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2 years ago

7

In order to increase his gravitational potential energy by an amount equal to his kinetic energy at full speed, how high would s

uch a sprinter need to climb

1 answer:

Ksju [112]2 years ago

4 0

Answer:

<h2>Hola eres nuevo yo nací en el 2021 del 4 de junio</h2>

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What is the function of eye lens of the human eye<br>

Artyom0805 [142]

Answer:

Lens of the human Eye is a important and one of that most complex sense organ.

Explanation:

Lens of the human eye it helps that in visualizing light and color perception and objects , glance on the human eye structure and function.

Sense organs are much pretty similar to the camera they help us see the objects clear.

A human eye is the 2.3 cm in diameter and all filled some fluid, and there are following parts in eye:- cornea, Retina , Lens ,Pupil , optic nerves.

cornea :- the cornea is the first transparent part of is called cornea, enters the light through the cornea.

Retina :-it is light sensitive layer that consists of nerve cells,then transmitted to the brain through nerves.

Lens:- behind the pupil there is a transparent structure called lens,it shape focus light on the retina.

Pupil:-it control the value of light that enters the human eye.

Optic nerves is are the two types :- (1) cones (2) Rods .

7 0

3 years ago

If a machine exerts a force of 250N on an object and no work is done, what must have occurred?

ss7ja [257]

The object did not move / was kept from moving.

7 0

3 years ago

In one sentence describe how the electricity in an action potential is generated

vova2212 [387]

Due to influx of potassium ions, electricity is generated in axon of a neuron.

<u>Explanation:</u>

Axon membrane is the semi permeable membrane that is full of potassium and sodium channels. There’s also Sodium Potassium ATPase pumps. When there’s an impulse coming through the synapse, the potassium channels open. This leads to influx of sodium from outside the membrane to inside it. Then the membrane becomes positive.

Then the electricity is generated and its conducted from one part to another. After the impulse is conducted, the sodium potassium pumps come in action which transports 3 sodium inside and 2 potassium outside in consumption to an ATP.

8 0

2 years ago

An asteroid is on a collision course with Earth. An astronaut lands on the rock to bury explosive charges that will blow the ast

forsale [732]

Answer:

The maximum radius the asteroid can have for her to be able to leave it entirely simply by jumping straight up is approximately 1782.45 meters

Explanation:

Whereby the height the astronaut can jump on Earth = 0.500 m, we have the following kinematic equation;

v² = u² - 2·g·h

Where;

v = The final velocity

u = The initial velocity

g = The acceleration due to gravity ≈ 9.8 m/s²

h = The height she jumps

At the maximum height, $h_{max}$ = 0.500 m, she jumps, v = 0, therefore, we have;

0² = u² - 2·g· $h_{max}$

u² = 2 × 9.8 × 0.5 = 9.8

u = √9.8 ≈ 3.13

u = 3.13 m/s

Her initial jumping velocity ≈ 3.13 m/s

Escape velocity, $v_e = \sqrt{\dfrac{2 \cdot G \cdot M}{r} }$

Where;

M = The mass of the asteroid

G = The Universal gravitational constant = 6.67408 × 10⁻¹¹ m³/(kg·s²)

r = The radius of the asteroid

The average density of the Earth = 5515 kg/m³

The mass of the asteroid, M = Density × Volume = 5515 kg/m³× 4/3 × π × r³

The escape velocity, she has, $v_e$ ≈ 3.13 m/s is therefore;

$3.13 = \sqrt{\dfrac{2 \times 6.67408 \times 10^{-11} \times 5515 \times \frac{4}{3} \times \pi \times r^3}{r} } = r \times \sqrt{3.084 \times 10^{-6}}$

$r = \dfrac{3.13}{ \sqrt{3.084 \times 10^{-6}}} \approx 1782.45$

Therefore, the maximum radius of the asteroid can have for her jumping velocity to be equal to the escape velocity for her to be able to leave it entirely simply by jumping straight up = r ≈ 1782.45 meters.

7 0

3 years ago

An air balloon is moving upward at a constant speed of 3 m/s. Suddenly a passenger realizes that she left her camera on the grou

frosja888 [35]

Answer:t=0.3253 s

Explanation:

Given

speed of balloon is $u=3\ m/s$

speed of camera $u_1=20\ m/s$

Initial separation between camera and balloon is $d_o=5\ m$

Suppose after t sec of throw camera reach balloon then,

distance travel by balloon is

$s=ut$

$s=3\times t$

and distance travel by camera to reach balloon is

$s_1=ut+\frac{1}{2}at^2$

$s_1=20\times t-\frac{1}{2}gt^2$

Now

$\Rightarrow s_1=5+s$

$\Rightarrow 20\times t-\frac{1}{2}gt^2 =5+3t$

$\Rightarrow 5t^2-17t+5=0$

$\Rightarrow t=\dfrac{17\pm \sqrt{17^2-4(5)(5)}}{2\times 5}$

$\Rightarrow t=\dfrac{17\pm 13.747}{10}$

$\Rightarrow t=0.3253\ s\ \text{and}\ t=3.07\ s$

There are two times when camera reaches the same level as balloon and the smaller time is associated with with the first one .

(b)When passenger catches the camera time is $t=0.3253\ s$

velocity is given by

$v=u+at$

$v=20-10\times 0.3253$

$v=16.747\ m/s$

and position of camera is same as of balloon so

Position is $=5+3\times 0.3253$

$=5.975\approx 6\ m$

8 0

3 years ago