A rectangular tank is filled to a depth of 10m with freshwater and open to air at atmospheric pressure.
1 answer:
Answer:
<em>1.</em> <em>39068.07 N</em>
<em>2. 19534.036 N</em>
Explanation:
depth of water h = 10 m
atmospheric pressure Patm = 101325 Pa
density of water p = 1000 kg/m^3
acceleration due to gravity g = 9.81 m/s^2
pressure due to depth of water = pgh
P = 1000 x 9.81 x 10 = 98100 Pa
total pressure on the bottom of the tank is Patm + p = 101325 + 98100 = 199425 Pa
Left plug has diameter = 50 cm = 0.5 m
radius = 0.5/2 = 0.25 m
height = 1 cm = 0.01 m
<em>height below tank surface = 10 - 0.01 = 9.99</em>
pressure at this depth = 1000 x 9.81 x 9.99 = 98001.9 Pa
<em>total pressure = Patm + P = 101325 + 98001.9 = 199326.9 Pa</em>
surface area of plug = π = 3.142 x
= 0.196
<em>force required to lift left plug = pressure x area</em>
F = 199326.9 x 0.196 = <em>39068.07 N</em>
<em>The right side is a hemisphere with the same diameter, therefore surface area is half of the left plug</em>
A = 0.196/2 = 0.098
force F required to lift right plug = 199326.9 x 0.098 =<em> 19534.036 N</em>
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Answer:
Option A = 1.
Explanation:
So, in order to solve this question we are given the Important infomation or data or parameters in the question above as;
(1). First, Both objects A and D represent fixed.
(2). Both objects A and D are negatively-charged particles of equal magnitude.
(3). "Object B represents a fixed, positively-charged particle (equal, but opposite charge from A and D)."
(4). "Object C shows a moving, positively-charged particle."
So, our mission is to determine the arrow that would correctly show the force of attraction or repulsion on object C caused by the other two objects.
We can do that by drawing out the forces of attraction and the resultants. Therefore, CHECK THE ATTACHED FILE/PICTURE FOR THE DRAWINGS.
The forces of attraction due to objects A and B on on object C will be towards themselves. Hence, the resultant is ONE(1).
