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Anna71 [15]
2 years ago
13

The amount of solute that can be dissolved in a specific amount of solvent at a given temperature is its what?

Physics
1 answer:
slavikrds [6]2 years ago
5 0
This is called super-saturation
hope this helps :)))))))
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Light traveling though air strikes the surface of the four different materials shown. Which material reflects light but does not
dangina [55]
The spoons .................
8 0
3 years ago
What is motivation in your own world
hjlf

Answer:

Motivation is the process that initiates, guides, and maintains goal-oriented behaviors.

5 0
2 years ago
A skydiver of mass 80.0 kg jumps from a slow-moving aircraft and reaches a terminal speed of 50.0 m/s. (a) What is her accelerat
kirill [66]

Answer:

6.22²

Explanation:

Given that

Mass of the skydiver, m = 80 kg

Terminal speed of the skydiver, v(f) = 50 m/s

Speed of the skydiver, v(i) = 30 m/s

Acceleration of the skydiver, a = ?

To solve this, we use the formula

W - k v² = ma, where

W = weight of the skydiver

k = constant

v = speed of the skydiver

m = mass of the skydiver

So, if we substitute the values into it we have

W = mg = 80 * 9.8 = 784 N

784 - k 50² = 80 *0

784 - 2500k = 0

784 = 2500k

k = 0.3136

Now, we use this value of k to find the needed acceleration using the same formula at a speed of 30 m/s

784 - 0.3136 * 30² = 80 * a

784 - 0.3136 * 900 = 80a

784 - 282.24 = 80a

497.76 = 80a

a = 497.76 / 80

a = 6.22 m/s²

Thus, we can conclude that the acceleration when the speed of the skydiver is 30 m/s, is 6.22 m/s²

4 0
3 years ago
.
beks73 [17]

Answer:

2 m/s²

Explanation:

the equations of motion are

S= ut +½at²

v² = u²+ 2as

v = u + at

s = (u+v)/2 × t

From the parameters given

u = 0m/s this is because it starts from rest

Distance (s)  = 9m

Time (t)  = 3s

Based on this the first equation would be used

s = ut + ½at²

Input values

9 = 0×3 + ½ × a x 3²

9 = 0 + 9a/2

9 = 4.5a

Divide both sides by 4.5

a = 9 / 4.5 m/s²

a = 2 m/s²

I hope this was helpful, please mark as brainliest

3 0
3 years ago
Banked Curves: A 600-kg car is going around a banked curve with a radius of 110 m at a steady speed of 24.5 m/s. What is the app
Aleks04 [339]

The angle of baking from the calculation is obtained as 30°.

<h3>What is banking?</h3>

The term banking refers to a means of preventing vehicles from skidding off the road at curves.

We know that the banking angle is obtained from;

θ = tan-1(v^2/rg)

v = 24.5 m/s

r = 110 m

g = 9.8 m/s^2

θ = tan-1(25^2/9.8 * 110)

θ = tan-1(625 /1078)

θ = 30°

Learn more about the banking angle:brainly.com/question/26759099?r

#SPJ1

8 0
2 years ago
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