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MissTica
2 years ago
7

A new planet is discovered beyond Pluto at a mean distance to the sun of 4004 million miles. Using Kepler's third law, determine

an estimate for the time T to travel around the sun in an orbit.
Physics
1 answer:
AVprozaik [17]2 years ago
7 0

Answer:

103239.89 days

Explanation:

Kepler's third law states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

                               a³ / T² = 7.496 × 10⁻⁶  (a.u.³/days²)

where,

a is the distance of the semi-major axis in a.u

T is the orbit time in days

Converting the mean distance of the new planet to astronomical unit (a.u.)

                       1 a.u = 9.296 × 10⁷ miles        

                                      \frac{4004 * 10^{6}}{9.296 * 10^{7}}  =  43.07\ a.u.

Substituting the values into Kepler's third law equation;

                                    \frac{(43.07)^{3}}{T^{2}}  =  7.496 * 10^{-6}  

                                    T^{2} = \frac{(43.07)^{3}}{7.496 * 10^{-6}} (days)²

                                    T^{2} = \sqrt{\frac{(43.07)^{3}}{7.496 * 10^{-6}}}

                                    T = 103239.89 days

An estimate time T for the new planet to travel around the sun in an orbit is 103239.89 days

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Two window washers, Bob and Joe, are on a 3.00 m long, 395 N scaffold supported by two cables attached to its ends. Bob weighs 8
WINSTONCH [101]

Answer:

- the forces on the left hand side is 1.038 kN

- the forces on the right hand side is 1.483 kN

Explanation:

Given the data in the question, as illustrated in the image below;

Length of the scaffold = 3 m

weight of the scaffold = 395 N

Weight of Bob = 805 N and stands 1 m from the left end

weight of washing equipment = 500N and on sits 2 m from the left end

Weight of Joe = 820 N and stand 0.500 m from the right end

so the force on the left cable will be;

T_{left = \frac{1}{3m}[ (805 N)( (3-1) m) + ( 395 N )( \frac{3}{2} m) + ( 500 N )(1m ) + ( 820 N)( 0.500m ) ]

T_{left =  \frac{1}{3m}[ 1610 + 592.5 + 500 + 410 ]

T_{left =  \frac{1}{3m}[ 3112.5 ]

T_{left =  1037.5 N

T_{left =  1.038 kN

Therefore, the forces on the left hand side is 1.038 kN

On the right hand side;

T_{Right =  ( 805 N + 395 N + 500 N + 820 N ) - 1037.5 N

T_{Right =  2520 N - 1037.5 N

T_{Right =  1482.5 N

T_{Right =  1.483 kN

Therefore, the forces on the right hand side is 1.483 kN

5 0
1 year ago
A particular material has an index of refraction of 1.25. What percent of the speed of light in a vacuum is the speed of light i
beks73 [17]

Answer:

80% (Eighty percent)

Explanation:

The material has a refractive index (n) of 1.25

Speed of light in a vacuum (c) is 2.99792458 x 10⁸  m/s

We can find the speed of light in the material (v) using the relationship

n = c/v, similarly

v = c/n

therefore v = 2.99792458 x 10⁸  m/s ÷ (1.25) = 239 833 966 m/s

v = 239 833 966 m/s

Therefore the percentage of the speed of light in a vacuum that is the speed of light in the material can be calculated as

(v/c) × 100 = (1/n) × 100 = (1/1.25) × 100 = 0.8 × 100 = 80%

Therefore speed of light in the material (v) is eighty percent of the speed of light in the vacuum (c)

3 0
2 years ago
Read 2 more answers
A 40-mH ideal inductor is connected in series with a 50 Ω resistor through an ideal 15-V DC power supply and an open switch. If
sergey [27]

Answer:i=300 mA

Explanation:

Given

inductance(L)=40 mH

Resistor(R)=50 \Omega

Voltage(V)=15 V

Time constant(\tau)=\frac{L}{R}

\tau =\frac{40\times 10^{-3}}{50}=8\times 10^{-4}

current i_0=\frac{V}{R}

i_0=\frac{15}{50}=0.3 A

Current as a function of time is given by

i=i_0\left ( 1-e^{-\frac{t}{\tau }}\right )

i=0.3\times 0.9998

i= 299.95 mA

6 0
2 years ago
Walt ran 5 kilometers in 25 minutes, going eastward. What was his average velocity?
storchak [24]
1km per 5 mins
PLEASE VERIFY WITH SOMEONE I MAY BE INCORRECT
7 0
1 year ago
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A cord is wrapped around the rim of a solid uniform wheel 0.280m in radius and of mass 8.80kg. A steady horizontal pull of 32N t
Grace [21]

Answer:25.97 rad/s^2

Explanation:

Given

radius of wheel r=0.28 m

mass of wheel m=8.80 kg

Force F=32 N

Moment of Inertia of solid wheel I=\frac{mr^2}{2}

I=\frac{8.8\times 0.28^2}{2}

I=0.344 kg-m^2

Torque is given by

\tau =F\times r=I\times \alpha

32\times 0.28=0.344\times \alpha

\alpha =25.97 rad/s^2

Force on the axle is 32 N since there is no linear acceleration of the system

using Third law F=32 N

Torque of the axle applied to the wheel is zero because force of axle imparted at the center of axle

3 0
2 years ago
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