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krek1111 [17]
4 years ago
10

To make salt water a heterogeneous mixture, you would have to A) add sand. B) add a small amount of sugar. C) boil the water. D)

remove the salt.
Physics
2 answers:
satela [25.4K]4 years ago
5 0
To make salt water a heterogeneous mixture, you would have to add sand and that's is how to make a salt water with heterogeneous mixture.
A is correct answer.
Hope it helped you.

-Charlie

Thanks!

-:)
Vladimir79 [104]4 years ago
3 0

Answer : The correct option is, (A) add sand

Explanation :

Heterogeneous mixtures : It is a mixture that has non-uniform composition throughout the solution and the particle size or shapes are also different.

There is a physical boundary between the dispersed phase and dispersion medium.

Homogeneous mixtures : It is a mixture that has uniform composition throughout the solution and the particle size or shapes are not different.

There is no physical boundary between the dispersed phase and dispersion medium.

As we know that salt water is a homogeneous mixture because its composition is the same throughout the solution.

To make salt-water a heterogeneous mixture, we would have to add sand in the solution.

If we add small amount of sugar then the mixture becomes homogeneous.

Hence, the correct option is, (A) add sand

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The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static fricti
Colt1911 [192]

This question is incomplete, the complete question;

The L-ft ladder has a uniform weight of W lb and rests against the smooth wall at B. θ = 60. If the coefficient of static friction at A is μ = 0.4.

Determine the magnitude of force at point A and determine if the ladder will slip. given the following; L = 10 FT, W = 76 lb

Answer:

- the magnitude of force at point A is 79.1033 lb

- since FA < FA_max; Ladder WILL NOT slip

Explanation:

Given that;

∑'MA = 0

⇒ NB [Lsin∅] - W[L/2.cos∅] = 0

NB = W / 2tan∅ -------let this be equation 1

∑Fx = 0

⇒ FA - NB = 0

FA = NB

therefore from equation 1

FA = NB = W / 2tan∅

we substitute in our values

FA = NB = 76 / 2tan(60°) = 21.9393 lb

Now ∑Fy = 0

NA - W = 0

NA = W = 76 lb

Net force at A will be

FA' = √( NA² + FA²)

= √( (W)² + (W / 2tan∅)²)

we substitute in our values

FA' = √( (76)² + (21.9393)²)

= √( 5776 + 481.3328)

= √ 6257.3328

FA' = 79.1033 lb

Therefore the magnitude of force at point A is 79.1033 lb

Now maximum possible frictional force at A

FA_max = μ × NA

so, FA_max = 0.4 × 76

FA_max = 30.4 lb

So by comparing, we can easily see that the actual friction force required for keeping the the ladder stationary i.e (FA) is less than the maximum possible friction available at point A.

Therefore since FA < FA_max; Ladder WILL NOT slip

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