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4vir4ik [10]
3 years ago
11

Sasha did an experiment to study the solubility of two substances. She poured 100 mL of water at 20 °C into each of two beakers

labeled A and B. She put 50 g of Substance A in the beaker labeled A and 50 g of Substance B in the beaker labeled B. The solution in both beakers was stirred for 1 minute. The amount of substance left undissolved in the beakers was weighed. The experiment was repeated for different temperatures of water and the observations were recorded as shown. (8 points)
Experimental Observations
Substance Mass of Undissolved Substance at Different Temperatures (gram)
20 °C 40 °C 60 °C 80 °C
A 12 11 9 8
B 45 40 34 31
Part 1: Which substance has a higher solubility?
Part 2: Explain your answer for Part 1.
Physics
1 answer:
chubhunter [2.5K]3 years ago
4 0

Answer:

b is the higher solubility then A

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A converging lens of focal length 20 cm is used to form a real image 1.0 m away from the lens. How far from the lens is the obje
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Answer:

0.25 m

Explanation:

We can solve the problem by using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem, we have

f = +20 cm=+0.20 m (the focal length is positive for a converging lens)

q = +1.0 m (the image distance is positive for a real image)

Solving the equation for p, we find

\frac{1}{p}=\frac{1}{f}-\frac{1}{q}=\frac{1}{0.20 m}-\frac{1}{1 m}=4 m^{-1}\\p=\frac{1}{4 m^{-1}}=0.25 m

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(a) Calculate the magnitude of the gravitational force exerted by Mars on a 80 kg human standing on the surface of Mars. (The ma
andrew11 [14]

Answer:

a) F=1.044\times 10^9\ N

b)F'=1.044\times 10^9\ N

c) F_p=1.0672\times10^{-7}\ N

d) Treat the humans as though they were points or uniform-density spheres.

Explanation:

Given:

  • mass of Mars, M=6.4\times 10^{23}\ kg
  • radius of the Mars, r=3.4\times 10^{6}\ m
  • mass of human, m=80\ kg

a)

Gravitation force exerted by the Mars on the human body:

F=G.\frac{M.m}{r^2}

where:

G=6.67 \times 10^{-11}\ m^3.kg^{-1}.s^{-2} = gravitational constant

F=6.67\times10^{-11}\times \frac{6.4\times 10^{23}\times 80}{(3.4\times 10^{6})^2}

F=1.044\times 10^9\ N

b)

The magnitude of the gravitational force exerted by the human on Mars is equal to the force by the Mars on human.

F'=F

F'=1.044\times 10^9\ N

c)

When a similar person of the same mass is standing at a distance of 4 meters:

F_p=6.67\times10^{-11}\times \frac{80\times 80}{4}

F_p=1.0672\times10^{-7}\ N

d)

The gravitational constant is a universal value and it remains constant in the Universe and does not depends on the size of the mass.

  • Yes, we have to treat Mars as spherically symmetric so that its center of mass is at its geometric center.
  • Yes, we also have to ignore the effect of sun, but as asked in the question we have to calculate the gravitational force only due to one body on another specific body which does not brings sun into picture of the consideration.
4 0
3 years ago
A race car drives one lap around a race track that is 500 meters in length.
pychu [463]

Explanation:

Check out the picture I drew for a minute before reading this...

B. Distance [the red line] is a scalar quantity reflecting how far an object has traveled. Displacement [the green line] is a vector quantity reflecting how far an object has moved from a point. The key difference is that distance can be any sort of path while displacement is always a vector (or a straight line) between a starting point and a finishing point. Sometimes distance and displacement are equal to one another. Sometimes you have a distance traveled, but zero displacement overall; which is what's going on in your question.

A. The distance that the racecar traveled is indeed 500m. But at the end of the lap, it is right back where it started. So overall, it has been displaced 0m.

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3 years ago
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