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3 years ago

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Physics

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hjlf3 years ago

7 0

Which of the following are characteristics of noble gases?

${ \bf{ \underbrace{Answer :}}}$

$\sf\red{B. \:They're\: inert.}$ ✅

An inert gas is one that does not undergo chemical reactions. The noble gases have complete outer shells, so they have no tendency to lose, gain, or share electrons. This is why they are said to be inert.

$\sf\purple{D.\: They \:don't \:react\: with\: other\: elements.}$ ✅

Noble gases are the least reactive of all elements. This is because they already have the desired eight total 's' and 'p' electrons in their outermost (highest) energy level.

$\circ \: \: { \underline{ \boxed{ \sf{ \color{green}{Happy\:learning.}}}}}∘$

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The kinetic energy of an object with a mass of 6.8 kg and a velocity of 5.0 m/s is [BLANK] J. (Report the answer to two signific

dmitriy555 [2]

<h2>Hello!</h2>

The answer is:

The kinetic energy of the object is equal to 85 J.

The kinetic energy involves the speed and the mass of an object in motion. We can calculate the following the work needed to speed an object (kinetic energy) using the equation:

$KineticEnergy=\frac{1}{2}mv^{2}$

Where,

m, is the mas of the object

v, is the speed of the object.

Now, we are given:

$mass=m=6.8kg\\speed=v=5\frac{m}{s}$

So, substituting and calculating the kinetic energy of the object, we have:

$KineticEnergy=\frac{1}{2}*6.8kg*(5\frac{m}{s})^{2}$

$KineticEnergy=\frac{1}{2}*6.8kg*(25\frac{m^{2}}{s^{2}})$

$KineticEnergy=\frac{1}{2}*170kg\frac{m^{2}}{s^{2}}$

$KineticEnergy=85kg\frac{m^{2}}{s^{2}}=85J$

We have that the kinetic energy of the object is equal to 85 J.

Have a nice day!

8 0

3 years ago

Read 2 more answers

The change in the internal energy of a system that releases 2,500 j of heat and that does 7,655 j of work on the surroundings is

g100num [7]

We are given that the system “releases” heat of 2,500 J, and that it “does work on the surroundings” by 7,655 J.

The highlighted words releases and does work on the surroundings all refers to that it is the system itself which expends energy to do those things. Therefore the action of releasing heat and doing work has both magnitudes of negative value. Therefore:

heat released = - 2, 500 J

work done = - 7, 655 J

Which means that the total internal energy change of the system is:

change in internal energy = heat released + work

<span>change in internal energy = - 2, 500 J + - 7, 655 J</span>

<span>change in internal energy = -10,155 J</span>

3 0

4 years ago

Will give 20 point plus brainlist

zloy xaker [14]

Answer:

O 50 squirrels

Explanation:

1 acre => 5 squirrels

10 acres => 50 squirrels

7 0

2 years ago

A passenger at an airport pulls a rolling suitcase by its handle. If the force used is 10 N and the handle makes an angle of 25

Deffense [45]

Answer:

W= 1812.6 J

Explanation:

Work (W) is defined as the scalar product of force F by the distance (d) the body travels due to this force.

W= F*d* cosα Formula ( 1)

Where:

F is the force in Newtons (N)

d is the displacemente in meters (m)

α : Angle formed between force and displacement

Data

F = 10 N

d = 200 m

α = 25°

Work done by the pulling force while the passenger walks 200 m

We replace data in the formula (1)

W= F*d* cosα

W= (10 N)*(200 m)* cos25°

W= 1812.6 (N*m)

W= 1812.6 J

6 0

3 years ago

A hollow conducting sphere with an outer radius of 0.295 m and an inner radius of 0.200 m has a uniform surface charge density o

IrinaK [193]

Answer:

a. $6.032\times10^{-6}C/m^2$

b. $6.816\times10^5N/C$

Explanation:

#Apply surface charge density, electric field, and Gauss law to solve:

a. Surface charge density is defined as charge per area denoted as $\sigma$

$\sigma=\frac{Q}{4\pi r_{out}^2}$ , and the strength of the electric field outside the sphere $E=\frac{\sigma _{new}}{\epsilon _o}$

Using Gauss Law, total electric flux out of a closed surface is equal to the total charge enclosed divided by the permittivity.

$\phi=\frac{Q_{enclosed}}{\epsilon_o}\\\\\sigma=\frac{Q}{4\pi r_{out}^2}\\\\\sigma=\frac{0.370\times 10^{-6}}{4\pi \times (0.295m)^2}\\\\=3.383\times10^{-7}C/m^2$ #surface charge outside sphere.

$\sigma_{new}=\sigma_{s}-\sigma\\\\\sigma_{new}=6.37\times10^{-6}C/m^2-3.383\times10^{-7}C/m^2\\\\\sigma_{new}=6.032\times10^{-6}C/m^2$

Hence, the new charge density on the outside of the sphere is $6.032\times10^{-6}C/m^2$

b. The strength of the electric field just outside the sphere is calculated as:

From a above, we know the new surface charge to be $6.032\times10^{-6}C/m^2$ ,

$E=\frac{\sigma _{new}}{\epsilon _o}\\\\=\frac{6.032\times10^{-6}C/m^2}{\epsilon _o}\\\\\epsilon _o=8.85\times10^{-12}C^2/N.m^2\\\\E=\frac{6.032\times10^{-6}C/m^2}{8.85\times10^{-12}C^2/N.m^2}\\\\E=6.816\times10^5N/C$

Hence, the strength of the electric field just outside the sphere is $6.816\times10^5N/C$

5 0

3 years ago