All categories

3 years ago

When you blink your eye, the upper lid goes from rest with your eye open to completely covering your eye in a time of 0.024 s.

Physics

1 answer:

dusya [7]3 years ago

4 0

a) Distance moved by the top lid during a blink: 1 cm (estimate)

b) The acceleration is $34.7 m/s^2$

c) The final speed is 0.83 m/s

Explanation:

For the purpose of this problem, we can estimate the size of the eye from the top lid to the bottom lid to be 1 cm, so this is the distance moved by the top lid during a blink.

Assuming the motion of the eyelid to be at constant acceleration, we can find the acceleration by using the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

For the eyelid, we have:

u = 0 (it starts from rest)

s = 1 cm = 0.01 m is the distance covered

t = 0.024 s is the time taken

Solving for a, we find the acceleration:

$a=\frac{2s}{t^2}=\frac{2(0.01)}{0.024^2}=34.7 m/s^2$

The final speed of the upper eyelid can be found by using another suvat equation:

$v=u+at$

where

v is the final speed

u is the initial speed

a is the acceleration

t is the time

For the eyelid in this problem, we have

u = 0

$a=34.7 m/s^2$

t = 0.024 s

Substituting, we find the final speed:

$v=0+(34.7)(0.024)=0.83 m/s$

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

You might be interested in

Light of wavelength 600 nm passes though two slits separated by 0.22mm and is observed on a screen 1.1m behind the slits. The lo

Ne4ueva [31]

Answer:

$y(m=1)=\frac{(1)(600*10^{-9}m)(1.1m)}{0.22m}=3*10^{-6}m\\\\y(m=1)=\frac{(-1)(600*10^{-9}m)(1.1m)}{0.22m}=-3*10^{-6}m$

Explanation:

We have to take into account the expression for the position of the fringes

$dsen\theta=m\lambda\\y=\frac{m\lambda D}{d}$

where m is the number of the maximum, d is the separation of the slits, D is the distance to the screen.

(a) By replacing we obtain

$y(m=1)=\frac{(1)(600*10^{-9}m)(1.1m)}{0.22m}=3*10^{-6}m\\\\y(m=1)=\frac{(-1)(600*10^{-9}m)(1.1m)}{0.22m}=-3*10^{-6}m$

(b) more information is required to solve this point. Please complete the information.

HOPE THIS HELPS!

4 0

3 years ago

Read 2 more answers

Two flat 4.0 cm × 4.0 cm electrodes carrying equal but opposite charges are spaced 2.0 mm apart with their midpoints opposite ea

serious [3.7K]

Answer:

1.77 x 10^-8 C

Explanation:

Let the surface charge density of each of the plate is σ.

A = 4 x 4 = 16 cm^2 = 16 x 10^-4 m^2

d = 2 mm

E = 2.5 x 10^6 N/C

ε0 = 8.85 × 10-12 C2/N ∙ m2

Electric filed between the plates (two oppositively charged)

E = σ / ε0

σ = ε0 x E

σ = 8.85 x 10^-12 x 2.5 x 10^6 = 22.125 x 10^-6 C/m^2

The surface charge density of each plate is ± σ / 2

So, the surface charge density on each = ± 22.125 x 10^-6 / 2

= ± 11.0625 x 10^-6 C/m^2

Charge on each plate = Surface charge density on each plate x area of each plate

Charge on each plate = ± 11.0625 x 10^-6 x 16 x 10^-4 = ± 1.77 x 10^-8 C

7 0

3 years ago

A fluid is any substance that can flow and take the shape of its container A) true B) false

Tpy6a [65]

Its true...water takes the shape of its container and it flows..please give me 5 stars and brainliest

7 0

4 years ago

Read 2 more answers

In the Rutherford model of hydrogen the electron (mass=9.11x10-31 kg) is in a planetary type orbit around the proton (mass=1.67x

Andrei [34K]

Answer:

(a) $F_g=1.62*10^{-48}N$

(b) $F_e=3.68*10^{-9}N$

Explanation:

(a) We use Newton's law of universal gravitation, in order to calculate the gravitational force between electron and proton:

$F_g=-G\frac{m_1m_2}{r^2}$

Where G is the Cavendish gravitational constant, $m_1$ and $m_2$ are the masses of the electron and the proton respectively and r is the distance between them:

$F_g=-6.67*10^{-11}\frac{N\cdot m^2}{kg^2}\frac{(9.11*10^{-31}kg)(1.67*10^{-27}kg)}{(2.5*10^{-10}m)^2}\\F_g=-1.62*10^{-48}N$

The minus sing indicates that the force is repulsive. Thus, its magnitude is:

$F_g=1.62*10^{-48}N$

(b) We use Coulomb's law, in order to calculate the electric force between electron and proton, here k is the Coulomb constant and e is the elementary charge:

$F_e=-k\frac{e^2}{r^2}\\F_e=-8.99*10^{9}\frac{N\cdot m^2}{C^2}\frac{(1.6*10^{-19}C)^2}{(2.5*10^{-10}m)^2}\\F_e=-3.68*10^{-9}N$

Its magnitude is:

$F_e=3.68*10^{-9}N$

6 0

3 years ago

In the last part of the experiment you need to investigate the behaviour of gamma radiation as it passes through air. A gamma so

andrew11 [14]

Answer:

linear cart C Vs 1/R² or log cart C Vs R

Explanation:

Gamma radiation is very high energy electromagnetic rays, but its behavior is the same as for all radiation. By the principle of conservation of energy after the radiation is emitted, it must be distributed on a spherical surface which determines the behavior of the inverse of the square.

In this experiment you are measuring the rate of counts by time (C), this must be the dependent variable since it is not controlled by the experimenter and on the other hand it measures the distance (X) this is the independent variable since it is the one that we can control.

To make a graph with this data, the counting rate must be plotted against the inverse of the squared distance (1/R²). On the Y axis the counts per second and the X 1 / R² axis, with this graph a line must be obtained.

Another graph that we can make on double logarithmic paper where the Y axis plotted the counting rate and on the X axis the distance, the slope should give -2.

C == A / R²

Log C = log A -2 log R

With either of the two graphs, the law of the inverse of the square is tested

7 0

3 years ago