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13 days ago

6

"As the Voyager spacecraft penetrated into the outer solar system, the illumination from the Sun declined. Relative to the situa

tion at Earth, how bright is the sunlight at Jupiter? Jupiter's semi-major axis is 5.2 AU. Round your answer to two decimal places."

1 answer:

serg [3.5K]13 days ago

3 0

Answer:

$\frac{I_{2}}{I_{1}} = 0.04$

Explanation:

The brightness of a star observed at a specific distance decreases inversely proportional to the square of the distance. Thus:

$I_{1}\cdot r_{1}^{2} = I_{2}\cdot r_{2}^{2}$

The brightness of sunlight on Jupiter as compared to Earth is:

$\frac{I_{2}}{I_{1}} = \frac{r_{1}^{2}}{r_{2}^{2}}$

$\frac{I_{2}}{I_{1}} = \left(\frac{1\,AU}{5.2\,AU} \right) ^{2}$

$\frac{I_{2}}{I_{1}} = 0.04$

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An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate

inna [3103]

Answer:

the time it takes after impact for the puck is 2.18 seconds

Explanation:

initially given information

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thickness = 0.1 mm = 1 × $10^{-4}$ m

dynamic viscosity = 1.75 × $10^{-5}$ Ns/m²

temperature of air = 15°C

to determine

time needed for the puck to reduce its speed by 10%

solution

we note that velocity changes from 0 to v

assuming initial velocity = v

therefore final velocity = 0.9v

implying a change in velocity is du = v

and clearance dy = h

shear stress acting on the surface is expressed as

= µ $\frac{du}{dy}$

therefore

= µ $\frac{v}{h}$ ............1

substituting the values

= 1.75 × $10^{-5}$ × $\frac{v}{10^{-4}}$

= 0.175 v

the area between the air and puck is given by

Area = $\frac{\pi }{4} d^{2}$

area = $\frac{\pi }{4} 0.1^{2}$

area = 7.85 × $\frac{v}{10^{-3}}$ m²

thus, the force on the puck can be represented as

Force = × area

force = 0.175 v × 7.85 × $10^{-3}$

force = 1.374 × $10^{-3}$ v

now applying Newton's second law

force = mass × acceleration

- force = $mass \frac{dv}{dt}$

- 1.374 × $10^{-3}$ v = $0.03 \frac{0.9v - v }{t}$

solving for t = $\frac{0.1 v * 0.03}{1.37*10^{-3} v}$

the time needed after impact for the puck is 2.18 seconds

3 0

1 month ago

A rocket is continuously firing its engines as it accelerates away from Earth. For the first kilometer of its ascent, the mass o

kicyunya [3294]

Answer:

The correct response is:

1. KE Increases, PE Increases, ME Increases.

Explanation:

In this context, kinetic energy refers to the energy associated with an object's motion. Kinetic energy can be defined as the energy required to accelerate a mass from rest to a specified velocity, which it maintains once that speed is reached:

KE = 1/2 mv².

This definition indicates that KE is on the rise.

Potential energy is the energy stored in a body due to its position in a gravitational field:

PE = mgh,

which increases as the object is elevated against gravitational pull.

Since both kinetic and potential energies are increasing, it follows that the total mechanical energy (ME) is also rising:

ME = PE + KE.

4 0

1 month ago

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Kai takes a hamburger off of the grill and puts a piece of cheese on it. Explain how and why the cheese melts when Kai puts it o

Keith_Richards [3271]

When Kai removes the burger from the grill and adds cheese, the cheese melts due to the heat retained by the burger from cooking. The warmth remains in the burger for a while, which explains the melting of the cheese.

4 0

16 days ago

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In a demonstration, a 4.00 cm2 square coil with 10 000 turns enters a larger square region with a uniform 1.50 T magnetic field

serg [3582]

Explanation:

It is stated that,

Area of square coil, $A=4\ cm^2=0.0004\ m^2$

Length of one side of the square, L = 0.02 m

Number of coils, N = 10000

Consistent magnetic field, B = 1.5 T

Velocity, v = 100 m/s

An electromotive force is produced in the coil calculated as:

$E=NBLv$

$E=10000\times 1.5\times 0.02\times 100$

E = 30000 V

Breakdown voltage of air, $V=4000\ V/cm=400000\ V/m$

Let d be the distance between the ends of the coil wires that can still produce a spark. Therefore,

Electric field, $E'=\dfrac{V}{d}$

$\dfrac{30000}{d}=400000$

d = 0.075 m

Thus, this forms the final answer.

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21 day ago

The filament of the light bulb is made of tungsten. The resistance of the light bulb at room temperature (20∘C), measured by an

Ostrovityanka [3204]

Explanation:

The formula illustrating the relationship between resistance and temperature is as follows:

R =

$R_{o} + \alpha [T_{2} - T_{1}]$

where, R = final resistance

= initial resistance $R_{o}$

= temperature coefficient of resistivity $\alpha$

= final temperature $T_{2}$

= initial temperature $T_{1}$

Given data as follows.

$T_{1} = (20 + 273) K = 293 K$

R = 36 ohm,

= 3 ohm $T_{2}$

= 0.0045 $R_{o}$

Substituting the provided values into the above formula gives us the following.

R = $\alpha$

36 =

$R_{o} + \alpha [T_{2} - T_{1}]$

=

$3 + 0.0045 \times [T_{2} - 293]$

= 7626.33 K

$T_{2}$ Thus, it can be concluded that $\frac{34.3185}{0.0045}$ the temperature of the light bulb at 12.0 V is 7626.33 K.

7 0

1 month ago