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Klio2033 [76]
3 years ago
12

What is 8200 N/m2 expressed in kilopascals?

Physics
1 answer:
natima [27]3 years ago
5 0
The S.I. unit for the measure of the pressure is the Pascal (Pa). 1 Pascal corresponds to
1 Pa = 1 N/m^2

We can convert the number given by the problem into Pascal:
8200 N/m^2 = 8200 Pa

And since 1 Pa = 10^{-3}kPa, we have
8200 Pa = 8200 \cdot 10^{-3}kPa =8.2 kPa
You might be interested in
Every Saros cycle (19 eclipse years): A. All total solar eclipses occur as total solar eclipses. B. All annular solar eclipses o
pickupchik [31]

Answer:

D. All lunar eclipses reoccur.

Explanation:

Every Saros cycle (19 eclipse years)

Saro was described by Edmond Halley arround the year 1691, which is a period of 223 synodic which can be interepreted as "repitition" it is used by the scientist for eclipse prediction either for the sun or moon eclipse. It is 18years and 11days and some hours.

It should be noted that with this All lunar eclipses reoccur.

4 0
3 years ago
A child’s toy rake is held so that its output arm is 0.75 meters. If the mechanical advantage is 0.33, what is
Masja [62]
Input= ?
Output= 0.75
MA= 0.33

So I just kept on Dividing numbers until I was close to "0.33" & I cam up with the answer of .......... 0.25.


So when you divide 0.25 by 0.75 you get the MA of 0.33 
4 0
3 years ago
What should you do if...
Hunter-Best [27]

Answer:

3. you need to ask your available lab instructors what to do.

4. You immediately have to drop down your cloth and roll it to extinguish the fire or move to the emergency shower if available

5. You have too keep calm and report to the lab instructor but do no shout.

6. Move immediately to the eye rinse basin if available and wash your eyes gently and thoroughly

6 0
2 years ago
A wire is formed into a circle having a diameter of 10.0cm and is placed in a uniform magnetic field of 3.00mT . The wire carrie
Paul [167]

The range of potential energies of the wire-field system for different orientations of the circle are -

θ                  U

0°             375 π x 10^{-7}

90°              0

180°        - 375 π x 10^{-7}

We have current carrying wire in a form of a circle placed in a uniform magnetic field.

We have to the range of potential energies of the wire-field system for different orientations of the circle.

<h3>What is the formula to calculate the Magnetic Potential Energy?</h3>

The formula to calculate the magnetic potential energy is -

U = M.B = MB cos $\theta

where -

M is the Dipole Moment.

B is the Magnetic Field Intensity.

According to the question, we have -

U = M.B = MB cos $\theta

We can write M = IA (I is current and A is cross sectional Area)

U = IAB cos $\theta

U = Iπr^{2}B cos $\theta

For $\theta = 0° →

U(Max) = MB cos(0) = MB =  Iπr^{2}B = 5 × π × ( 0.05 ) ^{2} × 3 × 10^{-3} =

375 π x 10^{-7}.

For $\theta = 90° →

U = MB cos (90) = 0

For $\theta = 180° →

U(Min) = MB cos(0) = - MB =  - Iπr^{2}B = - 5 × π × ( 0.05 ) ^{2} × 3 × 10^{-3} =

- 375 π x 10^{-7}.

Hence, the range of potential energies of the wire-field system for different orientations of the circle are -

θ                  U

0°             375 π x 10^{-7}

90°              0

180°        - 375 π x 10^{-7}

To solve more questions on Magnetic potential energy, visit the link below-

brainly.com/question/13708277

#SPJ4

3 0
1 year ago
Consider as a system the Sun with Saturn in a circular orbit around it. Find the magnitude of the change in the velocity of the
Doss [256]

Answer:

v_{su} = 19.44 m/s

Explanation:

m_{su}=5.68x10^{29}kg\\m_{sa}=5.68x10^{26}kg

T=9.29x10^8\\r_{o}=1.43x10^{12}

If the sun considered as x=0 on the axis to put the center of the mass as a:

m_{su}*r_{o}=(m_{sa}+m_{su})*r_{1}

solve to r1

r_1=\frac{m_{sa}*r_{o}}{m_{sa}+m_{su}}=\frac{5.68x10^{26}*1.43x10^{12}}{5.68x10^{26}+5.68x10^{26}}

r_1=1.428x10^9m

Now convert to coordinates centered on the center of mass.  call the new coordinates x' and y' (we won't need y').  Now since in the sun centered coordinates the angular momentum was  

L = \frac{m_{sa}*2*pi*r_1^2}{T}

where T = orbital period

then L'(x',y') = L(x) by conservation of angular momentum.  So that means

L_{sun}=\frac{m_{sa}*2*\pi *( 2r_{o}*r_1 -r_1^2)}{T}

Since

L_{su}= m_{su}*v_{su}*r_1

then

v_{su}=\frac{m_{sa}*2*pi*(2r_{o}*r_{1}-r_{1}^2)}{T*m_{sa}*r_1}

v_{su} = 19.44 m/s

7 0
3 years ago
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