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miv72 [106K]
3 years ago
13

What is a colloid? Are either of the two substances in the beakers in the chart above a colloid?

Physics
1 answer:
fenix001 [56]3 years ago
4 0

Answer:Colloids include gels and emulsions; the particles do not settle, and cannot be separated out by ordinary filtering or centrifugtion

Explanation:they are homogenous/non crystalline

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Does air resistance affect the motion of a falling object differently when the mass is greater?
mixas84 [53]
<span>NO.

Air resistance does not affect the motion of a falling object differently when the mass is greater because the mass of an object does not in any way affect the speed of falling due to gravity, and air resistance depends only on the speed of the object and its surface area.</span>
8 0
3 years ago
Question 1 (1 point)
deff fn [24]

Answer:

Travelled 18 km, they are 6 km from home.

Explanation:

12/2 (halfway) is 6km. So, 6 + 12 would be 18 km, total amount travelled. The total distance of the trip would be 24 km (12 km out, 12km back) if they travelled 12+6 (18km) then they only have 6 km more to go.

5 0
3 years ago
Why do the four outer planets have lower density than the four inner planets?
Marrrta [24]
Because the four outer planets are comprised of mostly gases give me a thanks or brainiest answer if this helps! 
4 0
3 years ago
A point charge (–5.0 µC) is placed on the x axis at x = 4.0 cm, and a second charge (+5.0 µC) is placed on the x axis at x = –4.
AveGali [126]

Answer:

The magnitude of electric force is  7.2\times10^{-3} N

Explanation:

Coulomb's Law:

The force of attraction or repletion is

  • directly proportional to the products of charges i.e F\propto q_1q_2
  • inversely proportional to the square of distance i.e F\propto \frac{1}{r^2}

\therefore F\propto \frac{q_1q_2}{r^2}

\Rightarrow F=k \frac{q_1q_2}{r^2}    [ k is proportional constant=9×10⁹N m²/C²]

There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C

Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C

and F₂ force  be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C

Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.

If we draw a line from q₁ to Q .

The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.

Let hypotenuse = r

Therefore, r=\sqrt{altitude^2+base^2} =\sqrt{3^2+4^2} =5

we know,

cos \theta = \frac{base }{hypotenuse}

\Rightarrow cos \theta = \frac{4 }{r}

Total force F_Q = 2.F_1 cos\theta \hat{i}

                         =2k\frac{Qq_1}{r^2} cos\theta \hat i

                         =2\ \frac{9\times1 0^9\times2.5 \times 5\times 10^{-12}}{r^2} \frac{4}{r} \hat i

                         =8\ \frac{9\times10^9\times2.5 \times 5\times 10^{-12}}{5^3} \hat i     [ r=5]

                         =7.2\times10^{-3}\hat i   N

The magnitude of electric force is  7.2\times10^{-3} N

                         

3 0
3 years ago
What property is an acid-base indicator used to measure?
Volgvan

Answer:

ph

Explanation:

4 0
3 years ago
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