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Anika [276]
3 years ago
15

The dispersed particles of a colloid exhibit brownian motion, in which they move _____.

Physics
2 answers:
Archy [21]3 years ago
8 0

The dispersed particles of a colloid exhibit brownian motion, in which they move _____. ... The dispersed particles of a colloid exhibit brownian motion, in which they move in a chaotic manner without a discernible pattern.

zaharov [31]3 years ago
5 0

Answer:

In a chaotic manner without a discernible pattern

Explanation:

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a uniform disc and hollow right circular cone have the same formula for their moment of inertia when rotating about the central
Romashka-Z-Leto [24]

Answer:

This is as a result that about the central axis a collapsed hollow cone is equivalent to a uniform disc

Explanation:

The integration of the differential mass of the hollow right circular cone yields

I=\int\limits   dmr^2 = \int\limits^a_b {\frac{2Mxr^2}{R^2 +H^2} } \, dx  = \frac{2MR^2dx}{(R^2 +H^2)^2} \frac{(R^2 +H^2)^2}{4} = \frac{1}{2}MR^2

and for a uniform disc

I = 1/2πρtr⁴ = 1/2Mr².

6 0
4 years ago
A positive charge of 18nC is evenly distributed along a straight rod of length 4.0 m that is bent into a circular arc with a rad
sergey [27]

Explanation:

Formula for angle subtended at the center of the circular arc is as follows.

           \theta = \frac{S}{r}

where,   S = length of the rod

              r = radius

Putting the given values into the above formula as follows.      

                \theta = \frac{S}{r}

                             = \frac{4}{2}

                             = 2 radians (\frac{180^{o}}{\pi})

                             = 114.64^{o}

Now, we will calculate the charge density as follows.

                 \lambda = \frac{Q}{L}

                            = \frac{18 \times 10^{-9} C}{4 m}

                            = 4.5 \times 10^{-9} C/m

Now, at the center of arc we will calculate the electric field as follows.

                 E = \frac{2k \lambda Sin (\frac{\theta}{2})}{r}

                     = \frac{2(9 \times 10^{9} Nm^{2}/C^{2})(4.5 \times 10^{-9}) Sin (\frac{114.64^{o}}{2})}{2 m}

                      = 34.08 N/C

Thus, we can conclude that the magnitude of the electric eld at the center of curvature of the arc is 34.08 N/C.

5 0
3 years ago
75kg man climbs a mountain 1000m high in 3hrs and uses 4100 joulse/min. (a) calculate the power consumption in watt, (b) what is
mr Goodwill [35]

Answer:

Explanation:

a) Power consumption is 4100 J/min / 60 s/min = 68.3 W(atts)

work done raised the potential energy

b) 75(9.8)(1000) / (3(3600)) = 68.055555... 68.1 W

c) efficiency is 68.1 / 68.3 = 0.99593... or nearly 100%

Not a very likely scenario.

3 0
3 years ago
Calculate the critical angle for light going from Glycerine to air.
Georgia [21]
The refractive index for glycerine is n_g=1.473, while for air it is n_a = 1.00.

When the light travels from a medium with greater refractive index to a medium with lower refractive index, there is a critical angle over which there is no refraction, but all the light is reflected. This critical angle is given by:
\theta_c = \arcsin ( \frac{n_2}{n_1} )
where n1 and n2 are the refractive indices of the two mediums. If we susbtitute the refractive index of glycerine and air in the formula, we find the critical angle for this case:
\theta_c = \arcsin ( \frac{1.00}{1.473} )=42.8^{\circ}
6 0
3 years ago
If an exit sign is hanging from the ceiling by three chains each of which has a tension of 15N what is the weight of the sign
kvasek [131]
The weight is 45 N, because the three chains hold the sign, and each contributes 15 N.

Notice that the mass would be the weight/acceleration of gravity, m = 45/9.8 kg. But they ask the weight (force, so Newtons)
3 0
4 years ago
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