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2 years ago

If we double only the amplitude of a vibrating ideal mass-and-spring system, the mechanical energy of the system:

Physics

1 answer:

Lelechka [254]2 years ago

8 0

Answer:

D. increases by a factor of 4.

Explanation:

General equation of SHM

Lets taken the general equation of the displacement given as

x = A sinω t

A=Amplitude ,t=time ,ω=natural frequency

We know that speed V

$V=\dfrac{dx}{dt}$

V= A ω cosωt

The mechanical energy of spring mass system

$U=\dfrac{1}{2}KA^2$

K=Spring constant

Now when Amplitude A become 2 times then the mechanical energy will become 4 times.

Therefore the answer is D.

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Which statement about electrons and atomic orbitals is NOT true?

balandron [24]

<em>An electron has the same amount of energy in all orbitals is not true</em>

$\boxed{\boxed{\bold{Further~explanation}}}$

In an atom there are levels of energy in the skin and sub skin.

This energy level is expressed in the form of electron configurations.

Writing electron configurations starts from the lowest to the highest sub-shell energy level.

So electrons that occupy the orbitals in the lowest sub-skin have the lowest energy level

In the principle of Pauli's prohibition it was stated that there are no two electrons in one atom that can have the same four quantum numbers.

So suppose that there are two electrons occupying one orbital can have the same main quantum number (n), azimuth (l) and magnetic (m), then the last quantum number that is the quantum spin number (s) must be different.

So that the two electrons are different from just the quantum spin number, even though the other quantum numbers are the same.

So in one orbital only a maximum of 2 electrons is occupied, because if there is a third electron, this third electron will have the same quantum spin number as the previous electron

The electron cloud is a visual representation of the location of electrons in an atom.

Orbital is the place around the nucleus where electrons may be found

Electron clouds show the state of electrons in their orbitals

So electron clouds can show the condition of all orbitals in an atom

The lowest energy level of an electron occupies a sub-skin of 1s which has only one orbital

Charging electrons in the sub skin uses the following sequence:

1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s², etc.

Statement about electrons and atomic orbitals is not true is An electron has the same amount of energy in all orbitals

the electron configuration for barium (Ba) in noble-gas notation brainly.com/question/11147367

the formation of a bond.

brainly.com/question/11311275

quantum number

brainly.com/question/2292596

Keywords: the electron configuration, orbitals, atoms, energy, skin, sub skin, electron clouds

8 0

2 years ago

Read 2 more answers

A small particle with positive charge q = +4.25 x 10^-4C and mass m = 5.00 x 10^-5 kg is moving in a region of uniform electric

Tcecarenko [31]

Answer:

a) r = (0.6 i- 2039 j ^ + 0.102 k⁾ m and b) vₓ = 30.0 m / s , v_{y} = 2.04 10⁵ m / s c) v_{z} = 1.02 10⁻¹m / s

Explanation:

a) To find the position of the particle at a given moment we must know the approximation of the body, use Newton's second law to find the acceleration

Fe + Fm = m a

a = (Fe + Fm) / m

the electric force is

Fe = q E k ^

Fe = 4.25 10-4 60 k ^

Fe = 2.55 10-2 k ^

the magnetic force is

Fm = q v x B

Fm = 4.25 10⁻⁴ $\left[\begin{array}{ccc}i&j&k\\30&0&0\\0&0&49\end{array}\right]$

fm = 4.25 10⁻⁴ (-j ^ 30 4)

fm 0 = ^ -5,10 10⁻² j

We look for every component of acceleration

X axis

aₓ = 0

there is no force

Axis y

ay = -5.10 10²/5 10⁻⁵ j ^

ay = -1.02 107 j ^ / s2

z axis

az = 2.55 10⁻² / 5 10⁻⁵ k ^

az = 5.1 10² k ^ m / s²

Having the acceleration in each axis we can encocoar the position using kinematics

