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3 years ago

After which action would the concentration of a solution remain constant?(1 point)

Physics

1 answer:

Alika [10]3 years ago

5 0

Answer:

C. Evaporating water from the container.

Explanation:

The concentration of solution changes when solvent or solute are added/removed from a solution.

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A skier descends a mountain at an angle of 35.0º to the horizontal. If the mountain is 235 m long, what are the horizontal and v

Masja [62]

Total displacement along the length of mountain is given as

L = 235 m

angle of mountain with horizontal = 35 degree

now we will have horizontal displacement as

x = L cos35

x = 235 cos35 = 192.5 m

similarly for vertical displacement we can say

y = L sin35

y = 235 sin35 = 134.8 m

7 0

2 years ago

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Which idea would most likely be dangerous for a student to think while entering a lab?

aleksley [76]

Your answer should be a i can figure out how the tools work as i go along

3 0

2 years ago

Spitting cobras can defend themselves by squeezing muscles around their venom glands to squirt venom at an attacker. Suppose a s

Alenkasestr [34]

Answer: 1.289 m

Explanation:

The path the cobra's venom follows since it is spitted until it hits the ground, is described by a parabola. Hence, the equations for parabolic motion (which has two components) can be applied to solve this problem:

<u>x-component: </u>

$x=V_{o}cos\theta t$ (1)

Where:

$x$ is the horizontal distance traveled by the venom

$V_{o}=3.10 m/s$ is the venom's initial speed

$\theta=47\°$ is the angle

$t$ is the time since the venom is spitted until it hits the ground

<u>y-component: </u>

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=0.44 m$ is the initial height of the venom

$y=0$ is the final height of the venom (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Let's begin with (2) to find the time it takes the complete path:

$0=0.44 m+3.10 m/s sin\theta(47\°)+\frac{-9.8m/s^{2} t^{2}}{2}$ (3)

Rewritting (3):

$-4.9 m/s^{2} t^{2} + 2.267 m/s t + 0.44 m=0$ (4)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula: