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stellarik [79]
3 years ago
8

Determine the pressure point in a liquid container within a pool. The height of the liquid column is 9 m, The density of the liq

uid is 997 kg/m^3, and gravity is 9.8 m/s^2.
Physics
1 answer:
7nadin3 [17]3 years ago
6 0

Absolute pressure: 1.89\cdot 10^5 Pa

Explanation:

The total pressure at a certain depth of a liquid is given by

p=p_0 + \rho g h

where

p_0 = 1.01\cdot 10^5 Pa is the atmospheric pressure

\rho is the density of the liquid

g is the acceleration of gravity

h is the depth of the liquid

For the liquid in this problem, we have

\rho=997 kg/m^3 is the density

g=9.8 m/s^2

h = 9 m is the column of liquid

Substituting, we find

p=1.01\cdot 10^5 + (997)(9.8)(9)=1.89\cdot 10^5 Pa

Learn more about pressure in a fluid:

brainly.com/question/9805263

#LearnwithBrainly

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Two 2.5 kg physical science textbooks on a bookshelf are 0.14 m apart. What is the magnitude of the gravitational attraction bet
finlep [7]

Answer:

Explanation:

According to Newton's laws of gravitation force of attraction is given by

F = GMm/R²

G = 6.67 X 10⁻¹¹

,M = 2.5 Kg ,

m = 2.5 Kg ,

R = .14 m

=

F = 6.67 X 10⁻¹¹ X 2.5 X 2.5 / 0.14²

= 2126.9 X 10 ⁻¹¹ N.

5 0
4 years ago
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Two identical conducting spheres A and B carry equal charge. They are separated by a distance much larger than their diameters.
VashaNatasha [74]
What a delightful little problem !
Here's how I see it:

When 'C' is touched to 'A', charge flows to 'C' until the two of them are equally charged.  So now, 'A' has half of its original charge, and 'C' has the other half.

Then, when 'C' is touched to 'B', charge flows to it until the two of <u>them</u> are equally charged.  How much is that ?  Well, just before they touch, 'C' has half of an original charge, and 'B' has a full one, so 1/4 of an original charge flows from 'B' to 'C', and then each of them has 3/4 of an original charge.

To review what we have now:  'A' has 1/2 of its original charge, and 'B' has 3/4 of it.

The force between any two charges is:

F = (a constant) x (one charge) x (the other one) / (the distance between them)².

For 'A' and 'B', the distance doesn't change, so we can leave that out of our formula.

The original force between them was  3 = (some constant) x (1 charge) x (1 charge).

The new force between them is  F = (the same constant) x (1/2) x (3/4) .

Divide the first equation by the second one, and you have a proportion:

3 / F  =  1 / ( 1/2 x 3/4 )

Cross-multiply this proportion:

3 (1/2 x 3/4)  = F

F  =  3/2 x 3/4  =  9/8  =  <em>1.125 newton</em>.


That's my story, and I'm sticking to it.
 
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3 years ago
Two children are pulling and pushing a 30.0 kg sled. The child pulling the sled is exerting a force of 12.0 N at a 45o angle. Th
irinina [24]
I'm not quite sure what happens to Fay so I didn't finish but hope it helps

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3 years ago
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Lucinda frequently finds herself getting upset and yelling at her husband, co-workers, and friends. What therapeutic technique w
IrinaK [193]

The therapeutic technique Lucinda would start is behavioral self-monitoring.

<u>Explanation:</u>

Behavioral self-monitoring, introduced by Mark Snyder in the 1970s, is an ability of an individual to regulate the behavior to accommodate in various social situations.

Here, Lucinda frequently gets upset and yells at her husband, co-workers, and friends. She wants to keep a record of where, when, or with whom she might lose her temper in order to evaluate her behavior.

The individuals with high levels of self-monitoring possess greater skill at bridging social situations and in relationships they seem to be more flexible and adaptable.

6 0
4 years ago
An aerobatic airplane pilot experiences weightlessness as she passes over the top of a loop the loop maneuver. The acceleration
Natali5045456 [20]

The radius of the loop is 18.9 km

Explanation:

When the airplane is at the top of the loop, the pilot experiences two forces:

  • The force of gravity, acting downward, of magnitude mg
  • The normal reaction exerted by the seat on the pilot, also acting  downward, N

Since the plane is moving in a circular motion, the net force on the pilot must be equal to the centripetal force, therefore we can write:

mg+N = m\frac{v^2}{r}

where

m is the mass of the pilot

g=9.8 m/s^2 is the acceleration of gravity

N is the normal reaction

v = 430 m/s is the speed of the plane

r is the radius of the loop

Here we are told that the pilot feel weightless at the top of the loop: this means that the normal reaction is zero,

N = 0

Therefore the equation becomes

mg=\frac{mv^2}{r}

And so we can find the radius of the loop:

r=\frac{v^2}{g}=\frac{430^2}{9.8}=18.9 \cdot 10^3 m = 18.9 km

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