel, stress is always proportional to strain with Young's modulus 20 × 10¹⁰ N/m² . The steel has
timurjin [86]

1.58\times 10^{-4}\ \mathrm{s}$ is the time interval elapses before the back end of the rod receives the message that it should stop.

Given:

Length of the rod, L = 80 cm = 0.800 m

Young's modulus, Y = 20 \times 10^{10}\;N/m^{2}

steel density, \rho = 7.86 \times 10^{3}\;kg/m^{3}

The speed of the wave in the rod is,

$v = \sqrt{\frac{Y}{\rho}} = \sqrt{\frac{20\times 10^{10}\ \mathrm{N/m^2}}{7.86\times 10^3\ \mathrm{kg/m^3}}} = 5044\ \mathrm{m/s}$

Consequently, the length of the rod's end is traveled by the wave in at

$t=\frac{L}{v} = \frac{0.800\ \mathrm{m}}{5044\ \mathrm{m/s}} = 1.58\times 10^{-4}\ \mathrm{s}$

Hence, 1.58\times 10^{-4}\ \mathrm{s}$ is the time interval elapses before the back end of the rod receives the message that it should stop.

<h3>What are Newtons Laws?</h3>

The three fundamental laws of classical mechanics known as Newton's laws of motion describe how an object's motion and the forces acting on it interact. The following paraphrase of these statutes is available

Unless a force acts upon a body, it remains at rest or in continual straight-line motion.

When a force acts on a body, the force is equal to the time rate at which the body's momentum changes.

When two bodies exert force on one another, the direction and amount of the force are opposed.

Isaac Newton first identified the three laws of motion in his 1687 book Philosophize Naturalis Principia Mathematica (Mathematical Principles of Natural Philosophy).

They served as the cornerstone for classical mechanics as Newton used them to examine and explain the motion of numerous physical objects and systems. The conceptual foundations of classical physics have been reconstructed in several ways since Newton, utilizing various mathematical techniques that have revealed insights that were hidden in the original, Newtonian formulation.

To know more about Newtons Laws, visit:

brainly.com/question/27573481

#SPJ4

7 0
1 year ago
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