# n a distant solar system, a planet of mass 5.0 x 1024 kg orbits a sun of mass 3.0 x 1030 kg at a constant distance of 2.0 x1011 m. How many earth days does it take for the planet ot execute one complete orbit about the sun

F = M2 ω^2 R       centripetal force of sun on planet

ω = (F / (M2 R))^1/2 = 2 pi f = 2 pi / P        where P is the period

P = 2 pi (M2 * R / F)^1/2

F = G M1 M2 / R^2        gravitational force on planet

P = 2 pi {R^3 / (G M1)]^1/2

P = 6.28 [(2.0E11)^3 / (6.67E-11 * 3.0E30)]^1/2

P = 6.28 (8 / 20)^1/2 E7 = 3.9E7 sec

1 yr = 3600 * 24 * 365 = 3.15E7 sec

P = 3.9 / 3.2 = 1.2 years

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Bailey wants to find out which frozen solid melts the fastest: soda, ice, or orange juice. She pours each of the three liquids into the empty cubes of an ice tray, and then places the ice tray in the freezer overnight. The next day, she pulls the ice tray out and sets each cube on its own plate. She then waits and watches for them to melt. When the last part of the frozen liquid melts, she records the time.

its 45 over 6

Explanation:the answer is in  the question

Answer: Only the melted cube's shape changed.

Explanation:

Box 1 and box 2 are whirling around a shaft with a constant angular velocity of magnitude ω. Box 1 is at a distance d from the central axis, and box 2 is at a distance 2d from the axis. You may ignore the mass of the strings and neglect the effect of gravity. Express your answer in terms of d, ω, m1 and m2, the masses of box 1 and 2. (a) Calculate TB, the tension in string B (the string connecting box 1 and box 2). (b) Calculate TA, the tension in string A (the string connecting box 1 and the shaft).

a) TB = m2 * w^2 * 2*d

b) TA = m1 * w^2 * d + m2 * w^2 * 2*d

Explanation:

The tension on the strings will be equal to the centripetal force acting on the boxes.

The centripetal force is related to the centripetal acceleration:

f = m * a

The centripetal acceleration is related to the radius of rotation and the tangential speed:

a = v^2 / d

f = m * v^2 / d

The tangential speed is:

v = w * d

Then

f = m * w^2 * d

For the string connecting boxes 1 and 2:

TB = m2 * w^2 * 2*d

For the string connecting box 1 to the shaft

TA = m1 * w^2 * d + m2 * w^2 * 2*d

A 0.134-A current is charging a capacitor that has square plates 6.00 cm on each side. The plate separation is 4.00 mm. (a) Find the time rate of change of electric flux between the plates. V·m/s (b) Find the displacement current between the plates. A

To find the time rate of change of electric flux between the plates of the capacitor, use the formula $$\frac{d\phi_E}{dt} = \frac{I}{A_\text{plate}}$$. The displacement current between the plates can be found using the formula $$I_d = \varepsilon_0 \kappa \frac{d\phi_E}{dt}$$.

### Explanation:

To find the time rate of change of electric flux between the plates of the capacitor, we can use the formula: $$\frac{d\phi_E}{dt} = \frac{I}{A_\text{plate}}$$, where $$\frac{d\phi_E}{dt}$$ is the time rate of change of electric flux, $$I$$ is the current, and $$A_\text{plate}$$ is the area of one plate. In this case, the area of each plate is $$(0.06 \,\text{m})^2$$ and the current is 0.134 A. Thus, the time rate of change of electric flux is $$\frac{0.134 \,\text{A}}{(0.06 \,\text{m})^2}$$ V·m/s.

The displacement current between the plates of a capacitor can be found using the formula: $$I_d = \varepsilon_0 \kappa \frac{d\phi_E}{dt}$$, where $$I_d$$ is the displacement current, $$\varepsilon_0$$ is the vacuum permittivity, $$\kappa$$ is the dielectric constant, and $$\frac{d\phi_E}{dt}$$ is the time rate of change of electric flux. In this case, $$\varepsilon_0$$ is a constant, $$\kappa$$ depends on the material between the plates (not provided), and we found $$\frac{d\phi_E}{dt}$$ to be $$\frac{0.134 \,\text{A}}{(0.06 \,\text{m})^2}$$ V·m/s. So the displacement current can be calculated once these values are known.

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The time rate of change of electric flux can be found using I/ε₀A. The displacement current can be found using ε₀A(dE/dt)

### Explanation:

The time rate of change of electric flux between the plates of the capacitor can be found using the formula:

dΦ/dt = I/ε₀A

Where dΦ/dt is the time rate of change of electric flux, I is the current, ε₀ is the permittivity of free space, and A is the area of one of the plates.

We are given that the current is 0.134 A and the area of each plate is (0.060 m)² = 0.0036 m². Plugging these values into the equation, we get: the time rate of change of electric flux between the plates is 37.22 V·m/s.

Similarly, the displacement current between the plates can be found using the formula:

Id = ε₀A(dE/dt)

Where Id is the displacement current, ε₀ is the permittivity of free space, A is the area of one of the plates, and dE/dt is the time rate of change of electric field intensity between the plates.

We are given that ε₀ is 8.854 × 10⁻¹² F/m and dΦ E/ dt is 37.22 V·m/s. Plugging these values into the equation, we get:

Id = (8.854 × 10⁻¹² F/m)(37.22 V·m/s) = 3.29 × 10⁻¹⁰ A

Therefore, the displacement current between the plates is 3.29 × 10⁻¹⁰ A.

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A water-skier is being pulled by a tow rope attached to a boat. As the driver pushes the throttle forward, the skier accelerates. A 76.2-kg water-skier has an initial speed of 5.0 m/s. Later, the speed increases to 10.4 m/s. Determine the work done by the net external force acting on the skier.

Work done will be equal to 3186.396 J

Explanation:

We have mass m = 76.2 kg

Initial velocity u = 5 m/sec

Final velocity v = 10.4 m/sec

We have to find the work done

From work energy theorem work done is equal to change in kinetic energy

w = 3168.396 J

So work done will be equal to 3186.396 J

An airplane is traveling 835 km/h in a direction 41.5 ∘ west of north. Find the components of the velocity vector in the northerly and westerly directions. How far north and how far west has the plane traveled after 2.20 h ?

I assume the graph is looking like in the picture bellow.

North component:
cos(41.5) * 835 = 625.37 km/h

West component of speed:
sin(41.5) * 835 = 553.29 km/h

After 2.2 hours plane will fly:
2.2*625.37 = 1375.81 km north
2.2*553.29 = 1217.23 km  west

To find the components of the velocity vector, you can use trigonometry. The north component is calculated using the sine function and the west component is calculated using the cosine function. After 2.20 hours, the distance traveled north and west can be found by multiplying the velocity components by the time.

### Explanation:

To find the components of the velocity vector in the northerly and westerly directions, we can use trigonometry. The velocity vector is 835 km/h and is traveling in a direction 41.5° west of north. To find the north component, we can use the sine function: North component = velocity * sin(angle). To find the west component, we can use the cosine function: West component = velocity * cos(angle).

After 2.20 hours, we can find the distance traveled north and west by multiplying the velocity components by the time: Distance north = North component * time and Distance west = West component * time.

Let's calculate the values:

1. North component = 835 km/h * sin(41.5°)
2. West component = 835 km/h * cos(41.5°)
3. Distance north = North component * 2.20 h
4. Distance west = West component * 2.20 h