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4 years ago

Cognitive maps allow organisms to

Physics

1 answer:

RoseWind [281]4 years ago

5 0

B. Mentally represent their environment

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a person is sitting on the last bench can see clearly see things written on book but cannot see them distinctly on board. what t

marysya [2.9K]

Answer:

I think concave

Explanation:

7 0

3 years ago

Compute the expected shell-model quadrupole moment of 209Bi () and compare with the experimental value, - 0.37 b

Over [174]

Answer:

0.22 b

Explanation:

Quadrupole moment of the nucleon is,

$Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2}$

And also,

$R^{2}=R^{2} _{0}A^{\frac{2}{3} }$

And, $R _{0}=1.2\times 10^{-15}m$

Now,

$Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2} _{0}A^{\frac{2}{3} }$

For Bismuth $j=\frac{9}{2}$ and A is 209.

$Q=-\frac{2\frac{9}{2} -1}{2(\frac{9}{2} +1)}\frac{3}{5}(1.2\times 10^{-15}) ^{2}(209)^{\frac{2}{3} }\\Q=0.628\times 35.28\times 10^{-30} \\Q=22.15\times 10^{-30} m^{2} \\Q=0.2215\times 10^{-28} m^{2} \\Q=0.22 barn$

Therefore, the expected value of quadrupole is 0.22 b which is quite related with experimental value which is 0.37 b

3 0

3 years ago

On a distance-time graph, time is shown on the y-axis.<br> A) True<br> B) False

Artist 52 [7]

Answer:

false : In distance time graph,time is shown on the x -axis

7 0

3 years ago

Spiders may "tune" strands of their webs to give an enhanced response at frequencies corresponding to the frequencies at which d

Bingel [31]

Answer:

0.0000167283525619 N

Explanation:

$\rho$ = Density of silk = 1300 kg/m³

d = Diameter = 0.002 mm

r = Radius = 0.001 mm

l = Length = 16 cm

f = Frequency = 200 Hz

Mass of the string is

$m=\rho V\\\Rightarrow m=\rho Al\\\Rightarrow m=1300\times \pi (0.001\times 10^{-3})^2\times 0.16\\\Rightarrow m=6.5345127195\times 10^{-10}\ kg$

Frequency is given by

$f=\dfrac{1}{2l}\sqrt{\dfrac{T}{m/l}}\\\Rightarrow T=4lf^2m\\\Rightarrow T=4\times 0.16\times 200^2\times 6.5345127195\times 10^{-10}\\\Rightarrow T=0.0000167283525619\ N$

The tension on the string is 0.0000167283525619 N

3 0

3 years ago

A 77.0 kg woman slides down a 42.6 m long waterslide inclined at 42.3. at the bottom, she is moving 20.3 m/s.how much work did f

zhuklara [117]

Answer:

5.791244495 KNm

Explanation:

The height h is given by, $h=42.6sin42.3^{o}$

Potential energy, PE is given by

PE=mgh where m is mass of the woman, g is acceleration due to gravity whose value is taken as $9.81 m/s^{2}$ and h is already given hence substituting 77 Kg for m we obtain

$PE=77*9.81*42.6sin42.3^{o}=21656.7095 Nm$

PE=21.6567095 KNm

We also know that Kinetic energy is given by $0.5mv^{2}$ where v is the velocity and substituting v for 20.3 we obtain

$KE=0.5*77*20.3^{2}=15865.465 Nm$

KE=15.865465 KNm

Friction work is the difference between PE and KE hence

Friction work=21.6567095 KNm-15865.465 Nm=5.791244495 KNm

8 0

3 years ago