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3 years ago

Why is vinegar considered a solution?

Physics

1 answer:

Natalija [7]3 years ago

3 0

Answer:

Vinegar is a homogenous mixture of acetic acid and water. As the mixture created has only one phase it is a solution. ... There are no chemical bonds created between water and the acid and it is possible to separate the two without breaking any chemical bonds.

Explanation:

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Question 2 A horizontal line on a position vs time graph means the object is O moving faster. O at rest O slowing down. O moving

Salsk061 [2.6K]

Explanation:

this question can also shown in figure

5 0

3 years ago

Suppose you are an astronaut on a spacewalk, far from any source of gravity. You find yourself floating alongside your spacecraf

erik [133]

Answer:

c. Throw the items away from the spaceship.

Explanation:

By the Principle of action and reaction yu can get back to your spacecraft throwing the items away from the spaceship.

7 0

4 years ago

Read 2 more answers

1. Boyle's law relates the pressure of a gas to its

KIM [24]

Answer:

volume is the correct answer

Explanation:

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2 years ago

A pool is to be filled with 60 m3 water from a garden hose of 2.5 cm diameter flowing water at 2 m/s. Find the mass flow rate of

stealth61 [152]

Answer:

Time= 6.12*10^4s

mass flow rate m=0.98kg/s

Explanation:

Given

Volume= 60m^3

diamter= 2.5cm= 0.025m

radius= 0.0125m

area A= πr^2

area A= 3.142*0.0125^2

area A= 4.9*10^-4m^2

the velocity of the flow 2m/s

volume flow rate

V=vA

V=2* 4.9*10^-4

V=9.82*10^-4 m^3/s

Time taken to fill the pool

time= volume/volume flow rate

time= 60/9.82*10^-4

time= 6.12*10^4s

Mass flow rate

m= density *volume flow rate

Assuming the density of water to be 997kg/m^3

m= 997*9.82*10^-4

m=0.98kg/s

6 0

3 years ago

An insect 1.1 mm tall is placed 1.0 mm beyond the focal point of the objective lens of a compound microscope. The objective lens

Pavlova-9 [17]

Answer:

Explanation:

For image formation in objective lens

object distance u = 14 +1 = 15 mm

focal length f = 14 mm .

image distance v = ?

lens formula

$\frac{1}{v} -\frac{1}{u} =\frac{1}{f}$

Putting the values

$\frac{1}{v} +\frac{1}{15} =\frac{1}{14}$

v = 210 mm .

B )

magnification = v / u

= 210 / 15

= 14

size of image = 14 x 1.1 mm

= 15.4 mm

= 15 mm approx

C )

For final image to be at infinity , image produced by objective lens must fall at the focal point of eye piece . so objective lens's distance from the image formed by objective must be equal to focal length of eye piece that is 21 mm .

21 mm is the answer .

D )

overall magnification =

$\frac{210}{15} \times \frac{D}{f_e}$

D = 25 cm , f_e = focal length of eye piece

= 14 x 250 / 21

= 166.67

= 170 ( in two significant figures )

7 0

3 years ago