A satellite that weighs 4900 N on the launchpad travels around the earth's equator in a circular orbit with a period of 1.667 h.

The light's wavelength absorbed during the transition is 459 nm. Energy difference between the 5-d and the 6-s sub-levels in gold is expressed as ΔE. Let the wavelength associated with the electron's transition from the 5-d to the 6-s state be λ. The relationship that describes the connection between energy and wavelength is defined as: E = hc/λ, where E stands for photon energy, h represents Planck's constant, c is the speed of light, and λ denotes the wavelength of the photon. Therefore, the absorption wavelength in this transition stands at 459 nm.

To tackle this question, we know the following:

1 Albert equals 88 meters.

1 A = 88 m.

Initially, we square both sides of the equation:

(1 A)^2 = (88 m)^2

1 A^2 = 7,744 m^2

<span>Since 1 acre equals 4,050 m^2, let’s divide both sides by 7,744 to find out how many acres match this value:</span>

1 A^2 / 7,744 = 7,744 m^2 / 7,744

(1 / 7,744) A^2 = 1 m^2

Then multiply both sides by 4,050.

(4050 / 7744) A^2 = 4050 m^2

0.523 A^2 = 4050 m^2

<span>Thus, one acre is approximately 0.52 square alberts.</span>

Answer:

b) TA = TB = TC

Explanation:

• When the blocks are brought into contact and isolated from the environment, they will exchange heat until they achieve thermal equilibrium.
• During this exchange, the hotter body will lose heat, which will be gained by the cooler body.
• The equilibrium state will be established once this equation is satisfied:

       

• Substituting the initial temperatures T₀A = 300º C and T₀B = 400ºC, while simplifying for equal block masses mA = mB, enables us to solve for the final temperature, Tfin:

       

• At equilibrium, when both blocks combine, they will yield a uniform final temperature of 350ºC.
• When block C, also at this temperature, makes contact, all three blocks will simultaneously reflect this final temperature of 350 ºC.
• Therefore, option b) is correct.

Answer:

The overall length of the spiral, designated as L, is calculated to be 5378.01 m

Explanation:

Provided information:

Inner radius R1=2.5 cm

and outer radius R2= 5.8 cm.

The thickness of the spiral winding is (d) =1.6 \mu m = 1.6x 10^{-6} m

The total length of the spiral can be computed as

= 5378.01 m

The distinction lies in the fact that the candle emits an emission spectrum, while the book reflects an absorption spectrum. In the case of the book, light is observed from all directions, causing its reflection to be diffuse. Explanation: The light emitted by a candle originates from the heat of the flame, composed of a combination of emissions from a black body at that temperature along with emissions from the chemical elements within the candle. On the contrary, the light reflected off a book cover consists of the incident light spectrum minus the wavelengths that trigger electronic transitions in the cover's elements, resulting in dark areas on the spectrum. Consequently, the difference stems from the candle producing an emission spectrum, whereas the book showcases an absorption spectrum. For a book's cover to reflect light specularly, incident rays would need to reflect uniformly, creating dark areas. However, since light is observed from all directions when reflecting off a book, the result is diffuse reflection.