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4 years ago

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Explain how VSEPR theory predicts the shapes of molecules. Match the words in the left column to the appropriate blanks in the s

entences on the right. Make certain each sentence is complete before submitting your answer.

1 answer:

vlabodo [156]4 years ago

8 0

Answer:

The VSEPR theory and how it predicts the shapes of molecules:

Explanation:

The Valence Shell Electron Pair Repulsion (VSEPR) theory is a model used in chemistry to predict the shapes of individual molecules by the number of electron pairs that they have in the center of the atom. This theory is also based on the notion that the electrons around the atom repel one another. The Valence electrons on the outermost layer of the molecule are the most important in defining the geometry as they are the first to interact with other atoms and will be involved in bonding.

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A weight lifter is trying to do a bicep curl with a weight of 300 N. At the "sticking point", the moment arm of this weight is 3

lesantik [10]

Answer:

The weight lifter would not get past this sticking point.

Explanation:

Generally torque applied on the weight is mathematically represented as

T = F z

To obtain Elbow torque we substitute 4000 N for F (the force ) and 2cm $= \frac{2}{100} = 0.02m$ for z the perpendicular distance

So Elbow Torque is $T_e= 4000 * 0.02$

$= 80Nm$

To obtain the torque required we substitute 300 N for F and 30cm $=\frac{30}{100} = 0.3 m$

So the Required Torque is $T_R = 300 *0.3$

$=90Nm$

Now since $T_e < T_R$ it mean that the weight lifter would not get past this sticking point

7 0

3 years ago

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km fr

mestny [16]

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.

What is the tension between her ears?

Would the astronaut find it difficult to keep from being torn apart by the gravitational forces?

Answer:

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

Explanation:

Given that:

Orbital radius of the spacecraft (R) = 120 Km = 120 × 10³ m

Mass of the black hole (m) = $5 \ * (M \ _{sun})$

where : $M_{sun} = 1.99*10^{33} \ kg$

Then; we have:

$m = 5*(1.99*10^{30} \ kg ) \\ = 9.95*10^{30} kg$

Schwarzchild radius of the black hole

r - 15.0 km

Mass of each ear $m_{ear} = 0.030 \ kg$

Farther distance between one ear and the black hole (d) = 6.0 cm

= 0.06 m

Closer distance between the other ear and the black home is (d) 6.0 cm

= 0.6 cm

NOW, If we assume that the tension force should be T; then definitely the two ears will posses the same angular velocity .

The net force on the ear closer to the black hole will be:

$\frac{GMm_{ear} }{(R-d)}- T = m_{ear} (R - d) \omega^2$

$\frac{GMm_{ear} }{(R-d)^2}- \frac{T}{(R-d)} = m_{ear} \omega^2 \ ----> \ (1)$

The net force on the ear farther to the black hole is :

$\frac{GMm_{ear} }{(R+d)}- T = m_{ear} (R + d) \omega^2$

$\frac{GMm_{ear} }{(R+d)^2}- \frac{T}{(R+d)} = m_{ear} \omega^2 \ ----> \ (2)$

Equating equation (1) and (2) & therefore making (T) the subject of the formula; we have:

$T = \frac{3GMm_{ear}d}{R^3}$

$T = \frac{3(6.67*10^{-11}N.m^2/kg^2)(1.95*10^{30}kg)(0.03kg)(0.06m)}{(120*10^3m)^3}$

$T = 2073.9 N\\T = 2.07 KN$

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

3 0

4 years ago

What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway? a plane accelerates from

ElenaW [278]

When is at the end of the runway the velocity of the plane is given by the equation $vf^{2}=0+2*a*s$ where s=1800 m is the runway length. Thus
$vf^{2}=2*5*1800=18000 (m/s)^{2}$
$vf =134.164 (m/s)$

At half runway the velocity of the plane is
$v^{2}=2*5* \frac{1800}{2}=9000 ( \frac{m}{s} )^{2}
$
$v= \sqrt{9000}=94.87 ( \frac{m}{s})$

Therefore at midpoint of runway the percentage of takeoff velocity is
‰ $P= \frac{v}{vf}= \frac{94.87}{134.164}=0.707$

6 0

3 years ago

Electricity does not commonly occur in nature except in the form of ( hydrocarbons, photosyntheis, lightning, thunder)

horrorfan [7]

Hydrocarbons are compounds of hydrogen and carbon.
Photosynthesis is a process of nutrition in plants.
Thunder is a sound.
Lightning and electric eels are electrical phenomena in nature.

7 0

4 years ago

Read 2 more answers

The newly formed xenon nucleus is left in an excited state. Thus, when it decays to a state of lower energy a gamma ray is emitt

nevsk [136]

Answer:3.87*10^-4

Explanation:

What is the decrease in mass, delta mass Xe , of the xenon nucleus as a result of this deca

We have been given the wavelength of the gamma ray, find the frequency using c = freq*wavelength.

C=f*lambda

3*10^8=f*3.44*10^-12

F=0.87*10^20 hz

Then with the frequency, find the energy emitted using equation

E=hf E = freq*Plank's constant

E=.87*10^20*6.62*10^-34

E=575.94*10^(-16)

With this energy, convert into MeV from joules.

With the energy in MeV, use E=mc^2 using c^2 = 931.5 MeV/u.

Plugging and computing all necessary numbers gives you

3.87*10^-4 u.

6 0

3 years ago