What is the orbital velocity of Jupiter around the sun

What must be the diameter of a cylindrical 120-m long metal wire if its resistance is to be 6.0 ω? the resistivity of this metal

Free_Kalibri [48]

The resistance of the cylindrical wire is $R=\frac{\rho l}{A}$ .

Here $R$ is the resistance, $l$ is the length of the wire and $A$ is the area of cross section. Since the wire is cylindrical $A=\frac{\pi d^2}{4}$ . Rearranging the above equation,

$A=\frac{\rho l}{R}\\ \frac{\pi d^2}{4}=\frac{\rho l}{R}\\ d=\sqrt{\frac{4\rho l}{\pi R}}$

Here $l=120, R=6, \rho=1.68(10^{-8})$ .

Substituting numerical values,

$d=\sqrt{\frac{4(1.68)(10^{-8}) (120)}{\pi (6)}}\\ d=0.0006541$

Te diameter of the wire is $0.6541 mm$

7 0

4 years ago

What type of nuclear decay occurs when the original nucleus loses two protons and two neutrons?

Igoryamba

THe loss of 2 protons and 2 neutrons (also called a helium nucleus) is defined as alpha decay.

3 0

3 years ago

A race car's velocity increases from +28 m/s to +36 m/s over a 2.0-s time interval. What is the car's

BigorU [14]

Answer:

$4 m/s^{2}$

Explanation:

Acceleration, $a=\frac {v-u}{t}$

Where v and u are the final and initial velocities of the race car respectively, t is the time taken for the race car to attain velocity of 36 m/s.

Substituting 36 m/s for v, 28 m/s for u and 2 s for t then

$a=\frac {36 m/s-28 m/s}{2}=4 m/s^{2}$

3 0

3 years ago

Is work done if you carry a book across the room at a constant velocity?

Stella [2.4K]

<span>No. Work is not done if you carry a book across the room
at a constant velocity?

The force applied is perpendicular to the direction of motion. (C)</span>

4 0

4 years ago

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Water evaporating from a pond does so as if it were diffusing across an air film 0.15 cm thick. The diffusion coefficient of wat

QveST [7]

Answer:

The water level will drop by about 1.24 cm in 1 day.

Explanation:

Here Mass flux of water vapour is given as

$j_{H_2O}=\frac{D}{l} \bigtriangleup c$

where

$j_{H_2O}$ is the mass flux of the water which is to be calculated.

D is diffusion coefficient which is given as $0.25 cm^2/s$

l is the thickness of the film which is 0.15 cm thick.
$\bigtriangleup c$ is given as

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT}$

In this

$P_{sat}$ is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa

$P_a$ is the air pressure which is given as 0.5 times of $P_{sat}$

R is the universal gas constant as $8.314 kJ/kmol-K$

T is the temperature in Kelvin scale which is $20+273= 293K$

By substituting values in the equation

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT} \\ \bigtriangleup c= \frac{P_{sat}-0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5 \times 2.34}{8.314 \times 293} \\\bigtriangleup c= 0.48 mol/m^3$