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Lady_Fox [76]
2 years ago
12

In 2004 astronomers reported the discovery of a large Jupiter-sized planet orbiting very close to the star HD 179949 (hence the

term "hot Jupiter"). The orbit was just 19 the distance of Mercury from our sun, and it takes the planet only 3.09 days to make one orbit (assumed to be circular). What is the mass of the star? Express your answer (a)in kilograms and (b)as a multiple of our sun's mass.
Physics
1 answer:
sp2606 [1]2 years ago
6 0
  <span>I'll tell you how to do it but you must crunch the numbers. 

Use Kepler's 3rd Law 

T^2 = k R^3 

where k = 4(pi)^2/ GM 
G =gravitational constant = 6.67300 × 10-11 m3 kg-1 s-2 
M = mass of this new planet 
pi = 3.14159265 
T =3.09 days = 266976 seconds 
R = (579,000,000km)/9 = 64333333.3 km 

a) 
Solve Kepler's 3rd Law for M. Your answer will be in kg 


b) 
mass of the sun = 1.98892 × 10^30 kilograms 
Form the ratio 
M(planet)/M(sun) </span>
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How many joules of electrical energy is transferred per second by 6V 0.5 A lamp?
Degger [83]

When you ask for "joules per second", you're asking for "watts".
The rate of energy "transfer" is 'power'.  In this case, the light bulb
transfers energy out of the electrical circuit and into the space
around it, in the form of light and heat radiation.

Electrical power = (voltage) x (current) =

                              (6 volts) x (0.5 ampere) =

                                          3 watts  =  3 joules per second.
 
4 0
2 years ago
How much energy is stored in a 2.80-cm-diameter, 14.0-cm-long solenoid that has 200 turns of wire and carries a current of 0.800
ozzi

Answer:

The energy stored in the solenoid is 7.078 x 10⁻⁵ J

Explanation:

Given;

diameter of the solenoid, d = 2.80 cm

radius of the solenoid, r = d/2 = 1.4 cm

length of the solenoid, L = 14 cm = 0.14 m

number of turns, N = 200 turns

current in the solenoid, I = 0.8 A

The cross sectional area of the solenoid is given as;

A = \pi r^2\\\\A = \pi (0.014)^2\\\\A = 6.16*10^{-4} \ m^2

The inductance of the solenoid is given by;

L = \frac{\mu_0 N^2A}{l} \\\\L =  \frac{(4\pi*10^{-7})(200^2)(6.16*10^{-4})}{0.14}\\\\L = 2.212*10^{-4} \ H

The energy stored in the solenoid is given by;

E = ¹/₂LI²

E = ¹/₂(2.212 x 10⁻⁴)(0.8)²

E = 7.078 x 10⁻⁵ J

Therefore, the energy stored in the solenoid is 7.078 x 10⁻⁵ J

8 0
3 years ago
A student pushes a 0.2 kg box against a spring causing the spring to compress 0.15 m. When the spring is released, it will launc
german

Answer:

The maximum height the box will reach is 1.72 m

Explanation:

F = k·x

Where

F = Force of the spring

k = The spring constant = 300 N/m

x  = Spring compression or stretch = 0.15 m

Therefore the force, F of the spring = 300 N/m×0.15 m = 45 N

Mass of box = 0.2 kg

Work, W, done by the spring = \frac{1}{2} kx^2 and the kinetic energy gained by the box is given by KE = \frac{1}{2} mv^2

Since work done by the spring = kinetic energy gained by the box we have

\frac{1}{2} mv^2 =  \frac{1}{2} kx^2  therefore we have v = \sqrt{\frac{kx^2}{m} } = x\sqrt{\frac{k}{m} } = 0.15\sqrt{\frac{300}{0.2} } = 5.81 m/s

Therefore the maximum height is given by

v² = 2·g·h or h = \frac{v^2}{2g} = \frac{5.81^{2} }{2*9.81} = 1.72 m

6 0
3 years ago
Vectors A and B lie in the xy-plane. Vector A has a magnitude of 19.1 and is at an angle of 125.5º counterclockwise from the +x-
Nady [450]

Answer:

a!aaaaaaaaaaaa!aaaaaa

Explanation:

l

8 0
3 years ago
The angular momentum of a flywheel having a rotational inertia of 0.140 kg ·m2 about its central axis decreases from 3.00 to 0.8
Rasek [7]

Answer

given,

I = 0.140 kg ·m²

decrease from 3.00 to 0.800 kg ·m²/s in 1.50 s.

a) \tau = \dfrac{\Delta L}{\Delta t}

   \tau = \dfrac{0.8-3}{1.5}

        τ = -1.467 N m

b) angle at which fly wheel will turn

   \theta= \omega t +\dfrac{1}{2}\alpha t^2

   \theta= \dfrac{L}{I} t +\dfrac{1}{2}\dfrac{\tau}{I}t^2

   \theta= \dfrac{3}{0.14}\times 1.5+\dfrac{1}{2}\dfrac{-1.467}{0.14}\times 1.5^2

        θ = 20.35 rad

c) work done on the wheel

     W = τ x θ

     W = -1.467 x 20.35 rad

    W = -29.86 J

d) average power of wheel

    P_{av} =-\dfrac{W}{t}

    P_{av} =-\dfrac{(-29.86)}{1.5}

     P_{av} =19.91\ W          

7 0
2 years ago
Read 2 more answers
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