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3 years ago

What diameter is the sun by the scale used? a0 mm

Physics

2 answers:

tino4ka555 [31]3 years ago

7 0

The answer is kilometers.

ehidna [41]3 years ago

7 0

6gms thats what I think so yes umm thats it pretty much

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todos los cuerpos pueden estar sobre la tierra por la fuerza de gravedad .cuando nos caemos ,saltamos y realizamos otras activid

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2 years ago

Which of the following is the best hypothesis

tekilochka [14]

Answer:

Melting and frezzing are physical changes

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2 years ago

How do the forces in nuclear and chemical reactions differ?

kvasek [131]

B. Chemical reactions must overcome the strong nuclear force

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3 years ago

Read 2 more answers

Indigenous people sometimes cooked in watertight baskets by placing enough hot rocks into the water to bring it to a boil. What

yaroslaw [1]

Answer:

The rock has a mass of 4.02 kg

Explanation:

Step 1: Data given

Mass of the rock = TO BE DETERMINED

Temperature of the rock = 500 °C

Mass of the water =4.24 kg

⇒ loses 0.044kg as vapor

Initial temperature of the water = 29°C

Final temperature = 100°C

Specific heat of rock = 0.20 kcal/kg °C

Specific heat of water = 1kcal/kg°C

Latent heat of vaporization = 539 kcal/kg

Step 2: formules

Qlost,rock + Qgained,water = 0

Qtotal,water = Qwater +Qvapor

Step 3: Calculate Qvapor

Qvapor = mass of vapor * Latent heat of vapor

Qvapor = 0.044kg * 539 kcal/kg = 23.716 kcal

Step 4: Calculate Qwater

Qwater = mass of water * specific heat * Δtemperature

Qwater = 4.196 kg * 1kcal/kg°C *( 100-29)

Qwater = 297.916 kcal

Step 5: Calculate Qwater,total

Qwater,total = Qwater + Qvapor

Qwater,total = 23.716 kcal + 297.916 = 321.632 kcal

Step 6: Calculate Qrock

Qrock = mass of rock * specific heat rock * Δtemperature

Qrock = mass of rock * 0.20 kcal/kg°C * (100-500)

Qrock = mass of rock * -80 kcal/kg

Step 7: Calculate mass of rock

Qlost,rock + Qgained,water = 0

Qlost,rock = -Qgained,water

mass of rock * -80 kcal/kg = -321.632 kcal

mass of rock = 4.02 kg

The rock has a mass of 4.02 kg

7 0

2 years ago

Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp

slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

$Q=Av$

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

$Q=1.20 m^3/s$

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

$A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2$

So we can re-arrange the equation to find the speed of the water:

$v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s$

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

$Q_1 = Q_2\\A_1 v_1 = A_2 v_2$

where we have

$A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s$

and where $A_2$ is the cross-sectional area of the pipe at the second point.

Solving for A2,

$A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2$

And finally we can find the radius of the pipe at that point:

$A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m$

6 0

3 years ago