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4 years ago

What is a solution that has the capacity to hold additional solute at a given temperature called

Physics

1 answer:

Cerrena [4.2K]4 years ago

7 0

Unsaturated solution. It has more capacity to hold more solute.

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Does Anybody Know The Answers?

ch4aika [34]

Answer:

I was going to give you the paper where I saw it but since you are not giving enough points I can not give you so I am only going to give you some of these that are here sorry

Explanation:

$9^{2} + 12^2 = x^2\\81 + 144= x^2\\\sqrt{225} = \sqrt{x} \\ 15=x\\\\ 2.\\x^2+12^2+=13^2\\x^2+144 =169\\x^2 = 25\\\sqrt{x^2 =\sqrt{25\\\\$

x=5

$3.\\12^2+32^2 = x^2\\34.176= x$

5,12,13

$\frac{x}{4} ,\frac{12}{4} ,\frac{20}{4}\\\\\frac{x}{4},3,5 \\\\x=16\\\\12. \\x^2 + 48^2=50^2\\\\x^2=196\\x=14$

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5 0

3 years ago

A ball is thrown off a cliff at a speed of 10 m/s in a horizontally direction. The ball reaches the ground 1.5 seconds. If the b

Tems11 [23]

I am pretty sure it is d

5 0

3 years ago

A proud new Jaguar owner drives her car at a speed of 25 m/s into a corner. The coefficients of friction between the road and th

ehidna [41]

Answer:

ac = 3.92 m/s²

Explanation:

In this case the frictional force must balance the centripetal force for the car not to skid. Therefore,

Frictional Force = Centripetal Force

where,

Frictional Force = μ(Normal Force) = μ(weight) = μmg

Centripetal Force = (m)(ac)

Therefore,

μmg = (m)(ac)

ac = μg

where,

ac = magnitude of centripetal acceleration of car = ?

μ = coefficient of friction of tires (kinetic) = 0.4

g = 9.8 m/s²

Therefore,

ac = (0.4)(9.8 m/s²)

5 0

3 years ago

A rocket burns fuel as it shoots into the sky. What happens to the mass and volume of the rocket?

diamong [38]

Answer:

mass goes down volume remains the same

6 0

3 years ago

A bird is about 6.26.2 in. long, with a thin, dark bill and a wide, white wing stripe. If the bird can fly 9292 mi with the w

Trava [24]

Answer:

$209$ mph

Explanation:

$V$ = Speed of bird in still air

$v$ = Speed of wind = 44 mph

Consider the motion of the bird with the wind

$D_{1}$ = distance traveled with the wind = 9292 mi

$t_{1}$ = time taken to travel the distance with wind

Time taken to travel the distance with wind is given as

$t_{1} = \frac{D_{1}}{V + v}$

$t_{1} = \frac{9292}{V + 44}$ eq-1

Consider the motion of the bird with the wind

$D_{2}$ = distance traveled against the wind = 6060 mi

$t_{2}$ = time taken to travel the distance against wind

Time taken to travel the distance against wind is given as

$t_{2} = \frac{D_{2}}{V + v}$

$t_{2} = \frac{6060}{V - 44}$ eq-2

As per the question,

Time taken with the wind = Time taken against the wind

$t_{1} = t_{2}$

$\frac{9292}{V + 44} = \frac{6060}{V - 44}$

$(9292) (V - 44) = (6060) (V + 44)$

$9292V - 408848 = 6060V + 266640$

$3232V = 675488$

$V = 209$ mph

5 0

3 years ago