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Tanya [424]
3 years ago
9

A semicircular plate with radius 6 m is submerged vertically in water so that the top is 2 m above the surface. Express the hydr

ostatic force against one side of the plate as an integral and evaluate it. (Round your answer to the nearest whole number. Use 9.8 m/s2 for the acceleration due to gravity. Recall that the weight density of water is 1000 kg/m3.)

Physics
1 answer:
dalvyx [7]3 years ago
5 0

Answer: 313920

Explanation:First, we’re going to assume that the top of the circular plate surface is 2 meters under the water. Next, we will set up the axis system so that the origin of the axis system is at the center of the plate.

Finally, we will again split up the plate into n horizontal strips each of width Δy and we’ll choose a point y∗ from each strip. Attached to this is a sketch of the set up.

The water’s surface is shown at the top of the sketch. Below the water’s surface is the circular plate and a standard xy-axis system is superimposed on the circle with the center of the circle at the origin of the axis system. It is shown that the distance from the water’s surface and the top of the plate is 6 meters and the distance from the water’s surface to the x-axis (and hence the center of the plate) is 8 meters.

The depth below the water surface of each strip is,

di = 8 − yi

and that in turn gives us the pressure on the strip,

Pi =ρgdi = 9810 (8−yi)

The area of each strip is,

Ai = 2√4− (yi) 2Δy

The hydrostatic force on each strip is,

Fi = Pi Ai=9810 (8−yi) (2) √4−(yi)² Δy

The total force on the plate is found on the attached image.

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Familiarize yourself with the map showing the DSDP Leg 3 drilling locations and the position of the mid-ocean ridge (Figure 1 to
Inga [223]

Answer:

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As IODP draws to a close in 2013, a new process for defining the scope of the next phase of scientific ocean drilling has begun. Illuminating Earth’s Past, Present, and Future: The International Ocean Discovery Program Science Plan for 2013-20231 (hereafter referred to as “the science plan”), which is focused on defining the scientific research goals of the next 10-year phase of scientific ocean drilling, was completed in June 2011 (IODP-MI, 2011). The science plan was based on a large, multidisciplinary international drilling community meeting held in September 2009.2 A draft of the plan was released in June 2010 to allow for additional comments from the broader geoscience community prior to its finalization. As part of the planning process for future scientific ocean drilling, the National Science Foundation (NSF) requested that the National Research Council (NRC) appoint an ad hoc committee (Appendix B) to review the scientific accomplishments of U.S.-supported scientific ocean drilling (DSDP, ODP, and IODP) and assess the science plan’s potential for stimulating future transformative scientific discoveries (see Box 1.1 for Statement of Task). According to NSF, “Transformative research involves ideas, discoveries, or tools that radically change our understanding of an important existing scientific or engineering concept or educational practice or leads to the creation of a new paradigm or field of science, engineering, or education. Such research challenges current understanding or provides pathways to new frontiers.”3 This report is the product of the committee deliberations on that review and assessment.

HISTORY OF U.S.-SUPPORTED SCIENTIFIC OCEAN DRILLING, 1968-2011

The first scientific ocean drilling, Project Mohole, was conceived by U.S. scientists in 1957. It culminated in drilling 183 m beneath the seafloor using the CUSS 1 drillship in 1961. During DSDP, Scripps Institution of Oceanography was responsible for drilling operations with the drillship Glomar Challenger. The Joint Oceanographic Institutions for Deep Earth Sampling (JOIDES), which initially consisted of four U.S. universities and research institutions, provided scientific advice. Among its numerous achievements, DSDP

Explanation:

7 0
3 years ago
If mechanical energy is conserved, then a pendulum that has a potential energy of 20 J at its highest point and 0.5 J at its low
liraira [26]
At the highest point: kinetic energy is 0 due to the speed  is 0

So the total mechanical energy is 20

Assume no frictions present, then the mechanical energy is conserved

So at the lowest point, kinetic energy = mechanical energy - potential energy

Answer will be 20 - 0.5 = 19.5 J
4 0
3 years ago
What do it mean by the value of gravitational constant is 6.67×10^-11Nm^2/kg^2<br>​
luda_lava [24]

Answer:

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8 0
1 year ago
A car engine applies a force of 65,000 N, how much work is done by the engine as it pushed a car a distance of 75 m?
kupik [55]

Answer:

workdone = force \times distance \\  = 65000 \times 75 \\  = 4,875,000 \: J

3 0
2 years ago
A car moves uphill at 40 km/h and then back downhill at 60 km/h. What is the average speed for the round trip?
jok3333 [9.3K]

Answer:

S_a_v_e_r_a_g_e=48km/h

Explanation:

Ok, the average speed can be calculate with the next equation:

S_a_v_e_r_a_g_e=\frac{Total\hspace{3}distance}{Total\hspace{3}time}   (1)

Basically the car cover the same distance "d" two times, but at different speeds, so:

Total\hspace{3}distance=2*d

and the total time would be the time t1 required to go from A to B plus the time t2 required to go back from B to A:

Total\hspace{3}time=t1+t2

From basic physics we know:

t=\frac{d}{S1}

so:

t1=\frac{d}{S1}

t2=\frac{d}{S2}

Using the previous information in equation (1)

S_a_v_e_r_a_g_e=\frac{2*d}{\frac{d}{S1} +\frac{d}{S2} }=\frac{2*d}{\frac{d*S2+d*S1}{S1+S2} }

Factoring:

S_a_v_e_r_a_g_e=\frac{2*S1*S2}{S1+S2}   (2)

Finally, replacing the data in (2)

S_a_v_e_r_a_g_e=\frac{2*40*60}{60+40} =48km/h

5 0
2 years ago
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