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11 days ago

The Sun orbits the center of the Milky Way galaxy once each 2.60 × 108 years, with a roughly circular orbit averaging 3.00 × 104

light years in radius. (A light year is the distance traveled by light in 1 y.)
Calculate the centripetal acceleration of the Sun in its galactic orbit. Does your result support the contention that a nearly inertial frame of reference can be located at the Sun?

Physics

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The magnitude of the Poynting vector of a planar electromagnetic wave has an average value of 0.939 W/m2. The wave is incident u

Yuliya22 [3333]

Answer:

The total energy can be expressed as $T = 169.02 \ J$

Explanation:

The problem states that

The Poynting vector, which measures energy flux, equals $k = 0.939 \ W/m^2$

The rectangle's length is represented by $l = 1.5 \ m$

The width of the rectangle is $w = 2.0 \ m$

The duration considered is $t = 1 \ minute = 60 \ s$

Mathematically, the overall electromagnetic energy incident on the area is given by

$T = k * A * t$

where A denotes the area of the rectangle, calculated as

$A= l * w$

By plugging in the respective values

$A= 2 * 1.5$

$A= 3 \ m^2$

Again substituting values

$T = 0.939 * 3 * 60$

$T = 169.02 \ J$

5 0

2 months ago

If a steady-state heat transfer rate of 3 kW is conducted through a section of insulating material 1.0 m2 in cross section and 2

Maru [3345]

Answer:

$\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K$

Consequently, the temperature difference across the material will be $\Delta T = 375 K$

Explanation:

In this case, we apply the Fourier Law of heat conduction expressed by the following equation:

$Q = -kA \frac{\Delta T}{\Delta x}$ (1)

Where k = thermal conductivity = 0.2 W/ mK

A= 1m^2 denotes the cross-sectional area

Q= 3KW signifies the heat transfer rate

$\Delta T$ is the temperature difference we need to determine

represents the thickness of the material $\Delta x=2.5 cm =0.025 m$

To isolate $\Delta T$ from equation (1), we obtain:

$\Delta T =\frac{Q \Delta x}{Ak}$

Initially, we convert 3KW to W, resulting in:

$Q= 3 KW* \frac{1000W}{1 Kw}= 3000 W$

With all variables accounted for, we can substitute and calculate:

$\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K$

Thus, the temperature difference across the material will be $\Delta T = 375 K$

5 0

1 month ago

A robot probe drops a camera off the rim of a 239 m high cliff on mars, where the free-fall acceleration is −3.7 m/s2 .

Maru [3345]

<span>a. To determine the velocity at which the camera strikes the ground: v^2 = (v0)^2 + 2ay = 0 + 2ay v = sqrt{ 2ay } v = sqrt{ (2)(3.7 m/s^2)(239 m) } v = 42 m/s The camera impacts the ground with a speed of 42 m/s. b. To calculate the duration it takes for the camera to reach the bottom: y = (1/2) a t^2 t^2 = 2y / a t = sqrt{ 2y / a } t = sqrt{ (2)(239 m) / 3.7 m/s^2 } t = 11.4 seconds

The camera descends for 11.4 seconds before hitting the ground.</span>

3 0

2 months ago