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3 years ago

What is the sequence of events for the evolution of the atmosphere in the correct order

Physics

1 answer:

Shtirlitz [24]3 years ago

8 0

Answer:

Big bang theory

Explanation:

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When the air resistance can be ignored the velocity of an object dropped initially from rest is given by the following equation

ad-work [718]

Answer:

I am confused of your question. Do you want final velocity? To get final velocity, use (initial V)+(Gravity*Time)

Explanation:

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4 years ago

Read 2 more answers

An artist working on a piece of metal in his forging studio plunges the hot metal into oil in order to harden it. The metal piec

lutik1710 [3]

Answer:

The temperature of the metal is $T_m = 376.8 ^o C$

Explanation:

From the question we are told that

The mass of the metal is $M = 60 \ kg$

The specific heat of the metal is $c_p = 0.1027 kcal/(kg \cdot ^oC)$

The mass of the oil is $M_o = 810 \ kg$

The temperature of the oil is $T_o = 35^oC$

The specific heat of oil is $c_o = 0.7167 kcal/(kg \cdot ^oC )$

The equilibrium temperature is $T_e = 39 ^oC$

According to the law of energy conservation

Heat lost by metal = heat gained by the oil

The quantity of heat lost by the metal is mathematically represented as

$Q = - Mc_p \Delta T$

=> $Q = -Mc_p (T_m - T_c)$

Where $T_ m$ the temperature of metal before immersion

The negative sign show heat lost

The quantity of gained t by the metal is mathematically represented as

$Q = M_o c_o \Delta T$

=> $Q = M_o c_o (T_c - T_o)$

$Mc_p (T_m - T_c) = M_o c_o (T_c - T_o)$

substituting values

$- 60 * 0.1027 (T_m - 39) = 810 * 0.7167 * (39 - 35)$

=> $T_m = 376.8 ^o C$

6 0

3 years ago

POSSIBLE POINTS: 1.92

gogolik [260]

Answer:

jnfal4u4ryhfsbjls5

Explanation:

duehdakjweyedufkbshegygfr

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3 years ago

Please help with this :(

Angelina_Jolie [31]

Answer:

I'm sorry I don't have a answer but I like your pfp

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3 years ago

Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has

faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

<u>Given:</u>

Charge on first charged particle, $q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.$
Charge on the second charged particle, $q_2=3.80\ nC=3.80\times 10^{-9}\ C.$

Position of the first charge = $(x_1=0.00\ m,\ y_1=0.600\ m).$
Position of the second charge = $(x_2=1.50\ m,\ y_2=0.650\ m).$

The electric field at a point due to a charge $q$ at a point $r$ distance away is given by

$\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.$

where,

$k$ = Coulomb's constant, having value $\rm 8.99\times 10^9\ Nm^2/C^2.$

$\vec r$ = position vector of the point where the electric field is to be found with respect to the position of the charge $q$ .

$\hat r$ = unit vector along $\vec r$ .

The electric field at the origin due to first charge is given by

$\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.$

$\vec r_1$ is the position vector of the origin with respect to the position of the first charge.

Assuming, $\hat i,\ \hat j$ are the units vectors along x and y axes respectively.

$\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.$

Using these values,

$\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.$

The electric field at the origin due to the second charge is given by

$\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.$

$\vec r_2$ is the position vector of the origin with respect to the position of the second charge.

$\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.$

Using these values,

$\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\ N/C.$

The net electric field at the origin due to both the charges is given by

$\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.$

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

4 0

3 years ago