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3 years ago

7

Why do your legs appear shorter when standing in water

1 answer:

Effectus [21]3 years ago

8 0

It is because of refraction caused in water. There is difference in the medium between water and air so because of that refraction takes place. Due to this, a person’s legs appear to be shorter to our eyes. Hope this helps!

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Refer to the attached image!!!

dimaraw [331]

The time of motion of the track star is determined as 0.837 s.

<h3>Time of motion of the track star</h3>

The time of motion of the track star is calculated as follows;

T = (2u sinθ)/g

where;

T is time of motion
g is acceleration due to gravity
θ is angle of projection

T = (2 x 12 x sin20)/9.8

T = 0.837 s

Learn more about time of motion here: brainly.com/question/2364404

#SPJ1

6 0

1 year ago

What is 8200 N/m2 expressed in kilopascals?

natima [27]

The S.I. unit for the measure of the pressure is the Pascal (Pa). 1 Pascal corresponds to
$1 Pa = 1 N/m^2$

We can convert the number given by the problem into Pascal:
$8200 N/m^2 = 8200 Pa$

And since $1 Pa = 10^{-3}kPa$ , we have
$8200 Pa = 8200 \cdot 10^{-3}kPa =8.2 kPa$

5 0

3 years ago

In a hydraulic lift, if the pressure exerted on the liquid by one piston is increased by 100 N/m2 , how much additional weight c

shepuryov [24]

Answer:

The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.

Explanation:

By means of the Pascal's Principle, the hydraulic lift can be modelled by the following two equations:

Hydraulic Lift - Before change

$P = \frac{F}{A}$

Hydraulic Lift - After change

$P + \Delta P = \frac{F + \Delta F}{A}$

Where:

$P$ - Hydrostatic pressure, measured in pascals.

$\Delta P$ - Change in hydrostatic pressure, measured in pascals.

$A$ - Cross sectional area of the hydraulic lift, measured in square meters.

$F$ - Hydrostatic force, measured in newtons.

$\Delta F$ - Change in hydrostatic force, measured in newtons.

The additional weight is obtained after some algebraic handling and the replacing of all inputs:

$\frac{F}{A} + \Delta P = \frac{F}{A} + \frac{\Delta F}{A}$

$\Delta P = \frac{\Delta F}{A}$

$\Delta F = A\cdot \Delta P$

Given that $\Delta P = 100\,Pa$ and $A = 25\,m^{2}$ , the additional weight is:

$\Delta F = (25\,m^{2})\cdot (100\,Pa)$

$\Delta F = 2500\,N$

The additional mass needed for the additional weight is:

$\Delta m = \frac{\Delta F}{g}$

Where:

$\Delta F$ - Additional weight, measured in newtons.

$\Delta m$ - Additional mass, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

If $\Delta F = 2500\,N$ and $g = 9.807\,\frac{m}{s^{2}}$ , then:

$\Delta m = \frac{2500\,N}{9.807\,\frac{m}{s^{2}} }$

$\Delta m = 254.92\,kg$

The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.

3 0

3 years ago

Why is it so important to follow all directions in an experiment?

natita [175]

If you repeat the experiment, you will want to do it the same way to see if the results change.

HOPE THIS HELPS!!!

5 0

3 years ago

What are the major features of the human eye and to what are they analogous in a camera?

adelina 88 [10]

Answer:

The human major features with the analogous of camera are mentioned.

Explanation:

There are the following major features of human eye which are analogue with the camera:

Camera will have shutters which controls light entering into it, human eye consists of Diaphragm which also functions same.

Both gives Real & inverted images.

In camera film records image, in eye image is focused on retina and it is converted into electrical impulses.

Pupil and iris acts as aperture of camera.

8 0

3 years ago