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3 years ago

When water in a lake freezes, the ice that forms floats on top of any water that is still liquid. Why does the ice float?

Physics

1 answer:

ddd [48]3 years ago

5 0

Answer:

it floats because the lake is cold at that moment so when part of the lake freezes it still remains solid and floats because of the lake and the surrounding of the lake is still cold

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A mass accelerates uniformly when the resultant force on it is?

Elis [28]

. . . . . not zero .

Note: "... unbalanced" would be a terrible answer.

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2 years ago

Tammy leaves the office, drives 34 km due north, then turns onto a second highway and continues in a direction of 35◦ north of e

Natasha2012 [34]

Answer:

The total displacement is 102 km $51^o$ north of east.

Explanation:

We can treat this problem as a trigonometric one, so we need to calculate the total displacement on the north and east.

$d_n=34km+79km*sin(35^o)=79km$

and

$d_e=79*cos(35^o)=65km$

The total displacement is given by:

$D=\sqrt{(79km)^2+(65km)^2}=102km$

with an agle of:

$\alpha =arctg(\frac{79km}{65km})=51^o$

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2 years ago

use what you have learned so far about heat transfer to explain how hot rocks can be used to make steam?

Anna [14]

If you throw water on anything hot enough, the liquid water turns to steam. With this method, you can create an energy source.

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2 years ago

When studying atoms, scientists can ignore the blank force

natulia [17]

Answer:

Gravitational Force

Explanation:

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2 years ago

Read 2 more answers

Carbon dioxide (CO2) gas in a piston-cylinder assembly undergoes three processes in series that begin and end at the same state

topjm [15]

Answer:

a) W =400 kJ

b) W = 0 kJ

c) W =-160.944 KJ

Explanation:

Given

Process 1 ---> 2

The relation of the process P = constant

Pressure of point (1) P1 = 10 bar = P2

Volume of point (1) V1 = 1 m^3

Volume of point (2) V2 =4 m^3

The relation of the process V = constant

Process 2 ---> 3

The relation of the process V = constant

V3 = V2

Pressure of point (3) P3 = 10 bar

Volume of point (3) V3 = 4 m^3

Process 3 ---> 1 

The relation of the process PV = constant

Required

Sketch the processes on the PV coordinates

The work for each process in kJ

Solution

The work is defined by

W= $\int\limits^a_b {x} \, dx$

a=V2

b=V1

x=P

dx=dV

Process 1 ---> 2

P3 = P4 = 5 bar

W= $\int\limits^a_b {x} \, dx$

a=V3

b=V2

x=4

dx=dV

putting the value of a, b, x, dx in above integral

W=400 kJ

Process 2 ---> 3 

V = constant Then there is no change in the volume,hence W = 0 kJ

Process 3 ---> 1

By substituting with point (1) --> 5 x .2 = C ---> C = 1 P = 5V^-1

W= $\int\limits^a_b {x} \, dx$

a=V1

b=V3

x=1V^-1

dx=dV

putting the value of a, b, x, dx in above integral

W=| ln V | limit a and b

= -160.944 KJ

5 0

3 years ago