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KIM [24]
3 years ago
6

Existing rocks can become sedimentary rocks when they are subjected to which conditions?

Physics
1 answer:
Svetllana [295]3 years ago
4 0
The correct answer is C. Since that really was a tricky question I made sure it was correct by checking on Google. Hope I helped! - Amber
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What is wind or wind currents? I'm on that subject and don't get what it is. for the subject, I chose "Physics" because I don't
Licemer1 [7]

Explanation:

Wind Currents. currents of the surface waters of oceans and seas resulting from the action of wind on the surface of the water: Wind currents arise owing to the combined influence of the forces of friction, turbulent viscosity, pressure gradient, deflecting force of the earth's rotation, and so on.

6 0
4 years ago
Read 2 more answers
To meet a U.S. Postal Service requirement, employees' footwear must have a coefficient of static friction of 0.5 or more on a sp
liq [111]

Answer:

Minimum time interval (t2)=0.90 SECONDS

Explanation:

  • coefficient of friction for employees footwear = 0.5
  • coefficient of friction for typical athletic shoe = 0.810
  • frictional force = coefficient of friction X acceleration due to gravity X mass of body
  • Acceleration due to gravity is a constant = 9.81 m/s
  • Let frictional force for employee footwear = FF1
  • Let frictional force for athletic footwear =FF2

                 FF1 = O.5 X 9.81 X mass of body

                         = 4.905 x mass of body

                  FF2 = 0.810 X 9.81 X mass of body

                          = 7.9461 x mass of body

The body started from rest there by making the initial velocity zero ( u = 0)

From d= ut + 1/2 a x t^{2}

  •      d = \frac{1}{2} x a x t^{2}  .....................................i  

            where d= distance and it is given as 3.25m

  •          F =ma  ...................................ii

making acceleration subject of the formula from equation ii

  •              a =\frac{F}{m}

         Making t subject of formula from equation (i)

  • t=\sqrt{\frac{2d}{(f/m} }

    where

  • \frac{FF1}{Mass of body} = 4.905
  • \frac{FF2}{Mass of body} =7.9461

  Let

  •            t1 = minimum time taken for frictional force for employee foot wear
  •                                 t1 = \sqrt{\frac{6.5}{4.905} } =1.15 seconds

  •                                  t2 = \sqrt{\frac{6.5}{7.9461} } = 0.90 seconds

 

THANK YOU

5 0
3 years ago
When bitten by an animal, one should:
Scrat [10]

Answer:

consider rabies injections if the animal is not found

hope it helps

4 0
3 years ago
Read 2 more answers
henry heavyweight weighs 1450 n and stands on a pair of bathroom scales so that one scale reads twice as much as the other.
nikdorinn [45]
One scale will have a weight of 1087.50 and the next will be 362.50
6 0
4 years ago
3. A sample of argon of mass 6.56 g occupies 18.5 dm? at 305 K. (a) Calculate the work done when the gas expands isothermally ag
aleksandr82 [10.1K]

Answer:

(a) W=-19.25J

(b) W=-52.8J

Explanation:

Hello.

(a) In this case, since the initial volume is 18.5 dm³ and the final volume is 21 dm³ (18.5 +2.5), we can compute the work at constant pressure as shown below:

W=-P\Delta V=-7.7kPa*\frac{1000Pa}{1kPa} (21dm^3-18.5dm^3)*\frac{1m^3}{1000dm^3}\\ \\W=-19.25J

Which is negative as it expands against the given pressure.

(b) Moreover, of the process is carried out reversibly, the pressure can change, therefore, we need to compute the work via:

W=nRTln(\frac{V_1}{V_2} )

Whereas the moles are computed from the given mass of argon:

n=6.56g*\frac{1mol}{39.95g}=0.164mol

Thus, the work is:

W=0.164mol*8.314\frac{J}{mol*K} *305Kln(\frac{18.5dm^3}{21dm^3} )\\\\W=-52.8J

Regards.

4 0
3 years ago
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