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mash [69]
3 years ago
13

Generally you must stir a can of paint before you use it.. is paint miscible or immiscible? explain

Physics
1 answer:
Wewaii [24]3 years ago
6 0

Paint is immiscible because the layer that forms on top of settled paint is a substance which cannot dissolve with the rest of the paint.  Stirring the paint is only temporarily in effort to form an even and thoroughly mixed solution by physically mixing the separate molecules of the solution.  However, for it to be miscible, it the solution has to be evenly distributed (chemically) on a molecular level as to not separate when left for a lengthy period of time.

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If two swimmers compete in a race, does the faster swimmer develop more power?
valkas [14]
Power is equal to energy per unit time. In this case, power is proportional to energy while is inversely proportional to time,on the other hand. Given the two swimmers exerts same amount of energy but the faster swimmer just does things in faster time, then the faster swimmer should develop more power from shorter time
7 0
3 years ago
How can a gas become a good conductor? Simpler answers would be helpful.
Alex_Xolod [135]
A gas has to become ionisied in order to become a conductor. It must have a chain reaction in which atoms in it became unstable, in which they loose stabile electronic configuration. In order for a gas to become a conductor, it must have free particles, and it can happen only in ionisied gas.
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3 years ago
A 1.50-mm-diameter glass sphere has a charge of + 1.60 nC. What speed does an electron need to orbit the sphere 1.60 mm above th
saveliy_v [14]

Answer :

Velocity will be 3.28\times 10^{-11}m/sec

Explanation:

We have given glass surface has a diameter of 1.5 mm

And charge q = 1.60 nC

Radius of electrons orbit r = height of electron above surface + radius of sphere  = =1.6+\frac{1.5}{2}=2.35mm = 0.00235m

Force on electron is given by F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}, here q is charge on sphere and e is charge on electron

F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}=\frac{kqe}{r^2}=\frac{9\times 10^9\times 1.6\times 10^{-9}\times 1.6\times 10^{-19}}{0.00235^2}=4.172\times 10^{-13}N

This force work as centripetal force

So F=\frac{mv^2}{r}

4.172\times 10^{-13}=\frac{9.11\times 10^{-31}v^2}{0.00235}

v = =0.0328\times 10^{-9}=3.28\times 10^{-11}m/sec

   

6 0
3 years ago
A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)
Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is U  =  \int\limits^T_0 {P(t)} \, dt

Compute U if the bulb remains on for 5h

Answer:

The value is  U  =  7.563 *10^{5} \  J

Explanation:

From the question we are told that

   The power rating of the bulb is P  =  100 \  W

   The resistance is   R =  143 \ \Omega

   The  voltage is  V  =  V_o  sin [2 \pi ft]

   The  energy expanded is U  =  \int\limits^T_0 {P(t)} \, dt

   The  voltage  V_o  =  110 \  V

   The frequency is  f =  60 \  Hz

    The  time considered is  t =  5 \  h  =  18000 \  s

Generally power is mathematically represented as

             P =  \frac{V^2}{ R}

=>          P =  \frac{( 110  sin [2 \pi * 60t])^2}{ 144}

=>           P =  \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}

So  

     U  =  \int\limits^T_0 { \frac{ 110^2*  [sin [120 \pi t])^2}{ 144}} \, dt

=>  U  =  \frac{110^2}{144} \int\limits^T_0 { (   sin^2 [120 \pi t]} \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | T} \atop {0}} \right.

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | 18000} \atop {0}} \right.

U =  \frac{110^2}{144} [\frac{18000}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi (18000))}{240 \pi} ] ]

=>   U  =  7.563 *10^{5} \  J

7 0
3 years ago
How much larger is the diameter of Jupiter compared to the diameter of Saturn?
allsm [11]
Really long we’ll not long but far in distance
5 0
3 years ago
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