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ankoles [38]
2 years ago
3

Draw the resonance structure that would result from the movement of electrons shown by the fishhook notation. Include all nonbon

ding electrons.

Physics
1 answer:
lara [203]2 years ago
7 0

Answer:

PLZ BRINLY ME

Explanation:

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You and your friend throw balloons filled with water from the roof of a several story apartment house. You simply drop a balloon
Aleks [24]

Answer:

Height = 53.361 m

Explanation:

There are two balloons being thrown down, one with initial speed (u1) = 0 and the other with initial speed (u2) = 43.12

From the given information we make the following summary

u_{1} = 0m/s

t_{1} = t

u_{2} = 43.12m/s

t_{2} = (t-2.2)s

The distance by the first balloon is

D = u_{1} t_{1}  + \frac{1}{2} at_{1}^2

where

a = 9.8m/s2

Inputting the values

D = (0)t + \frac{1}{2} (9.8)t^2\\ D = 4.9t^2

The distance traveled by the second balloon

D = u_{2} t_{2}  + \frac{1}{2} at_{2}^2

Inputting the values

D = (43.12)(t-2.2)  + \frac{1}{2} (9.8)(t-2.2)^2

simplifying

D = 4.9t^2 + 21.56t -71.148

Substituting D of the first balloon into the D of the second balloon and solving

4.9t^2 = 4.9t^2 + 21.56t -71.148 \\ 21.56t = 71.148\\ t = 3.3s

Now we know the value of t. We input this into the equation of the first balloon the to get height of the apartment

D = 4.9(3.3)^2\\ D = 53.361 m

7 0
2 years ago
A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size
Margarita [4]

Answer:

The final size is approximately equal to the initial size due to a very small relative increase of 1.055\times 10^{- 7} in its size

Solution:

As per the question:

The energy of the proton beam, E = 250 GeV =250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J

Distance covered by photon, d = 1 km = 1000 m

Mass of proton, m_{p} = 1.67\times 10^{- 27} kg

The initial size of the wave packet, \Delta t_{o} = 1 mm = 1\times 10^{- 3} m

Now,

This is relativistic in nature

The rest mass energy associated with the proton is given by:

E = m_{p}c^{2}

E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J

This energy of proton is \simeq 250 GeV

Thus the speed of the proton, v\simeq c

Now, the time taken to cover 1 km = 1000 m of the distance:

T = \frac{1000}{v}

T = \frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s

Now, in accordance to the dispersion factor;

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}

Thus the increase in wave packet's width is relatively quite small.

Hence, we can say that:

\Delta t_{o} = \Delta t

where

\Delta t = final width

3 0
2 years ago
A lens of focal length 15.0 cm is held 10.0 cm from a page (the object ). Find the magnification .
nevsk [136]

Answer:

Magnification, m = 3

Explanation:

It is given that,

Focal length of the lens, f = 15 cm

Object distance, u = -10 cm

Lens formula :

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}

v is image distance

\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-10)}\\\\v=-30\ cm

Magnification,

m=\dfrac{v}{u}\\\\m=\dfrac{-30}{10}\\\\m=3

So, the magnification of the lens is 3.

3 0
2 years ago
An electric buzzer is activated, then sealed inside a glass chamber. When all of the air is pumped out of the chamber, how is th
Aleksandr-060686 [28]
The sound is increased because sound waves are in fact mech. waves which means the  that they can't travel through empty space and thus need a medium to travel through
4 0
2 years ago
The absolute pressure in water at a depth of 5m is read to be 145 kPa. Determine (a) the local atmospheric pressure, and (b) the
irga5000 [103]

Answer:

a) 95950 pascals

b) 137642.5 pascals

Explanation:

The absolute pressure (Pabs) on a fluid is:

P_{abs}=P_{gauge}+P_{atm} (1)

With Pgauge the pressure due depth on the fluid and Patm the atmospheric pressure. Pgauge is equal to:

P_{gauge}=\rho gh (2)

with ρ the fluid density, g the gravitational acceleration and h the depth on the fluid. Using (2) on (1) and solving for Patm:

P_{atm}=P_{abs}-P_{gauge}=P_{abs}-\rho_{water} gh

P_{atm}=(145000Pa)-(1000\frac{kg}{m^{3}})(9.81\frac{m}{s^{2}})(5m)

P_{atm}=95950Pa

b) Here we're going to use again (1) but now we have another value of density because it's other liquid, to know that value we should use the fact that specific gravity (S.G) for liquids is the ratio between fluid density and water density:

S.G=\frac{\rho_{fluid}}{\rho_{water}}

\rho_{liquid}=S.G*\rho_{water}

\rho_{liquid}=(0.85)*(1000\frac{kg}{m^{3}})=850\frac{kg}{m^{3}}

so:

P_{abs}=\rho_{liquid} gh+P_{atm}=(850\frac{kg}{m^{3}})(9.81\frac{m}{s})(5m)+95950Pa

P_{abs}=137642.5 Pa

3 0
2 years ago
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