X axis

the initial velocity is vo = 30 m / s and an initial position xo = 0

x = vo t + ½ aₓ t₂2

x = 30 0.02 + 0

x = 0.6m

Axis y

acceleration is ay = -1.02 10⁷ m / s², a starting position of i = 1m

y = I + go t + ½ ay t²

y = 1 + 0 + ½ (-1.02 10⁷) 0.02²

y = 1 - 2.04 10³

y = -2039 m j ^

z axis

acceleration is aza = 5.1 10² m / s², the position and initial speed are zero

z = zo + v₀ t + ½ az t²

z = 0 + 0 + ½ 5.1 10² 0.02²

z = 1.02 10⁻¹ m k ^

therefore the position of the bodies is

r = (0.6 i- 2039 j ^ + 0.102 k⁾ m

b) x axis

since there is no acceleration the speed remains constant

vₓ = 30.0 m / s

Axis y

let's use the equation v = v₀ + $a_{y}$ t

$v_{y}$ = 0 + -1.02 10⁷ 0.02

v_{y} = 2.04 10⁵ m / s

z axis

v_{z} = vo + az t

v_{z} = 0 + 5.1 10² 0.02

v_{z} = 1.02 10⁻¹m / s

8 0

2 years ago

Richard needs to fly from san diego to halifax, nova scotia and back in order to give an important talk about mathematics. on th

kondor19780726 [428]

When plane is going towards Halifax the speed of wind is in the direction of fly

so overall the net speed of the plane will increase

while when he is on the way back the air is opposite to flight so net speed will decrease

now the total time of the journey is 13 hours

out of this 2 hours he spent in mathematics talk

so total time of the fly is 13 - 2 = 11 hours

now we have formula to find the time to travel to Halinex

$t_1 = \frac{d}{v + 50}$

time taken to reach back

$t_2 = \frac{d}{v - 50}$

now we have total time

$T = t_1 + t_2$

$11 = \frac{d}{v - 50} + \frac{d}{v + 50}$

here d= 3000 miles

$11 = \frac{3000}{v - 50} + \frac{3000}{v + 50}$

$3.67 * 10^{-3} = \frac{2v}{v^2 - 2500}$

$v^2 - 2500 = 545.45v$

solving above quadratic equation we will have

$v = 550 mph$

so speed of plane will be 550 mph

3 0

2 years ago

Read 2 more answers

Inna Hurry is traveling at 6.8 m/s, when she realizes she is late for an appointment. She accelerates at 4.5 m/s^2 for 3.2 s. Wh

Alborosie

Answer:

1) v = 21.2 m/s

2) S = 63.33 m

3) s = 61.257 m

4) Deceleration, a = -4.32 m/s²

Explanation:

1) Given,

The initial velocity of Inna, u = 6.8 m/s

The acceleration of Inna, a = 4.5 m/s²

The time of travel, t = 3.2 s

Using the first equation of motion, the final velocity is

v = u + at

= 6.8 + 4.5 x 3.2

= 21.2 m/s

The final velocity of Inna is, v = 21.2 m/s

2) Given,

The initial velocity of Lisa, u = 12 m/s

The final velocity of Lisa, v = 26 m/s

The acceleration of Lisa, a = 4.2 m/s²

Using the III equations of motion, the displacement is

v² = u² +2aS

S = (v² - u²) / 2a

= (26² -12²) / 2 x 4.2

= 63.33 m

The distance Lisa traveled, S = 63.33 m

3) Given,

The initial velocity of Ed, u = 38.2 m/s

The deceleration of Ed, d = - 8.6 m/s²

The time of travel, t = 2.1 s

Using the II equations of motion, the displacement is

s = ut + 1/2 at²

=38.2 x 2.1 + 0.5 x(-8.6) x 2.1²

= 61.257 m

Therefore, the distance traveled by Ed, s = 61.257 m

4) Given,

The initial velocity of the car, u = 24.2 m/s

The final velocity of the car, v = 11.9 m/s

The time taken by the car is, t = 2.85 s

Using the first equations of motion,

v = u + at

∴ a = (v - u) / t

= (11.9 - 24.2) / 2.85

= -4.32 m/s²

Hence, the deceleration of the car, a = = -4.32 m/s²

5 0

2 years ago

Read 2 more answers

The hammer throw was one of the earliest Olympic events. In this event, a heavy ball attached to a chain is swung several times

Aleonysh [2.5K]

Answer:

Given that

T= 0.43 s

Radius of the ball path's , r=2.1 m

We know that

f= 1/T

Here f= frequency

T= Time period

Now by putting the values

f= 1/T

T= 0.43 s

f= 1/0.43

f=2.32 Hz

We know that

V= ω r

ω = 2 π f

ω=Angular speed

V= Linear speed

ω = 2 π f=ω = 2 x π x 2.32 =14.60 rad/s

V= ω r= 14.60 x 2.1 = 30.66 m/s

Acceleration ,a

a =ω ² r

a= 14.6 ² x 2.1 = 447.63 m/s²

We know that g = 10 m/s²

a= a/g= 447.63/10 = 44.7 g m/s²

a= 44.7 g m/s²

7 0

2 years ago