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Andreyy89
2 years ago
14

Animal studies indicate that

Physics
1 answer:
notka56 [123]2 years ago
3 0

Answer: mammals may be relatively better at solving problems than birds.

You might be interested in
Two objects, M = 15.3 ks and m = 8.29 kg are connected with an ideal string and suspended by a pulley (which rotates with no fri
Scilla [17]

Answer:

(a) 7 m/s

(b) 931 rad/s

(c) 0.716 s

Explanation:

Gravity would be exerting on the 2 masses

G_1 = Mg = 15.3*9.81 = 150.093N

G_2 = mg = 8.29*9.81= 81.32N

Since heavier, mass 1 (M) would be the one pulling down, while mass 2 is being pulled up.

So the net force on mass 1 is

F = G_1 - G_2 = 68.77N

This force would generate torque on the solid pulley

T = FR = 68.77 * 0.0075 = 0.5158 Nm

We can also calculate the pulley moments of inertia, with it being solid

I = 0.5MpR^2 = 0.5*14.1*0.0075^2 = 0.000396563kgm^2

From there we can calculate the angular acceleration of the pulley, which generates the entire system motion

\alpha = \frac{T}{I} = \frac{0.5158}{0.000396563} = 1300.58 rad/s^2

Since the system is moved by a distance of d = 2.5m, the pulley would have turn an angle of

\theta = \frac{d}{R} = \frac{2.5}{0.0075} = 333 rad

(c)The time it takes to get to this distance is

\theta = \frac{\alpha t^2}{2}

t^2 = \frac{2\theta}{\alpha} = \frac{2*333}{1300.58} =0.513

t = \sqrt{0.513} = 0.716s

(b)The final angular speed of the disk is

\omega = \alpha t = 1300.58*0.716 = 931 rad/s

(a) And so the perimeter speed of the pulley, which is also speed of mass 1 when it comes to d = 2.5 m is

v = \omega R = 931*0.0075 \approx 7m/s

5 0
2 years ago
If the range of the projectile is 4.3 m, the time-of-flight is T = 0.829 s, and air resistance is negligible, determine the foll
ankoles [38]

Answer

given,

range of the projectile = 4.3 m

time of flight = T = 0.829 s

v =\dfrac{d}{T}

v =\dfrac{4.3}{0.829}

     v = 5.19 m/s

vertical component of velocity of projectile

v_y = gt'

v_y = 9.8 \times {\dfrac{T}{2}}

v_y = 9.8 \times {\dfrac{0.829}{2}}

v_y =4.06\ m/s

a) Launch angle

 \theta = tan^{-1}(\dfrac{v_y}{v})

 \theta = tan^{-1}(\dfrac{4.06}{5.19})

    θ = 38°

b) initial speed of projectile

  v = \sqrt{v_x^2 + v_y^2}

  v = \sqrt{5.19^2 + 4.06^2}

         v = 6.59 m/s

c) maximum height reached by the projectile

     y_{max}=v_{avg}t'

     y_{max}=\dfrac{1}{2}v_y\dfrac{T}{2}

     y_{max}=\dfrac{1}{2}\times(g\dfrac{T}{2})\times\dfrac{T}{2}

     y_{max}=\dfrac{gT^2}{8}          

7 0
3 years ago
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:

For more than 40 years, results from scientific ocean drilling have contributed to global understanding of Earth’s biological, chemical, geological, and physical processes and feedback mechanisms. The majority of these internationally recognized results have been derived from scientific ocean drilling conducted through three programs—the Deep Sea Drilling Project (DSDP; 1968-1983), the Ocean Drilling Program (ODP; 1984-2003), and the Integrated Ocean Drilling Program (IODP; 2003-2013)—that can be traced back to the first scientific ocean drilling venture, Project Mohole, in 1961. Figure 1.1 illustrates the distribution of drilling and sampling sites for each of the programs, and Appendix A presents tables of DSDP, ODP, and IODP legs and expeditions. Although each program has benefited from broad, international partnerships and research support, the United States has taken a leading role in providing financial continuity and administrative coordination over the decades that these programs have existed. Currently, the United States and Japan are the lead international partners of IODP, while a consortium of 16 European countries and Canada participates in IODP under the auspices of the European Consortium for Ocean Research Drilling (ECORD). Other countries (including China, Korea, Australia, New Zealand, and India) are also involved.

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
A 35.8 kg box initially at rest is pushed 2.38 m along a rough, horizontal floor with a constant applied horizontal force of 108
tiny-mole [99]

Answer:

The work done by the applied force is 259.22 J.

Explanation:

The work done by the applied force is given by:

W = F*d

Where:

F: is the applied horizontal force = 108.915 N

d: is the distance = 2.38 m  

Hence, the work is:

W = F*d = 108.915 N*2.38 m = 259.22 J

Therefore, the work done by the applied force is 259.22 J.

I hope it helps you!                                                

6 0
2 years ago
A truck is carrying a refrigerator as shown in the figure. The height of the refrigerator is 158.0 cm, the width is 60.0 cm. The
Arada [10]

The maximum acceleration the truck can have so that the refrigerator does not tip over is 4.15 m/s².

<h3>What will be the maximum acceleration of the truck to avoid tipping over?</h3>

The maximum acceleration is obtained by taking clockwise moments about the tipping point of rotation.

Clockwise moment = Anticlockwise moment

Ft * 1.58 m = F * 0.67 m

where

  • Ft is tipping force = mass * acceleration, a
  • F is weight = mass * acceleration due to gravity, g

m * a * 1.58 = m * 9.81 * 0.67

a = 4.15 m/s²

The maximum acceleration the truck can have so that the refrigerator does not tip over is 4.15 m/s².

In conclusion, the acceleration of the truck is found by taking moments about the tipping point.

Learn more about moments of forces at: brainly.com/question/27282169

#SPJ1

3 0
1 year ago
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