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Assoli18 [71]
3 years ago
7

If the mass of the Earth somehow increased with no change in radius, your weight would

Physics
2 answers:
Margaret [11]3 years ago
5 0

Answer:

increase also

Explanation:

The weight of a person is equal to the gravitational pull exerted by the Earth on the person:

F=G\frac{mM}{R^2}

where

G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2} is the gravitational constant

M is the mass of the Earth

m is the mass of the person

R is the Earth's radius

We notice that the weight is directly proportional to the mass of the Earth. Therefore, if the mass of the Earth M increases, and the radius R does not change, the weight of the person increases as well.

Sever21 [200]3 years ago
5 0

<u>Answer</u>

Increases also.

The force due to gravity is given by,

F = GM.m/r²

Where G is a constant of proportionality

∴ F ∝ M.m/r²

When r remains constant, force due to gravity, F, will be;

F ∝ M.m

Where M is the mass of the earth and m is your mass

Since your mass does not change, we are going to have;

F ∝ M.

This means the weight F is directly proportional to the mass of the earth. when it increases the your weight also increases.

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Answer: D. 2.6

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3 years ago
Imagina que compras una placa rectangular de metal de 2mm de alto, 10mm de ancho x 50mm de largo, y una masa de 0.02kg. El vende
shusha [124]

Answer:

Densidad de la placa = 20 g/cm³.

La placa no es de oro.

Explanation:

Para encontrar la densidad de la placa rectangular primero debemos hallar su volumen:

V = 2 mm*10 mm*50 mm = 1000 mm^{3}*\frac{1 cm^{3}}{(10 mm)^{3}} = 1 cm^{3}      

Ahora, encontremos al densidad de la placa:

d = \frac{m}{V} = \frac{20 g}{1 cm^{3}} = 20 g/cm^{3}

Dado que la densidad del oro es 19.32 g/cm³ y que la densidad de la placa rectangular calculada es 20 g/cm³, podemos decir que dicha placa no es de oro.                  

Espero que te sea de utilidad!

6 0
3 years ago
At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2(g)+O2(g)→2SO3(g) At equilibrium, the partial
elena55 [62]

Answer : The partial pressure of SO_3 is, 67.009 atm

Solution :  Given,

Partial pressure of SO_2 at equilibrium = 30.6 atm

Partial pressure of O_2 at equilibrium = 13.9 atm

Equilibrium constant = K_p=0.345

The given balanced equilibrium reaction is,

2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)

The expression of K_p will be,

K_p=\frac{(p_{SO_3})^2}{(p_{SO_2})^2\times (p_{O_2})}

Now put all the values of partial pressure, we get

0.345=\frac{(p_{SO_3})^2}{(30.6)^2\times (13.9)}

p_{SO_3}=67.009atm

Therefore, the partial pressure of SO_3 is, 67.009 atm

6 0
3 years ago
Two facing surfaces of two large parallel conducting plates separated by 8.5 cm have uniform surface charge densities such that
elena-s [515]

Answer:

positive plate

E = 5.764 KV / m

W = 490eV or 7.85 * 10^-17 J

E_p = 4.74 *10^(-12) eV

E_k = 490 eV

Explanation:

part a

The potential difference between two plates = 490 V

Distance between two plates = 8.5 cm

Answer: The positive plate is at higher potential because of convention.

part b

Electric Field between the plates

E = V / d

E = 490 / 0.085 = 5.764 KV / m

Answer: Electric Field between the plates E = 5.764 KV / m

part c

Work done by electric field

W = V*q

W = 490 * 1.602*10^-19

W = 7.85 * 10^-17 J

or W = 490 eV

Answer: Work done by electric field W = 490eV or 7.85 * 10^-17 J

part d

Potential Energy of an electron gained:

E_p = m_e * g * d / (1.602*10^-19)

E_p =  9.109*10^-31* 9.81 * 0.085 / (1.602*10^-19)

E_p = 4.74 *10^(-12) eV

Very very small E_p approximately 0

Answer: Potential Energy of an electron gained E_p = 4.74 *10^(-12) eV or 0.

part e

Kinetic Energy of an electron gained:

W - E_p = E_k

E_k = 490eV - 4.74*10^(-12)eV

E_k = 490 eV

Answer: Kinetic Energy of an electron gained E_k = 490 eV

7 0
3 years ago
Which of the following is incorrect for nuclear forces
densk [106]

Answer:

c they obey inverse square law

4 0
3 years ago
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