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3 years ago

Do you know when the next full moon will beç

Physics

1 answer:

FromTheMoon [43]3 years ago

4 0

The most technically precise answer to that question is "Yes".

The date that provides the actual information for the next 10 days after I write this is "October 24th".

The actual information for ANY day ... past, present, or future ... is

"29days 12hours 43minutes after the LAST Full Moon" .

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Identify the law, write the equation and calculate the answer to the problem below.

lyudmila [28]

Find refractive index first

$\\ \rm\Rrightarrow \mu=\dfrac{c}{v}$

$\\ \rm\Rrightarrow \mu=\dfrac{1.0003}{1.33}$

$\\ \rm\Rrightarrow \mu =0.75$

Now

$\\ \rm\Rrightarrow \dfrac{sini}{sinr}=\mu$

$\\ \rm\Rrightarrow \dfrac{sin45}{sinr}=0.75$

$\\ \rm\Rrightarrow \dfrac{sin45}{0.75}=sinr$

$\\ \rm\Rrightarrow sinr=0.94$

$\\ \rm\Rrightarrow r=sin^{-1}(0.94)$

$\\ \rm\Rrightarrow r=70^{\circ}$

5 0

2 years ago

Radiant heat makes it impossible to stand close to a hot lava flow. Calculate the rate of heat loss by radiation from 1.00 m^2 o

VARVARA [1.3K]

The rate of heat loss by radiation is equal to -207.5kW

Why?

To calculate the heat loss rate (or heat transfer rate) by radiation, from the given situation, we can use the following formula:

$HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )$

Where,

E, is the emissivity of the body.

A, is the area of the body.

T, are the temperatures.

S, is the Stefan-Boltzmann constant, which is equal to:

$5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }$

Now, before substitute the given information, we must remember that the given formula works with absolute temperatures (Kelvin), so, we need to convert the given values of temperature from Celsius degrees to Kelvin.

We know that:

$K=Celsius+273.15$

So, converting we have:

$T_{1}=1110\°C+273.15=1383.15K\\\\T_{2}=36.2\°C+273.15=309.35K$

Therefore, substituting the given information and calculating, we have:

$HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )$

$HeatLossRate=1*5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }*1m^{2} *((309.35K)^{4} -(1383.15})^{4} )\\\\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(95697.42K^{4} -3.66x10^{12}K^{4})\\ \\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(-3.66x10^{12} K^{4})=-207522W=-207.5kW$

Hence, we have that the rate of heat loss is equal to -207.5kW.

8 0

3 years ago

LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.

Svetradugi [14.3K]

Answer:

i) E = 269 [MJ] ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

$E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.$

$W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]$

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

$E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]$

8 0

3 years ago

Tenses of write= read=

PilotLPTM [1.2K]

Answer:

present

Explanation:

read doesn't change but write is in present tense

7 0

3 years ago

Help it’s actually physical science but plz help

Butoxors [25]

The correct answer is B

3 0

3 years ago