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Ghella [55]
2 years ago
5

Find relation for pressure at a depth h in a liquid of density d​

Physics
2 answers:
klemol [59]2 years ago
5 0
Can you elaborate more
skad [1K]2 years ago
5 0

Explanation:

Hey there!!

  • Pressure is directly proportional to depth of a liquid as when there is more depth then it exerts more pressure and when there is less depth there is less pressure.
  • Pressure is also directly proportional to density as more the density there more pressure and less the density less the pressure.

<em><u>Hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

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A 600-kg car traveling at 30.0 m/s is going around a curve having a radius of 120 m that is banked at an angle of 25.0°. The coe
andrey2020 [161]

Answer:

The magnitude of force is 1593.4N

Explanation:

The sum of the horizontal components of the friction and the normal force will be equal to the centripetal force on the car. This can be represented as

fcostheta + Nsintheta = mv^2/r

Where F = force of friction

Theta = angle of banking

N = normal force

m = mass of car

v = velocity of car

r = radius of curve

The car has no motion in the vertical direction so the sum of forces = 0

The vertical component of the normal force acts upwards whereas the weight of the car and the vertical component friction acts downwards.

Taking the upward direction to be positive,rewrite the equation above to get:

Ncos thetha = mg - fsintheta =0

Ncistheta = mg + fain theta

N = mg/cos theta + sintheta/ costheta

fcostheta +[mg/costheta + ftan theta] sin theta = mv^2/r

Substituting gives:

f = (1/(costheta + tanthetasintheta) + mgtantheta = mv^2/r - mgtantheta)

Substituting given values into the above equation

f = 1/(cos25 + tan 25 )(sin25)[ 600×30/120 - (600×9.81)tan

f = 1593.4N25

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Explanation:

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