Which type of front causes many hours of steady rain before that front passes a location?

Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las person

Darina [25.2K]

Answer:

82.76m

Explanation:

In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.

You use the formula for the circumference of the steel ring:

$C=2\pi r$ (1)

C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)

you solve for r in the equation (1):

$r=\frac{C}{2\pi}=\frac{4*10^7m}{2\pi}=6,366,197.724m$

Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:

$L=Lo[1+\alpha \Delta T]$ (2)

L: final length of the tube = ?

Lo: initial length of the tube = 4*10^7m

ΔT = change in the temperature of the steel tube = 1°C

α: thermal coefficient expansion of steel = 13*10^-6 /°C

You replace the values of the parameters in the equation (2):

$L=(4*10^7m)(1+(13*10^{-6}/ \°C)(1\°C))=40,000,520m$

With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:

$r'=\frac{C}{2\pi}=\frac{40,000,520m}{2\pi}=6,366,280.484m$

Finally, you compare both r and r' radius:

r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m

Hence, the distance to the ring from the ground is 82.76m

4 0

2 years ago

Variations in the resistivity of blood can give valuable clues about changes in various properties of the blood. Suppose a medic

Elena-2011 [213]

Answer:

1.1 x 10^9 ohm metre

Explanation:

diameter = 1.5 mm

length, l = 5 cm

Potential difference, V = 9 V

current, i = 230 micro Ampere = 230 x 10^-6 A

radius, r = diameter / 2 = 1.5 / 2 = 0.75 x 10^-3 m

Let the resistivity is ρ.

Area of crossection

A = πr² = 3.14 x 0.75 x 0.75 x 10^-6 = 1.766 x 10^-6 m^2

Use Ohm's law to find the value of resistance

V = i x R

9 = 230 x 10^-6 x R

R = 39130.4 ohm

Use the formula for the resistance

$R=\rho \frac{l}{A}$

$\rho =\frac{RA}{l}$

$\rho =\frac{39130.4\times 0.05}{1.766\times 10^{-6}}$

ρ = 1.1 x 10^9 ohm metre

Explanation:

7 0

2 years ago

Seven seconds after a brilliant flash of lightning, thunder shakes the house. approximately how far was the lightning strike fro

tangare [24]

Very roughly 7,700 feet ... about 1.5 miles.

8 0

2 years ago

If the wire is replaced by an infinite current sheet with density Js = 0.40 A/m, what would be the magnetic flux (in T · m2) thr

oksian1 [2.3K]

Answer:

$\phi _{B} =0.855 T-m^{-2}$

Explanation:

given data

density of current sheet = 0.40 A/m

length a = 0.27 m

width b = 0.63 m

For infinite sheet, magnetic field is given as

$B = \mu _{O}J$

magnetic flux is given as

$\phi _{B} = BA$

$= \mu _{O}Jab$

$= 4\pi *0.40*0.27*0.63$

$\phi _{B} =0.855 T-m^{-2}$

6 0

2 years ago

A 35-g block of ice at -14°C is dropped into a calorimeter (of negligible heat capacity) containing 400 g of water at 0°C. When

kompoz [17]

Answer:

Total mass of ice = 38.06g

Explanation:

Since the heat capacity of calorimeter is negligible.

The water is already at 0°C, so the heat loss can no longer reduce the temperature of the water. It is used for fusion and forming more ice.

The equilibrium temperature will be 0°C, because the heat gain by ice is only enough to bring it down to 0°C.

Heat gained by ice = heat loss by water

Heat gained by ice (from -14°C to 0°C) = heat lost to fusion by water (heat of fusion of some amount of the water present in the calorimeter)

mi Ci ∆Ti = mw . L ......1

Where;

mi = mass of ice = 35g = 0.035 kg

Ci = specific heat capacity of ice = 2090 J/kg ∙ K

∆Ti = change in temperature of ice = 0-(-14) = 14 K

mw = the mass of water that have gained enough heat for fusion ( mass of water converted to ice)

L = latent heat of fusion of water = 33.5 × 10^4 J/kg.

From equation 1;

mw = (mi Ci ∆Ti )/L

mw = (0.035×2090×14)/335000

mw = 0.00306 kg

mw = 3.06 g

Therefore, 3.06 g of water has been converted to ice.

When combined with the initial amount of ice initially in the calorimeter (at 0°C)

Total mass of ice = mi + 3.06g = 35g + 3.06g = 38.06g

Total mass of ice = 38.06g

6 0

2 years ago

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