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ivolga24 [154]
3 years ago
6

Okno okrętu podwodnego ma powierzchnię 0,3 m2 i znajduje się na głębokości, na której ciśnienie wywierane przez słup wody wynosi

500 kPa. Dokończ zdanie. Wartość siły parcia wody na okno wynosi.
Physics
1 answer:
OLEGan [10]3 years ago
4 0

Answer:

The value of the water pressure force on the window is 150,000 N.

Wartość siły nacisku wody na okno wynosi 150000 N.

Explanation:

English Translation

The submarine window has an area of ​​0.3 m2 and is located at a depth at which the pressure exerted by the water column is 500 kPa. Finish the sentence. The value of the water pressure force on the window is.

Solution.

Pressure exerted by a force on a surface area perpendicular to the line of action of that force is given by the formula

P = (F/A)

where F = applied force, in this case, water pressure force on the window = ?

A = Area upon which the force is acting = Area of the window = 0.3 m²

P = 500 kPa = 500,000 Pa

500,000 = (F/0.3)

F = 500,000 × 0.3 = 150,000 N

In Polish/Po Polsku

Ciśnienie wywierane przez siłę na pole powierzchni prostopadłe do linii działania tej siły jest określone wzorem

P = (F / A)

gdzie F = przyłożona siła, w tym przypadku siła nacisku wody na okno =?

A = Obszar, na który działa siła = Powierzchnia okna = 0,3 m²

P = 500 kPa = 500000 Pa

500,000 = (F/0.3)

F = 500,000 × 0.3 = 150,000 N

Hope this Helps!!!

Mam nadzieję że to pomoże!!!

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Answer:

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Explanation:

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The number of protons equals the atomic number. TRUE OR FALSE?
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Answer:

true

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Answer:

A. v_{3}=12.17m/s

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Explanation:

From the exercise we know that the car and the truck are traveling eastward. I'm going to name the car 1 and the truck 2

v_{1}=5.79m/s\\m_{1}=102kg\\v_{2}=18.5m/s\\m_{2}=103kg

A. Since the two vehicles become entangled the final mass is:

m_{3}=102kg+103kg=205kg

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p_{1}=p_{2}

m_{1}v_{1}+m_{2}v_{2}=m_{3}v_{3}

v_{3}=\frac{m_{1}v_{1}+m_{2}v_{2}}{m_{3} }=\frac{(102kg)(5.79m/s)+(103kg)(18.5m/s)}{(205kg)}

v_{3}=12.17m/s

B. The change in velocity of both vehicles are:

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v_{car}=v_{f}-v_{o}=12.17m/s-5.79m/s=6.38m/s

For the truck

v_{truck}=12.17m/s-18.5m/s=-6.3m/s

C. The change in kinetic energy is:

ΔK=K_{2}-K_{1} =\frac{1}{2}m_{3}v_{3}^{2}-(\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2})

ΔK=\frac{1}{2}(205)(12.17)^{2}-(\frac{1}{2}(102)(5.79)^{2}+\frac{1}{2}(103)(18.5)^{2})=-4.13x10^{3}J

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Answer:

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Momentum = 2.59×10¯²² Kgm/s

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