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Tems11 [23]
2 years ago
14

In the photo below, astronaut Alan Bean works at the Apollo 12 lander. Describe the horizon and the surface you see. What kind o

f terrain did they land on for this, the second human Moon landing, and why ?

Physics
1 answer:
Marianna [84]2 years ago
3 0

Answer:

moon

Explanation:

I dont KNow

You might be interested in
-
evablogger [386]

Answer:

b

Explanation:

3 0
3 years ago
A light ray travels in the +x direction and strikes a slanted surface with an angle of 62° between its normal and the ty axis. T
Elena-2011 [213]

Answer:

- 0.6

Explanation:

Given that angle between normal y axis is 62° so angle between  normal

and x axis will be 90- 62 = 28 °. Since incident ray is along x axis , 28 ° will be the angle between incident ray and normal ie it will be angle of incidence

Angle of incidence = 28 °

angle of reflection = 28°

Angle between incident ray and reflected ray = 28 + 28 = 56 °

Angle between x axis and reflected ray = 56 °

x component of reflected ray

= - cos 56 ( it will be towards - ve x axis. )

- 0.6

6 0
3 years ago
What were some benefits to the american television from analog broadcast to digital broadcast?
Elanso [62]

Answer:

nothing no no no

Explanation:

nothing cause no

8 0
3 years ago
Read 2 more answers
Suppose you wish to fabricate a uniform wire out of 1.10 g of copper. If the wire is to have a resistance R = 0.390 Ω, and if al
Citrus2011 [14]

To solve this problem we will apply the concepts related to volume, as a function of length and area, as of mass and density. Later we will take the same concept of resistance and resistivity, equal to the length per unit area. Once obtained from the known constants it will be possible to obtain the area by matching the two equations:

Mass of copper wire(m) = 1.10g = 1.10*10^{-3} kg

Density (\rho)= 8.92*10^3kg/m^3

Resistively of copper (\gamma) = 1.7*10^{-8}\Omega \cdot m

Resistance (R) = 0.390\Omega

Volume is defined as,

V= lA \text{ and }\frac{m}{\rho}

lA= \frac{1.10*10^{-3}}{8.92*10^3}

lA = 1.233*10^{-7} m^3 (1)

We know that,

\frac{l}{A} = \frac{R}{\gamma}

\frac{l}{A}= \frac{0.390\Omega}{1.7*10^{-8}\Omega m}

\frac{l}{A} = 2.2941*10^7 m^{-1} (2)

Multiplying equation we have

l^2 = (1.233*10^{-7})( 2.2941*10^7)

l^2 = 2.8286m^2

l =\sqrt{2.8286m^2}

l = 1.68m

Therefore the length of the wire is 1.68m

6 0
3 years ago
A damped mass/spring system takes 14.0 s for its amplitude of the oscillator to decrease by a factor of 9. By what factor does t
fiasKO [112]

Answer:

The correct answer is "0.246".

Explanation:

Given that the amplitude is decreased by a factor of 9, then

A \rightarrow (A-\frac{A}{9} )

A \rightarrow \frac{8A}{9}

As we know,

Energy will be:

⇒  E_{1}=\frac{1}{2}KA^2

and,

⇒  E_{2}=\frac{1}{2}K(\frac{8A}{9} )^2

          =\frac{64KA^2}{162}

⇒  \Delta E=E_1-E_2

On putting the estimated values, we get

           =\frac{1}{2}KA^2-\frac{64KA^2}{162}

⇒  \frac{\Delta E}{E}=\frac{\frac{20}{162}KA^2}{\frac{1}{2}KA^2}

          =\frac{40}{162}

          =0.246

3 0
3 years ago
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