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3 years ago

how do the retinas of the eyes of night hunting animals differ from the retinas of animals that hunt during the day time

2 answers:

5 0

Hello,

Night-hunting animals have more rod cells than day-hunting animals. This is most likely because night-hunting animals have to be able to see better than day-hunting animals.

maksim [4K]3 years ago

4 0

Night hunting animals have more rod cells in their retinas which allows them to see better in the dark.

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A small sphere attached to a light rigid rod rotates about an axis perpendicular to and fixed to the other end of the rod. relat

Crank

d.rotating counterclockwise and slowing down This is a matter of understanding the notation and conventions of angular rotations. Positive rotations are counter clockwise and negative rotations are clockwise. An easy way to remember this is the "right hand rule". Make a closed fist with your right hand and have the thumb sticking outwards. If you orient your thumb such that it's pointing in the direction of the positive value along the axis, your fingers will be curled in the positive rotational direction. So in the described scenario, the sphere is rotating in the positive direction (counter clockwise) and decelerating due to the negative angular acceleration. That immediately indicates that options "a", "b", and "e" are wrong since they mention the sphere going clockwise at the beginning. Of the two remaining options "c" and "d", we can discard option "c" since it has the rotation speeding up, and that leaves us with option "d" where the sphere is rotating counter clockwise and slowing down.

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3 years ago

a car goes from a velocity zero to a velocity of 15 meters per second East in 2.1 seconds. What is the car's acceleration?

leva [86]

31.5 would be the acceleration.

3 0

3 years ago

This is my exam question be serious

elena-s [515]

Answer:

hey answer in the comment section

6 0

3 years ago

The Hubble Space Telescope orbits the Earth at approximately 612,000m altitude. Its mass is 11,100 kg and the mass of earth is 5

nexus9112 [7]

Answer:

7.55 km/s

Explanation:

The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:

$G\frac{mM}{R^2}=m\frac{v^2}{R}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$m=11,100 kg$ is the mass of the telescope

$M=5.97\cdot 10^{24} kg$ is the mass of the Earth

$R=r+h=6.38\cdot 10^6 m+612,000 m=6.99\cdot 10^6 m$ is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)

v = ? is the orbital velocity of the Hubble telescope

Re-arranging the equation and substituting numbers, we find the orbital velocity:

$v=\sqrt{\frac{GM}{R}}=\sqrt{\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24} kg)}{6.99\cdot 10^6 m}}=7548 m/s=7.55 km/s$

6 0

3 years ago

"Giant Swing", the seat is connected to two cables as shown in the figure (Figure 1) , one of which is horizontal. The seat swin

Bingel [31]

The horizontal force is m*v²/Lh, where m is the total mass. The vertical force is the total weight (233 + 840)N.

Fx = [(233 + 840)/g]*v²/7.5 

v = 32.3*2*π*7.5/60 m/s = 25.37 m/s 

The horizontal component of force from the cables is Th + Ti*sin40º and the vertical component of force from the cable is Ta*cos40º 

Thh horizontal and vertical forces must balance each other. First the vertical components: 

233 + 840 = Ti*cos40º 

solve for Ti. (This is the answer to the part b) 

Horizontally 

[(233 + 840)/g]*v²/7.5 = Th + Ti*sin40º 

Solve for Th 

Th = [(233 + 840)/g]*v²/7.5 - Ti*sin40º 

using v and Ti computed above.

3 0

3 years ago