All categories

2 years ago

Why does the moon appear in the daytime

Physics

1 answer:

Brut [27]2 years ago

4 0

Answer:

You can see the moon when it is in the perfect spot and it's reflecting the right amount of light. :)

Hope this helps! If you don't mind, please mark as brainliest! Thx!

You might be interested in

Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.

Leni [432]

Answer:

$T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}$

Explanation:

The given data :-

Diameter of rod AB ( d₁ ) = 200 mm.

Diameter of rod BC ( d₂ ) = 150 mm.

The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C

The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
Consider the required temperature increase to close the gap at C = T °C
Consider the change in length of the rod = бL

Solution :-

$\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}$

$\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}$

5 0

3 years ago

The barometric pressure at sea level is 30 in of mercury when that on a mountain top is 29 in. If the specific weight of air is

stealth61 [152]

To solve this problem we will apply the concepts related to pressure, depending on the product between the density of the fluid, the gravity and the depth / height at which it is located.

For mercury, density, gravity and height are defined as

$\rho_m = 846lb/ft^3$

$g = 32.17405ft/s^2$

$h_1 = 1in = \frac{1}{12} ft$

For the air the defined properties would be

$\rho_a = 0.0075lb/ft^3$

$g = 32.17405ft/s^2$

$h_2 = ?$

We have for equilibrium that

$\text{Pressure change in Air}=\text{Pressure change in Mercury}$

$\rho_m g h_1 = \rho_a g h_2$

Replacing,

$(846)(32.17405)(\frac{1}{12}) = (0.0075)(32.17405)(h_2)$

Rearranging to find $h_2$

$h_2 = \frac{(846)(32.17405)(\frac{1}{12}) }{(0.0075)(32.17405)}$

$h = 9400ft$

Therefore the elevation of the mountain top is 9400ft

7 0

2 years ago

What are the unit for acceleration

Flura [38]

<h3>Answer</h3>

m/s^2 (meter per sec square)

Explanation:

acc = change in velocity/time

= distance/time

----------------

time

= m/s

------

=m/s^2

7 0

3 years ago

A beam of electrons is accelerated from rest through a potential difference of 0.400 kV and then passes through a thin slit. Whe

Oksi-84 [34.3K]

Answer:

v= 103.5 V; energy =1.65 x 10^-17

Explanation:

the deflected energy eV sin θ

8 0

2 years ago

Read 2 more answers

The mass of the Moon is 7.35 x 1022 kg, while that of Earth is 5.98 x 1024 kg. The average distance from the center of the Moon

Kryger [21]

Answer:

aaa

Explanation:

$m_e$ = Mass of the Earth = 5.98 × 10²⁴ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

$r_1$ = Distance from the center of the Moon to the center of Earth = 6371000 m

$r_2$ = Distance from the center of the earth center to sun center

$m_m$ = Mass of moon = $7.35\times 10^{22}\ kg$

M = Mass of sun = $1.989\times 10^{30}\ kg$

$F_1=G\frac{m_em_m}{r_1^2}\\\Rightarrow F_1=6.67\times 10^{-11}\frac{5.98\times 10^{24}\times 7.35\times 10^{22}}{(384000000)^2}\\\Rightarrow F_1=1.988\times 10^{20}\ N$

$F_2=G\frac{Mm_e}{r_1^2}\\\Rightarrow F_2=6.67\times 10^{-11}\frac{5.98\times 10^{24}\times 1.989\times 10^{30}}{(149.6\times 10^9+6371000+695.51\times 10^6)^2}\\\Rightarrow F_2=3.511\times 10^{22} N$

$\frac{F_1}{F_2}=\frac{1.988\times 10^{20}}{3.511\times 10^{22}}\\\Rightarrow \frac{F_1}{F_2}=0.00566\\\Rightarrow F_1=F_20.00566$

Hence the force of moon on earth is 0.00566 times the force of earth on moon center to center

4 0

3 years ago