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Marianna [84]
3 years ago
9

To practice tactics box 13.1 hydrostatics. in problems about liquids in hydrostatic equilibrium, you often need to find the pres

sure at some point in the liquid. this tactics box outlines a set of rules for thinking about such hydrostatic problems.
part a) find the pressures pa and pb at surfaces a and b in the tube, respectively. use patmos to denote atmospheric pressure.

express your answers, separated by a comma, in terms of one or both of the variables pgas and patmos.

part b) as stated in rule 3 in the tactics box, it is always convenient to use horizontal lines in hydrostatic problems. in each one of the following sketches, a different horizontal line is considered. which sketch would be more useful in solving the problem of finding the gas pressure?
Physics
1 answer:
OLEGan [10]3 years ago
5 0

Answer:

A. The pressure denoted as Pa and Pb at the surfaces of A and B in the tube is

PA= Pgas

PB= Patmos

B. The second sketch

C. The gas pressure is

Pgas= Patmos+ rho.g(h2-h1)

= 1atm + rho.g (h2-h1)

Explanation:

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A solid 0.4550 kg ball rolls without slipping down a track toward a vertical loop of radius R = 0.6750 m . What minimum translat
Talja [164]

Answer:

u = 3.35 m/s

Explanation:

given,

mass , m = 0.455 kg  

R = 0.675 m

Height of Loop = 1.021 m

the speed required at the top of loop be v

equating the force vertically

m g =\dfrac{mv^2}{r}

9.81 =\dfrac{v^2}{0.675}

v² = 6.622

v = 2.57 m/s

Let the initial speed of ball be u

using conservation of energy

\dfrac{1}{2}mu^2 + \dfrac{1}{2}I\omega^2 + m g h = \dfrac{1}{2}I\omega^2 + \dfrac{1}{2}mv^2+ m g (2 R)

where, I =\dfrac{2}{5}mr^2

\dfrac{1}{2}mu^2 + \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{u}{r})^2 + m g h = \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{v}{r})^2 + \dfrac{1}{2}mv^2+ m g (2 R)

0.7 u^2 + g H = 0.7 v^2 + g(2R)

0.7 u^2 +9.81 \times 1.021= 0.7\times 2.57^2 + 9.81 \times 2\times 0.675)

0.7 u² = 7.85092

u² = 11.2156

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8 0
3 years ago
If 28.0 J of work is done by an external force to move a charge from a potential of 8.0 V 8.0 V to a potential of 4.0 V , 4.0 V,
jonny [76]

Answer:

Change in electric potential energy is -28.0 J

Explanation:

Electric potential energy is defined as the work is done to move a charge particle from one position to another in space in the presence of other charge particle or electric potential.

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According to the problem, Work Done is equal to 28 J. Thus,

Change in electric potential energy = -28 J  

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3 years ago
Which nucleus completes the following equation?
siniylev [52]
A is the answer I really don’t know the answer to that but if u can u can help me on my work
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3 years ago
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When Bella pushes a box with a mass of 5.25 kilograms with a force of 15.75 newtons, it accelerates at a rate of 2.5 meters/seco
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Sure !

Start with Newton's second law of motion:

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This formula is so useful, and so easy, that you really
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Now, watch:

The mass of the box is 5.25 kilograms, and the box is
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                    Net Force = (mass) x (acceleration)

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But hold up, hee haw, whoa !  Wait a second !
Bella is pushing with a force of 15.75 newtons, but the box
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What happened to the rest of Bella's force ? ?

==>  Friction is pushing the box in the opposite direction,
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How much ?

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The correct answer is - A. Plants store solar energy; the plants die; the plants are compressed; solar energy is released;

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3 years ago
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