Conclusions ____

At a depth of about 10 meters the water pressure on you is equal to the pressure of 1 atmosphere. the total pressure on you is t

MrRa [10]

It would be 22 depths inside the glass of water.

8 0

3 years ago

How do H-R diagrams help scientists compare new stars with the Sun? In your answer, name one type of data that H-R diagrams do n

damaskus [11]

Answer

Hertzsprung-Russell (HR) diagram is an essential tool used in stellar evolution. In the universe, there are several hundreds of billions of stars. Scientists use the tool, in differentiation, the billions of stars in the world from the sun. In the HR tool, there is plotting of the luminosity or energy output of a star, which is plotted on the X-axis of a graph against the absolute magnitude. The sun's magnitude is an absolute of +48, which, when plotted against its luminosity, helps in setting an apparent variance between the sun and any other star. Additionally, the sun has been identified as the primary star with a very high temperature. Hence the tool can locate the sun from other forms of stars. HR diagrams outline data such as temperature and luminosity or energy. However, star distance from the Erath is not a type of data represented in the charts.

Explanation:

Hope this helped you!

4 0

2 years ago

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A metal disk of radius 6.0 cm is mounted on a frictionless axle. Current can flow through the axle out along the disk, to a slid

Galina-37 [17]

Answer:

0.09 N

Explanation:

We are given that

Radius of disk,r=6 cm= $\frac{6}{100}=0.06 m$

1 m=100 cm

B=1 T

Current,I=3 A

We have to find the frictional force at the rim between the stationary electrical contact and the rotating rim.

$dF=IBdr$

$\tau=rdF=IBrdr$

$\tau=\int_{0}^{R}IBr dr$

$\tau=IB(\frac{R^2}{2}$

Torque due to friction

$\tau=R\times F$

Where friction force=F

$R\times F=\frac{IBR^2}{2}$

$F=\frac{IBR}{2}$

Substitute the values

$F=\frac{3\times 1\times 0.06}{2}$

$F=0.09 N$

7 0

3 years ago

Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if

klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:

"The electrostatic force $F_{E}$ between two point charges $q_{1}$ and $q_{2}$ is proportional to the product of the charges and inversely proportional to the square of the distance $d$ that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

$F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}$ (1)

Where:

$F_{E}=60 N$ is the electrostatic force

$K=8.99(10)^{9} Nm^{2}/C^{2}$ is the Coulomb's constant

$q_{1}$ and $q_{2}$ are the electric charges

$d=3 m$ is the separation distance between the charges

Then:

$60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}}$ (2)

Isolating $q_{1}$ and $q_{2}$ :

$q_{1}q_{2}=6(10)^{-8} C^{2}$ (3)

Now, if we keep the same charges but we decrease the distance to $d_{1}=1 m$ , (1) is rewritten as:

$F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}}$ (4)

Then, the new electrostatic force will be:

$F_{E}= 539.4 N$ (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

4 0

2 years ago

How does a lever work?(awarding 24pts for immediate answer!!!!)

vlada-n [284]

It's a simple machine, consisting of a rigid bar that rotates about a fixed point which is known as "Fulcrum" (most important part), It <span>affects the effort, or force and do the amount of work

Hope this helps!</span>

6 0

3 years ago

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Conclusions _____.