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3 years ago

Write a reaction equation to show HCO3 acting as an acid

Physics

1 answer:

aniked [119]3 years ago

6 0

Answer: H2CO3

Explanation:

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Un the way to the moon, the Apollo astro-

kherson [118]

Answer:

Distance = 345719139.4[m]; acceleration = 3.33*10^{19} [m/s^2]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

<u>Second part</u>

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The distance between the Earth and this point is calculated as follows:

re = 3.84 108 - 38280860.6 = 345719139.4[m]

Now the acceleration can be found as follows:

$a = G*\frac{m_{e} }{r_{e} ^{2} } \\a = 6.67*10^{11} *\frac{5.98*10^{24} }{(345.72*10^{6})^{2} } \\a=3.33*10^{19} [m/s^2]$

6 0

4 years ago

Vector B has x, y, and z components 4, 6, and 3, respectively.

nalin [4]

Bbbbbbbbbbbbbb gygffdd

8 0

3 years ago

A rubber ball with a mass of 0.145 kg is dropped from rest. From what height (in m) was the ball dropped, if the magnitude of th

Elena-2011 [213]

Answer:

1.55 m

Explanation:

Momentum: This can be defined as the product of mass of a body and it velocity. the S.I unit of momentum is kgm/s.

Mathematically,

Momentum can be represented as,

M = mv................................. Equation 1

Where m = mass of the body, v = velocity of the body, M = momentum.

Making v the subject of the equation,

v = M/m........................................... Equation 2

Given: M = 0.80 kg.m/s, m = 0.145 kg.

Substituting into equation 2,

v = 0.8/0.145

v = 5.52 m/s.

Using the equation of motion,

v² = u² + 2gs ....................... Equation 3.

Where v = final velocity of the rubber ball, u = initial velocity of the rubber ball, s = distance, g = acceleration due to gravity.

Given: v = 5.52 m/s, u = 0 m/s, g = 9.81 m/s².

Substituting into equation 2

5.52² = 0² + 2(9.81)s

30.47 = 19.62s

s = 30.47/19.62

s = 1.55 m.

Thus the ball was dropped from a height of 1.55 m

8 0

4 years ago

What is the escape velocity from a planet with a mass that is 10.8% Earths mass and a size that is 53% Earths radial size?

DENIUS [597]

The escape velocity from the planet is 5052 m/s

Explanation:

The formula to calculate the escape velocity from a planet is:

$v=\sqrt{\frac{2GM}{R}}$

where

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

M is the mass of the planet

R is the radius of the planet

For Earth,

$M_E=5.98\cdot 10^{24} kg$ is the mass

$R_E=6.37\cdot 10^6 m$ is the radius

Here we have a planet that:

- its mass is 10.8% of that of Earth, so

$M=0.108M_E$

- its radius is 53% of that of Earth, so

$R=0.53 R_E$

Substituting everything into the first equation, we find the escape velocity from this planet:

$v=\sqrt{\frac{2G(0.108M_E)}{0.53R_E}}=\sqrt{\frac{2(6.67\cdot 10^{-11})(0.108)(5.98\cdot 10^{24})}{(0.53)(6.37\cdot 10^6)}}=5052 m/s$

#LearnwithBrainly

4 0

3 years ago

An object with a density of 0.85 g/cc is dropped into each of the two beakers shown below. Beaker 1 has a density of 0.5 g/cc. B

Oliga [24]

Answer:

The object will sink in the liquid in beaker 1.

The object will float in the liquid in beaker 2

Explanation:

The density of an object relative to the density of a fluid determines if the object floats or sink in a fluid. The density of a material is the measure of the amount of mass of that material packed into a unit volume of that material.

For the beaker 1, the liquid in this beaker has a density of 0.5 g/cc, which is lesser than the density of the object (0.85 g/cc). This means that the object will add more mass than there should be to the volume of the space it displaces within the field. This results in the object sinking in the fluid.

For beaker 2, the liquid in this beaker has a density of 1 g/cc, which is more than the density of the object (0.85 g/cc). This means that the object will add less mass than there should be to the volume of the space it displaces within the field. This results in the object floating in the fluid.

8 0

4 years ago

Write a reaction equation to show HCO3 acting as an acid​

Write a reaction equation to show HCO3 acting as an acid