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sertanlavr [38]

3 years ago

11

A submarine is 20m below the surface of the sea. the pressure due to the water at this depth is P. on another day, the submarine

is 26m below the surface of fresh water. the density of sea water is 1,3 times the density of fresh water. what is the pressure due to the fresh water at a depth of 26m?

1 answer:

satela [25.4K]3 years ago

8 0

Answer:

1.7p

Explanation:

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Juan measured the temperature of salt water. He then added 273 to the measured value. Which conversion is Juan most likely doing

Semmy [17]

If Juan used a Celsius thermometer, it would tell him the Celsius temperature.

If he added 273 to that number, he'd have the "absolute" or Kelvin temperature.

7 0

3 years ago

Use Newton’s first law to explain why a book on a desk does not move

rewona [7]

Answer:

nothing moves if anything other doesn't move it.

natural position of things is the resting

Explanation:

yeah that's it

5 0

2 years ago

Read 2 more answers

An insect 5.25 mm tall is placed 25.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the righ

Zigmanuir [339]

Answer:

(A) therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

(B) therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

Explanation:

height of the insect (h) = 5.25 mm = 0.525 cm

distance of the insect (s) = 25 cm

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = -12.5 cm (because it is a planoconvex lens with the radius in the direction of the incident rays)

index of refraction (n) = 1.7

(A) we can find the location of the image by applying the formula below

$\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$ where

s' = distance of the image
f = focal length

but we first need to find the focal length before we can apply this formula

$\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )$

$\frac{1}{f} =(1.7-1)(\frac{1}{∞} -\frac{1}{-12.5} )$

$\frac{1}{f} =(0.7)(0 + \frac{1}{12.5} )$

$\frac{1}{f} =\frac{0.7}{12.5}$

f = $\frac{12.5}{0.7}$

f = 17.9 cm

now that we have the focal length we can apply $\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$

$\frac{1}{f} - \frac{1}{s}$ =\frac{1}{s'}

$\frac{1}{17.9} - \frac{1}{25}$ =\frac{1}{s'}

$\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}$

$\frac{7.1}{447.5} =\frac{1}{s'}$

s' = \frac{447.5}{7.1}[/tex] = 63 cm to the right of the lens

magnification = $\frac{-s'}{s} =\frac{y'}{y}$ where y' is the height of the image, therefore

$\frac{-s'}{s} =\frac{y'}{y}$

$\frac{-63}{25} =\frac{y'}{52.5}$

y' = $\frac{-63}{25}$ x 0.525 = -13.22 cm

therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

(B) if the lens is reversed, the radius of curvatures would be interchanged

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = 12.5 cm

we can find the location of the image by applying the formula below

$\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$ where

s' = distance of the image
f = focal length

but we first need to find the focal length before we can apply this formula

$\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )$

$\frac{1}{f} =(1.7-1)(\frac{1}{12.5} -\frac{1}{∞} )$

$\frac{1}{f} =(0.7)( \frac{1}{12.5} - 0)$

$\frac{1}{f} =\frac{0.7}{12.5}$

f = $\frac{12.5}{0.7}$

f = 17.9 cm

now that we have the focal length we can apply $\frac{1}{f} =\frac{1}{s'} +\frac{1}{s}$

$\frac{1}{f} - \frac{1}{s}$ =\frac{1}{s'}

$\frac{1}{17.9} - \frac{1}{25}$ =\frac{1}{s'}

$\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}$

$\frac{7.1}{447.5} =\frac{1}{s'}$

s' = \frac{447.5}{7.1}[/tex] = 63 cm to the right of the lens

magnification = $\frac{-s'}{s} =\frac{y'}{y}$ where y' is the height of the image, therefore

$\frac{-s'}{s} =\frac{y'}{y}$

$\frac{-63}{25} =\frac{y'}{52.5}$

y' = $\frac{-63}{25}$ x 0.525 = -13.22 cm

therefore the image is

63 cm to the right of the lens
the image size is -13.22 cm
it is real
it is inverted

7 0

3 years ago

Which of the following scenarios represents the best use of a scientific model?

valkas [14]

Answer:

Predicting weather patterns is the answer :-)

Explanation:

4 0

3 years ago

A ball player catches a ball 3.2 s after throwing it vertically upward. what height did it reach?

cluponka [151]

At the top of the height, the velocity is zero and acceleration is negative of acceleration due to gravity ( i.e $-9.8 m/s^2$ ).

The time of the ball in air is 3.2 s, so ascending time is $3.2/2=1.6 s$ .

Therefore from kinematic equation,

$v = u + gt$

Substituting the values we get,

$0= u - 9.8 (1.6)\\\\u=15.7 m/s$ , Here v = 0 at top.

Now from equation,

$h=ut+\frac{1}{2} gt^2$ , here h is the height .

So,

$h=(15.7 m/s) (1.6s)-\frac{1}{2} 9.8m/s^2(1.6)^2\\\\ h=25.12m-12.54m\\\\h=12.48 m$ .

Thus, the ball reached at its maximum height of 12.48 m.

8 0

3 years ago