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den301095 [7]
3 years ago
5

A pump is partially submerged in oil and is supported by 4 springs. The oil has a specific gravity of 0.90, the weight of the pu

mp is 14.6 lb, and the submerged volume is 40 in3. What force is needed in the springs so that the pump will remain in static equilibrium
Physics
1 answer:
muminat3 years ago
7 0

Answer:

Explanation:

submerged volume of pump = 40 x 12⁻³ ft³

density of oil = .9 x 62.4 lb ft⁻³ ( specific gravity x density of water )

mass of displaced oil = 40 x 12⁻³ x  .9 x 62.4 lb

= 1.3 lb

buoyant force = 1.3 lb

weight of pump = 14.6 lb

extra force needed for balancing pump by spring

= 14.6 - 1.3

= 13.3 lb .

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Answer:

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3 years ago
1. The planet Jupiter completes a revolution of the sun in 11.5 years. Express it in seconds. Given that one year= 3.154 × 10^7
xenn [34]

Answer:

The planet Jupiter completes one revolution of the sun in 362710000 seconds. Long time, right?

Explanation:

3.154x10^7=3.154x10000000=31540000

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7 0
1 year ago
The motion of a particle is described by the position function s(t) = 2t - 15t +33t+17,t>0 , where is measured in seconds and
8090 [49]

The time when the particle is at rest is at 1.63 s or 3.36 s.

The velocity is positive at when the time of motion is at 0.

The total distance traveled in the first 10 seconds is 847 m.

<h3>When is a particle at rest?</h3>
  • A particle is at rest when the initial velocity of the particle is zero.

The time when the particle is at rest is calculated as follows;

s(t) = 2t³ - 15t² + 33t + 17

v = \frac{ds}{dt} = 6t^2 -30t + 33\\\\at \ rest, \ v = 0\\\\6t^2 - 30t + 33 = 0\\\\6(t- \frac{5}{2} )^2- \frac{9}{2} = 0\\\\t = 1.63\ s \ \ or \ 3.36 \ s

The velocity is positive at when the time of motion is as follows;

0.

The total distance traveled in the first 10 seconds is calculated as follows;

2(10)^3 - 15(10)^2 + 33(10) + 17 = 847 \ m

Learn more about motion of particles here: brainly.com/question/11066673

4 0
2 years ago
The si base unit for length is the
Naya [18.7K]
The meter is the S.I.unit for length. 
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3 years ago
A truck initially traveling at a speed of 22 meters per second increases speed at a constant rate of 2.4 meters per second^2 for
Usimov [2.4K]
Thank you for posting your question here. The total distance traveled by the truck during the 3.2 seconds interval is 83 m. Below is the solution:

d = vit + 1/2 at^2
d = (22m/ s) (3.2s) + 1/2 (2.4m/ s^2) (3.2s)^2
d = 83 m 
Hope the answer helps. 
8 0
3 years ago
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