State of motion is described by

In January 2006, astronomers reported the discovery of a planet comparable in size to the earth orbiting another star and having

Orlov [11]

Answer:

R = 5.28 103 km

Explanation:

The definition of density is

ρ = m / V

V = m /ρ

Where m is the mass and V the volume of the body

The volume of a sphere is

V = 4/3 π r³

Let's replace

4/3 π r³ = m / ρ

R =∛ ¾ m / ρ π

The mass of the planet is

M = 5.5 Me

R = ∛ ¾ 5.5 Me /ρ π

Let's reduce the density to SI units

ρ = 1.76 g / cm³ (1 kg / 10³ g) (10² cm / 1 m)³

ρ = 1.76 10³ kg / m³

Let's calculate

R = ∛ ¾ 5.5 5.97 10²⁴ / (1.76 10³ pi)

R = ∛ 0.14723 10²¹

R = 0.528 10⁷ m

R = 0.528 104 km

R = 5.28 103 km

8 0

3 years ago

An electric dipole is formed from ±1.00nC charges spaced 3.00 mm apart. The dipole is at the origin, oriented along the x-axis.

Ronch [10]

Answer:

Value of electric field along the axis and equitorial axis $E=31.25\ N/c$ and $E = 15.625\ N/c$ respectively.

Explanation:

Given :

Distance between charges , $d = 3 \ mm =\dfrac{3}{1000}\ m=3\times 10^{-3}\ m.$

Magnitude of charges , $q=1\ nC = 10^{-9}\ C.$

Dipole moment , $p=qL=10^{-9}\times 3\times 10^{-3}=3\times 10^{-12} \ C\ m.$

Case A) (x,y) = (12.0 cm, 0 cm) :

Electric field of dipole in its axis ,

$E=\dfrac{2kp}{r^3}$

Putting all values and $r=12\times 10^{-2}\ m.$

We get , $E=31.25\ N/c.$

Case B) (x,y) = (0 cm, 12.0 cm) :

Electric field of dipole on equitorial axis ,

$E = \dfrac{kp}{r^3}$

Putting all values and $r=12\times 10^{-2}\ m.$

We get , $E = 15.625\ N/c.$

Hence , this is the required solution.

7 0

2 years ago

You wish to cool a 1.83 kg block of tin initially at 88.0°C to a temperature of 57.0°C by placing it in a container of kerosene

uranmaximum [27]

Answer:

0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is $150\,^{\circ}C$ . According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

$\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T)$ (1)

Where:

$\rho_{k}$ - Density of kerosene, measured in kilograms per cubic meter.

$V_{k}$ - Volume of kerosene, measured in cubic meters.

$c_{k}$ , $c_{t}$ - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

$T_{k,o}$ , $T_{t,o}$ - Initial temperatures of kerosene and tin, measured in degrees Celsius.

$T$ - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

$V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}$

If we know that $m_{t} = 1.83\,kg$ , $c_{t} = 218\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{t,o} = 88\,^{\circ}C$ , $T_{k,o} = 24.0\,^{\circ}C$ , $T = 57\,^{\circ}C$ , $c_{k} = 2010\,\frac{J}{kg\cdot ^{\circ}C}$ and $\rho_{k} = 820\,\frac{kg}{m^{3}}$ , then the volume of the liquid needed to accomplish this task without boiling is:

$V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}$