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Naddika [18.5K]
2 years ago
6

Identify the solutions (homogeneous mixtures) in the list below. check all that apply. lead, an alloy of tin and lead ammonia, a

pure compound of nitrogen and hydrogen window cleaner, a mixture of ammonia and coloring dissolved in water gasoline, a mixture of organic liquids with a fixed composition throughout distilled water, pure with no metals or particles
Physics
2 answers:
Goryan [66]2 years ago
4 0

The solutions and homogeneous mixtures in the given list are:

A. Lead solder, an alloy of tin and lead.

D. Window cleaner, a mixture of ammonia and coloring dissolved in water.

E. Gasoline, a mixture of organic liquids with a fixed composition throughout.

<h3>What is a solution?</h3>

A solution can be defined as a special type of homogeneous mixture that comprises a solute and a solvent.

A homogeneous mixture can be defined as any solid, liquid, or gaseous mixture which has an identical (uniform) composition and properties throughout any given sample of the mixture.

In Science (Physics), all solutions are considered to be a homogeneous mixture because their constituents are uniformly (evenly) distributed.

In conclusion, the solutions and homogeneous mixtures in the given list are:

  • Lead solder, an alloy of tin and lead.
  • Window cleaner, a mixture of ammonia and coloring dissolved in water.
  • Gasoline, a mixture of organic liquids with a fixed composition throughout.

To know more about solution follow

brainly.com/question/22070951

Sveta_85 [38]2 years ago
3 0

Answer:

Brainliest pls

Explanation:

lead solder, an alloy of tin and lead

window cleaner, a mixture of ammonia and coloring dissolved in water

gasoline, a mixture of organic liquids with a fixed composition throughout

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A double-slit experiment is set up using red light (λ = 706 nm). A first order bright fringe is seen at a given location on a sc
Elanso [62]

Answer:

λ = 470.66 nm

Explanation:

for bright fringey_m = \frac{m\lambda D}{d}

D= distance between slit and screen

d= distance between the slits

for first order bright fringe m = 1,

        y_1 = \frac{1\lambda D}{d}

         y_1 = {706*D}{d}

for dark fringe,we have

y_m = {(m + 1/2)\lambda D}{d}    

Now to get the dark fringes at the same location we should have;

(706)D/d = (m + 1/2)λD/d    

put m = 1

(1 + 1/2)λ = (706)

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6 0
3 years ago
A spring gun is made by compressing a spring in a tube and then latching the spring at the compressed position. A 4.97-g pellet
dimaraw [331]

Answer:

v  = 2.8898 \frac{m}{s}

Explanation:

This is a problem easily solve using energy conservation. As there are no non-conservative forces, we know that the energy is conserved.

When the spring is compressed downward, the spring has elastic potential energy. When the spring is relaxed, there is no elastic potential energy, but the pellet will have gained gravitational potential energy and kinetic energy. Lets see what are the terms for each of this.

<h3>Elastic potential energy</h3>

We know that a spring following Hooke's Law has a elastic potential energy:

E_{ep} = \frac{1}{2} k (\Delta x)^2

where \Delta x is the displacement from the relaxed length and k is the spring's constant.

To obtain the spring's constant, we know that Hooke's law states that the force made by the spring is :

\vec{F} = - k \Delta \vec{x}

as we need 9.12 N to compress 4.60 cm, this means:

k = \frac{9.12 \ N}{4.6 \ 10^{-2} \ m}

k = 198.26 \ \frac{ N}{m}

So, the elastic energy of the compressed spring is:

E_{ep} = \frac{1}{2} 198.26 \ \frac{ N}{m} (4.6 \ 10^{-2} \ m)^2

E_{ep} = 0.209759 \ Joules

And when the spring is relaxed, the elastic potential energy will be zero.

<h3>Gravitational potential energy</h3>

To see how much gravitational potential energy will the pellet win, we can use

\Delta E_{gp} = m g \Delta h

where m is the mass of the pellet, g is the acceleration due to gravity and \Delta h is the difference in height.

Taking all this together, the gravitational potential energy when the spring is relaxed will be:

\Delta E_{gp} = 4.97 \ 10^{-3} kg \ 9.8 \frac{m}{s^2} 4.6 \ 10^{-2} m

\Delta E_{gp} = 0.00224 \ Joules

<h3>Kinetic Energy</h3>

We know that the kinetic energy for a mass m moving at speed v is:

E_k = \frac{1}{2} m v^2

so, for the pellet will be

E_k = \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2

<h3>All together</h3>

By conservation of energy, we know:

E_{ep} = \Delta E_{gp} + E_k

0.209759 \ Joules = 0.00224 \ Joules + \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2

So

\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2  = 0.209759 \ Joules - 0.00224 \ Joules

\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2  = 0.207519 \ Joules

v  = \sqrt{ \frac{ 0.207519 \ Joules}{ \frac{1}{2} \ 4.97 \ 10^{-3} kg } }

v  = 2.8898 \frac{m}{s}

7 0
3 years ago
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