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Xelga [282]
3 years ago
10

What are the 5 steps in the process of air circulation in order

Physics
1 answer:
creativ13 [48]3 years ago
4 0
Yes,Physics 5pts just now<span>are the 5 steps in the process of air circulation in order</span>
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A 0.50-kg block attached to an ideal spring with a spring constant of 80 N/m oscillates on a horizontal frictionless surface. Th
Taya2010 [7]

Answer:

The greatest extension of the spring is \bf{0.055~m} and the maximum speed of the block is \bf{0.695~m/s}.

Explanation:

Given:

The mass of the block is, m = 0.50~kg

The spring constant of the spring is, k = 80~N/m

The mechanical energy of the block is, E = 0.12~J

When a particle is oscillating in a simple harmonic way, its total energy is given by

E = \dfrac{1}{2}m\omega^{2}a^{2}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)

where \omega is the angular velocity of the mass and a is the amplitude of its motion.

The relation between angular velocity and spring constant is given by

\omega = \sqrt{\dfrac{k}{m}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)

Substituting equation (2) in equation (1), we have

~~~~~~&& E = \dfrac{1}{2}ka^{2}\\&or,& a = \sqrt{\dfrac{2E}{k}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)

Substituting 0.12~J for E and 80~N/m for k in equation (3), we can write

a &=& \sqrt{\dfrac{2(0.12~J)}{80~N/m}}\\~~~&=& 0.055~m

The relation between the maximum velocity and the amplitude is given by

v_{m} &=& \omega a\\~~~~&=& \sqrt{\dfrac{k}{m}}a~~~~~~~~~~~~~~~~~~~~~~~~~~~~(4)

Substituting 80~N/m for k , 0.50~kg for m and 0.055~m for a in equation (4), we have

v_{m} &=& \sqrt{\dfrac{80~N/m}{0.50~kg}}(0.055~m)\\~~~&=& 0.695~m/s

8 0
3 years ago
What do you call a group of sea turtles?
Valentin [98]

Answer:

a bale

Explanation:

a bale is a group of turtles

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Think of a skateboarder riding down a hill. Once the skateboarder reaches the bottom of the hill, if she does not manually add m
ivanzaharov [21]

From first law of motion, law of inertia, we know that an object remains in the state of rest or motion unless acted upon by an unbalanced external force. The skateboard riding down a hill when reaches the bottom slows down due to <u>frictional force </u>acting in the opposite direction. Frictional force is a contact force which acts between two surfaces against the direction of  motion of one surface over the other. Here, frictional force acts <u>between skateboard and ground</u>. This reduces the speed of the skateboard and it eventually stops.

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When an object is stationary, all of the forces acting on it are balanced
Leviafan [203]
With utmost clarity, that is truest of the truths

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Two thin concentric spherical shells of radii r1 and r2 (r1 &lt; r2) contain uniform surface charge densities V1 and V2, respect
Lyrx [107]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a)  0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b)  r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 \pi r^{2}  = \frac{Q1}{E_{0} }

So,

Rearranging the above equation to get Electric field, we will get:

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   }

Multiply and divide by r1^{2}

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } x \frac{r1^{2} }{r1^{2} }

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x r1^{2}) /(E_{0} x r^{2})

c) r > r2 :

Electric Field = ?

E x 4 \pi r^{2}  = \frac{Q1 + Q2}{E_{0} }

Rearranging the above equation for E:

E = \frac{Q1+Q2}{E_{0} . 4 \pi. r^{2}   }

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

As we know from above, that:

\frac{Q1}{E_{0} . 4 \pi. r^{2}   } =  (σ1 x r1^{2}) /(E_{0} x r^{2})

Then, Similarly,

\frac{Q2}{E_{0} . 4 \pi. r^{2}   } = (σ2 x r2^{2}) /(E_{0} x r^{2})

So,

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

Replacing the above equations to get E:

E = (σ1 x r1^{2}) /(E_{0} x r^{2}) + (σ2 x r2^{2}) /(E_{0} x r^{2})

Now, for

d) Under what conditions,  E = 0, for r > r2?

For r > r2, E =0 if

σ1 x r1^{2} = - σ2 x r2^{2}

4 0
3 years ago
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