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4 years ago

What is the relationship between temperature and depth inside earth

Physics

1 answer:

Gekata [30.6K]4 years ago

4 0

In general the temperature increases with depth below the earths surface

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Three liquids are at temperatures of 6 ◦C, 23◦C, and 38◦C, respectively. Equal masses of the first two liquids are mixed, and th

kupik [55]

The equilibrium temperature is T13=3.12 ◦C

Explanation:

Given

The temperature of liquids: T1=6◦C, T2=23◦C, T3=38◦C

The temperature of 1+2 liquids mix: T12= 13◦C.

The temperature of 2+3 liquids mix: T23=26.8 ◦C.

The temperature of 1+3 liquids mix: T13= ??

1.When the first two liquids are mixed:

mC1(T1-T12)+mC2(T2-T12)=0
C1(6-13)=C2(23-13)=0
7C1=10C2
C1=1.42C2

2.When the second and third liquids are mixed:

mC2(T2-T23)+mC3(T3-T23)=0
C2(23-26.8)=C3(38-26.8)=0
3.8C2=12.8C3
C2=3.36C3

3.When the first and third liquids are mixed:

mC1(T1-T13)+mC3(T3-T13)=0
C1(6-T13)+C3(38-T13)=0
C1=1.42C2 C2=3.36C3
C1=1.42C2(3.36C3)
C1=4.77C3
C1(6-T13)+C3(38-T13)=0
4.77C3(6-T13)+C3(38-T13)=0
By solving the equation we get,
T13=3.12 ◦C
The equilibrium temperature is T13=3.12 ◦C

7 0

3 years ago

Before beginning an exercise program a person should always

swat32

Answer:

I do believe it is A it makes the most since

8 0

4 years ago

When have you experienced an increase in kinetic energy within a system?

Mars2501 [29]

Answer:

If a man starts running on a boat with an acceleration a with respect to the boat, there is no external force that acts on the Boat+Man system

8 0

3 years ago

A hypothesis is

Feliz [49]

The answer is C, an educated guess

3 0

3 years ago

A rock is thrown upward with a velocity of 18 meters per second from the top of a 38 meter high cliff, and it misses the cliff o

Roman55 [17]

The rock would be at a point 12 m from water at a time 4.8 s.

Take the origin of the coordinate system at the top of the cliff. It is thrown upwards with a velocity u. When the rock is at a point 12 m from water, calculate the vertical displacement of the rock from the origin.

$y= -38m- (-12 m)=-26 m$

Use the equation of motion,

$y=ut +\frac{1}{2} at^2$

The rock falls under the acceleration due to gravity, directed down wards.

Substitute 18 m/s for u, -26 m for y and -9.8 m/s² for a=g.

$y=ut +\frac{1}{2} at^2\\ (-26 m)=(18m/s)t+\frac{1}{2}(-9.8m/s^2)t^2$

Solve the quadratic equation for t.

$t=\frac{(18m/s)(+/-)\sqrt{(18m/s)^2-4(4.9m/s^2)(-26m)} }{2(4.9m/s^2)}$

Taking only the positive value,

$t=\frac{(18 m/s)+(28.87 m/s)}{9.8 m/s^2)} \\ t=4.78 s$

After a time of 4.8 s the rock would be at a distance of 12 m from water.

8 0

3 years ago