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Firlakuza [10]
3 years ago
10

A saturated solution of lead(II) nitrate was made at 50 °C using 100 g of water, and this was then allowed to cool to 30 °C. Pre

dict what would happen as the solution cooled?
Physics
2 answers:
olchik [2.2K]3 years ago
8 0
The solubility of the solution would decrease and some of the lead(II) nitrate would precipitate out 
Sunny_sXe [5.5K]3 years ago
4 0

Answer: The temperature of solution will decease along with the precipitation of solid lead nitrate in solution.

Explanation:

Solubility of the substance is directly effect by the temperature of the solvent.

This is because the kinetic energy of the solvent increase as temperature increase due to which solvent easily breaks apart the solute particles easily and faster which will result in solubility of more solute in solvent and vice versa.

So, when prepared saturated solution of lead nitrate at 50°C was allowed to cool down to 30°C one will observe that the precipitation of solid lead nitrate in solution. And this means that solubility of lead nitrate with decrease in temperature decreases as well.

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A small 24 kilogram canoe is floating downriver at a speed of 2 m/s. What is the canoe's kinetic energy?
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J can get answer on this way:
Ek=m*V*V/2= (24kg*2m/s*2m/s)/2=48 Ј
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What is light energy
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Light energy is defined as how nature moves energy at an extremely rapid rate, and it makes up about 99% of the body's atoms and cells, and signal all body parts to carry out their respective tasks. An example of light energy is the movement of a radio signal.
8 0
3 years ago
A hunter aims at a deer which is 40 yards away. Her cross- bow is at a height of 5ft, and she aims for a spot on the deer 4ft ab
shutvik [7]

Answer:

a)  θ₁ = 0.487º , b)   t = 0.400 s ,        x = 11.73 ft

Explanation:

For this exercise let's use the projectile launch relationships.

The initial height is I = 5 ft and the final height y = 4 ft

            y = y₀ + v_{oy} t - ½ g t²

The distance to the band is x = 40 yard (3 ft / 1 yard) = 120 ft

            x = v₀ₓ t

            t = x / v₀ₓ

We replace

             y –y₀ = v_{oy} x / v₀ₓ - ½ g x² / v₀ₓ²

             v_{oy} = v₀ sin θ

             v₀ₓ = vo cos θ

             

             y –y₀ = x tan θ - ½ g x² / v₀² cos² θ

                5-4 = 120 tan θ - ½ 32 120 / (300 2 cos2 θ)

                1 = 120 tan θ - 0.0213 sec² θ

Let's use the trigonometry relationship

               Sec² θ = 1 - tan² θ

                 1 = 120 tan θ - 0.0213 (1 –tan²θ)

                 0.0213 tan²θ + 120 tanθ -1.0213 = 0

                 

We change variables

          u = tan θ

          u² + 5633.8 u - 48.03 = 0

We solve the second degree equation

          u = [-5633.8 ±√(5633.8 2 + 4 48.03)] / 2

          u = [- 5633.8 ± 5633.82] / 2

           u₁ = 0.0085

           u₂= -5633.81

           u = tan θ

           θ = tan⁻¹ u

For u₁

           θ₁ = tan⁻¹ 0.0085

           θ₁ = 0.487º

For u₂

           θ₂ = -89.99º

The launch angle must be 0.487º

b) let's look for the time it takes for the arrow to arrive

         x = v₀ₓ t

         t = x / v₀ cos θ

         

         t = 120 / (300 cos 0.487)

         t = 0.400 s

The deer must be at a distance of

           v = 20 mph (5280 ft / 1 mi) (1 h / 3600s) = 29.33 ft / s

           x = v t

           x = 29.33 0.4

           x = 11.73 ft

3 0
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Answer:

...

Explanation:

................................................

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sveticcg [70]

by cosine law we know that

c^2 = a^2 + b^2 - 2 abcos\theta

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now using above equation

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4942.03 = b^2 - 85.34 b

b^2 - 85.34b - 4942.03 = 0

by solving above quadratic equation we have

b = 124.9 m

so it is at distance 124.9 m from deer a

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