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3 years ago

Ice sinks in methanol, but floats on water. From this we can conclude ?

Physics

1 answer:

-Dominant- [34]3 years ago

4 0

Answer:

Density:water>ice>methanol

Explanation:

ice is lighter than methanol hence it is able to float on it However ice is denser than water hence it sinks

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An automobile traveling 95 km/h overtakes a 1.30-km-long train traveling in the same direction on a track

BartSMP [9]

1) 234 s, 6.18 km

The position of the car at time t is described as

$x_c(t) = v_c t$

where

$v_c = 95 km/h$ is the velocity of the car

The position of the head of the train instead is given by

$x_t(t) = d + v_t t$

where

d = 1.30 km is the initial distance between the car and the head of the train

$v_t = 75 km/h$ is the velocity of the train

The car overtakes the train when

$x_c =x_t$

Substituting and solving for t,

$v_c t = d + v_t t\\t(v_c-v_t)=d\\t=\frac{d}{v_c-v_t}=\frac{1.30}{95-75}=0.065 h = 234 s$

And the distance travelled by the car is

$x_c = v_c t = (95 km/h)(0.065 h)=6.18 km$

2) 27.5 s, 0.72 km

In this case, the train is travelling in opposite direction, so we can write

$v_t = -75 km/h$

Again, we can use the same equation as before

$x_c =x_t$

And solving for t, we find

$v_c t = d + v_t t\\t(v_c-v_t)=d\\t=\frac{d}{v_c-v_t}=\frac{1.30}{95+75}=0.0076 h = 27.5 s$

And the distance travelled by the car is

$x_c = v_c t = (95 km/h)(0.0076 h)=0.72 km$

5 0

3 years ago

A street musician sounds the A string of his violin, producing a tone of 440 Hz, What frequency does a bicyclist hear as he a) a

DanielleElmas [232]

Answer:

a) $f_o=454.11Hz$

b) $f_o=425.89Hz$

Explanation:

Let´s use Doppler effect, in order to calculate the observed frequency by the byciclist. The Doppler effect equation for a general case is given by:

$f_o=\frac{v\pm v_o}{v\pm v_s} *f_s$

where:

$f_o=Observed\hspace{3}frequency$

$f_s=Actual\hspace{3}frequency$

$v=Speed\hspace{3}of\hspace{3}the\hspace{3}sound\hspace{3}waves$

$v_s=Velocity\hspace{3}of\hspace{3}the\hspace{3}source$

$v_o=Velocity\hspace{3}of\hspace{3}the\hspace{3}observer$

Now let's consider the next cases:

$+v_o\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}observer\hspace{3}moves\hspace{3}towards\hspace{3}the\hspace{3}source$

$-v_o\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}observer\hspace{3}moves\hspace{3}away\hspace{3}from\hspace{3}the\hspace{3}source$

$-v_s\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}source\hspace{3}moves\hspace{3}towards\hspace{3}the\hspace{3}observer$

$+v_s\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}source\hspace{3}moves\hspace{3}away\hspace{3}from\hspace{3}the\hspace{3}observer$

The data provided by the problem is:

$f_s=440Hz\\v_o=11m/s$

The problem don't give us aditional information about the medium, so let's assume the medium is the air, so the speed of sound in air is:

$v=343m/s$

Now, in the first case the observer alone is in motion towards to the source, hence:

$f_o=\frac{v+v_o}{v}*f_s=\frac{343+11}{343} *440=454.1107872Hz$

Finally, in the second case the observer alone is in motion away from the source, so:

$f_o=\frac{v-v_o}{v}*f_s=\frac{343-11}{343} *425.8892128Hz$

6 0

4 years ago

At a distance of 0.220 cm from the axis of a very long charged conducting cylinder with radius 0.100cm, the electric field is 49

Scorpion4ik [409]

Answer:

At the distance of 0.220cm from the axis.

r = 0.220cm = 0.0022m, E = 490N/C, e0 = 8.854 x 10^-12F/m

Linear charge density = 2*π*e0*r*E = 2 x 3.142 x 8.854x10^-12 x 0.0022 x 490 = 5.998 x 10^-11C/m

Thus, To Calculate the Electric field at the distance r = 0.616cm from the cylinder axis, we substitute the calculated linear change density in the equation

E = (linear charge density)/2*π*e0*r

Here, r = 0.616cm = 0.00616m

E = [(5.998 x 10^-11)/(2 x 3.142 x 8.854 x 10-12 x 0.00616)]

E = 175N/C

Explanation:

The Electric field of a charged conducting cylinder obey the Gauss Law.

Therefore, the Electric field is given as:

E = (linear charge density)/4πe0r,

Where e0 is the permittivity of free space with constant value of 8.854 x 10^-12F/m, r is the radial distance from the axis.

3 0

3 years ago

Convert 1 mm to meters using scientific notation. Include units in your answer.

irinina [24]

Answer: The answer in scientific notation with the unit included is (1 × 10^-3m.)

Explanation:

In the field of physics, when working with too large or too small numbers, such can be expressed with the use of scientific notation. This helps to eliminate huge or ambiguous numbers with significant units. It is written is such a way that the first number is greater than or equal to 1 and less than 10 which is then multiplied by a factor of 10.

Units are symbols that are used to represent a measurement For example the unit of meters is (m).

Converting 1mm to m:

1000millimeters = 1meter

1 millimetre = X

Making X the subject of formula,

X = 1 ×1 /1000

X = 0.001m

Therefore expressing X in scientific notation is

1 × 10^-3meter (m)

7 0

3 years ago

Naturally occurring magnets are called

mixas84 [53]

Lodestone is a naturally magnetized piece of the mineral magnetite

7 0

4 years ago

Ice sinks in methanol, but floats on water. From this we can conclude ?​

Ice sinks in methanol, but floats on water. From this we can conclude ?