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svet-max [94.6K]

3 years ago

7

Presence of _______- in retina gives us night vision that is help us to see in dim light.Required to answer. Single choice. Imme

rsive Reader (1 Point) cornea retina lens rods

1 answer:

qwelly [4]3 years ago

7 0

Answer: Rods

Explanation:

The rod cells in the retina are the reason we are able to see at night and in dim light. They exist on the edges of the retina which is why they are also very useful for the peripheral vision of a human.

Rod cells number over 90 million in the eyes and although very useful for seeing in dimmer light, they are not very useful for color vision which is why humans see less colors in the dark.

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A trolley is moving at constant speed in a friction compensated track. Some plasticine is dropped on the trolley and sticks on i

Maurinko [17]

Answer:

Since the momentum of the body remains constant ( conserved) the trolley slows down (its velocity reduces) since its mass increases.

5 0

2 years ago

Compare the strong and weak megnetic field

garri49 [273]

The strong magnetic fields is Long rang attractive power Reuther then week magnetic field.

7 0

3 years ago

An object of volume 0.0882 m3 is

vfiekz [6]

The density of the fluid is 776.3 $m^{-3}$

<u>Explanation:</u>

Buoyant force is the upward pushing force whenever an object is trying to get immersed in fluid. So this is the force given by the fluid on the object which is trying to get immersed. The buoyant force is found to be directly proportional to the product of density of the object, volume of the object. And here the acceleration due to gravity will be acting as proportionality constant.

$Buoyant force = Density \times Volume \times Acceleration$

As, buoyant force is given as 671 N and volume is 0.0882 $m^{3}$ and acceleration is known as 9.8 m/ $s^{2}$ . Then density is

$\text { Density }=\frac{\text { Buoyant force }}{\text {Volume } \times \text {Acceleration}}$

Thus,

$\text { Density }=\frac{671}{0.0882 \times 9.8}=\frac{671}{0.86436}=776.296 \mathrm{kg} / \mathrm{m}^{3}$

Density is 776.3 kg $m^{-3}$ .

7 0

3 years ago

If the average velocity during the athlete's walk back

goblinko [34]

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

$T=\frac{d}{1.50}$

From the question we are told

If the average velocity during the athlete's walk back to the starting line in Guided Example 2.5 is – 1.50 m/s,

Generally the equation Time spent is mathematically given as

T=\frac{d}{v}

Therefore

$T=\frac{d}{1.50}$

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

$T=\frac{d}{1.50}$

For more information on this visit

brainly.com/question/22271063

4 0

3 years ago

A high-pass filter consists of a 1.66 μF capacitor in series with a 80.0 Ω resistor. The circuit is driven by an AC source with

Julli [10]

Explanation:

Given that,

Capacitor $C=1.66\ \mu F$

Resistor $R=80.0\ \Omega$

Peak voltage = 5.10 V

(A). We need to calculate the crossover frequency

Using formula of frequency

$f_{c}=\dfrac{1}{2\pi R C}$

Where, R = resistor

C = capacitor

Put the value into the formula

$f_{c}=\dfrac{1}{2\pi\times80.0\times1.66\times10^{-6}}$

$f_{c}=1198.45\ Hz$

(B). We need to calculate the $V_{R}$ when $f = \dfrac{1}{2f_{c}}$

Using formula of $V_{R}$

$V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})$

Put the value into the formula

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=2.280\ Volt$

(C). We need to calculate the $V_{R}$ when $f = f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=3.606\ Volt$

(D). We need to calculate the $V_{R}$ when $f = 2f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=4.561\ Volt$

Hence, This is the required solution.

8 0

3 years ago