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3 years ago

11

Please solve this worksheet for me It’s very important please don’t write unnecessary things or else I’ll report please help me

i request

1 answer:

Sonja [21]3 years ago

3 0

Answer:

1. 231.25 mL

2. 218.75 kPa

3. 6.21 L

4. 427.5 mL

5. 83.01 mL

6. 307.6 mL

7. 1.15 L

8. 440.4 mL

Explanation:

This questions describes Boyle's law. Using Boyle's law equation to solve the questions as follows:

P1V1 = P2V2

Where;

P1 = initial pressure

P2 = final pressure

V1 = initial volume

V2 = final volume

Please find the answers to each question as an attachment.

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The space shuttle releases a satellite into a circular orbit 630 km above the Earth.

solniwko [45]

Answer:

7,539 m/s

Explanation:

Let's use this equation to find the gravitational acceleration of this space shuttle:

$\displaystyle g=\frac{GM}{r^2}$

We know that G is the gravitational constant: 6.67 * 10^(-11) Nm²/kg².

M is the mass of the planet, which is Earth in this case: 5.972 * 10^24 kg.

r is the distance from the center of Earth to the space shuttle: radius of Earth (6.3781 * 10^6 m) + distance above the Earth (630 km → 630,000 m).

Plug these values into the equation:

$\displaystyle g=\frac{(6.67\cdot 10^-^1^1 \ Nm^2kg^-^2)(5.972\cdot 10^2^4 \ kg)}{[(6.3781\cdot 10^6 \ m)+(630000 \ m)]^2}$

Remove units to make the equation easier to read.

$\displaystyle g=\frac{(6.67\cdot 10^-^1^1 )(5.972\cdot 10^2^4 )}{[(6.3781\cdot 10^6)+(630000 )]^2}$

Multiply the numerator out.

$\displaystyle g=\frac{(3.983324\cdot 10^1^4)}{[(6.3781\cdot 10^6)+(630000 )]^2}$

Add the terms in the denominator.

$\displaystyle g=\frac{(3.983324\cdot 10^1^4)}{[(7008100)]^2}$

Simplify this equation.

$\displaystyle g=8.11045189 \ \frac{m}{s^2}$

The acceleration due to gravity g = 8.11045189 m/s². Now we use the equation for acceleration for an object in circular motion which contains v and r.

$\displaystyle a = \frac{v^2}{r}$

a = g, v is the velocity that the space shuttle should be moving (what we are trying to solve for), and r is the radius we had in the previous equation when solving for g.

Plug these values into the equation and solve for v.

$\displaystyle 8.11045189 \ \frac{m}{s^2} = \frac{v^2}{7008100 \ m}$

Remove units to make the equation easier to read.

$\displaystyle 8.11045189 = \frac{v^2}{7008100}$

Multiply both sides by 7,008,100.

$56838857.89=v^2$

Take the square root of both sides.

$v=7539.154985$

The shuttle should be moving at a velocity of about 7,539 m/s when it is released into the circular orbit above Earth.

5 0

3 years ago

Please select the word from the list that best fits the definition Visualizes where items are located

Amanda [17]

Answer:I’m pretty sure it’s spatial

Explanation:

3 0

3 years ago

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8-30-2018 what phase change occurs as the substance at 0 degree c has its pressure dropped from 0.50 at to 0.25 atm

madam [21]

Answer:

at T = 0ºC the change of state is from the solid state to the gaseous state

Explanation:

In this exercise we are asked about the changes of state, from the data we will assume that the material is water.

Water can exist in three solid states, liquid and gas, in a graph of pressure ℗ against temperature (T) there is a point called triple at T = 0.01ºC, below this point the curve has two states at high pressure solid and low pressure gas.

As a result of the previous ones at T = 0ºC the change of state is from the solid state to the gaseous state

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3 years ago

In the model of the hydrogen atom due to Niels Bohr, the electron moves around the proton at a speed of 3.3 × 106 m/s in a circl

irga5000 [103]

Answer:

$1.5048\times 10^{-23}\ Am^2$

Explanation:

q = Charge of proton = $1.6\times 10^{-19}\ C$

r = Radius of circle = $5.7\times 10^{-11}\ m$

v = Velocity of proton = $3.3\times 10^6\ m/s$

Magnetic moment is given by

$M=\frac{1}{2}qrv\\\Rightarrow M=\frac{1}{2}1.6\times 10^{-19}\times 5.7\times 10^{-11}\times 3.3\times 10^6\\\Rightarrow M=1.5048\times 10^{-23}\ Am^2$

The magnetic moment associated with this motion is $1.5048\times 10^{-23}\ Am^2$

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3 years ago

Imagine that an electron in an excited state in a nitrogen molecule decays to its ground state, emitting a photon with a frequen

mash [69]

Since energy cannot be created nor destroyed, the change in energy of the electron must be equal to the energy of the emitted photon.

The energy of the emitted photon is given by:
$E=hf$
where
h is the Planck constant
f is the photon frequency
Substituting $f=8.88 \cdot 10^{14}Hz$ , we find
$E=hf=(6.6 \cdot 10^{-34} Js)(8.88 \cdot 10^{14} Hz)=5.86 \cdot 10^{-19} J$

This is the energy given to the emitted photon; it means this is also equal to the energy lost by the electron in the transition, so the variation of energy of the electron will have a negative sign (because the electron is losing energy by decaying from an excited state, with higher energy, to the ground state, with lower energy)
$\Delta E= -5.86 \cdot 10^{-19} J$

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3 years ago

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