How much energy is needed to move an electron in a hydrogen atom from the ground state (n = 1) to n = 3?

Answers

Answer 1

The energy needed to move an electron in a hydrogenatome from the ground state (n=1) to n=3 will be 1.93 *10^-18J and 12.09 eV.

How to compute the value?

The following can be deduced:

Energy of electron in hydrogen atom is

En = -13.6 /n2 eV

where n is principal quantum number of orbit.

Energy of electron in first orbit = E1 = -13.6 / 12 = - 13.6eV

Energy of electron in third orbit = E3 = -13.6 /32 = -1.51 eV

Energy required to move an electron fromfirst to thirdorbit ΔE = E3- E1

ΔE = -1.51 - ( 13.6) = 12.09 eV

Energy in Joule = 12.09 *l/× 1.6 × 10^-19 = 1.93 × 10^-18 J.

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Complete question:

How much energy is needed to move an electron in a hydrogenatome from the ground state (n=1) to n=3? Give theanswer (a) in joules and (b) in eV.


Related Questions

NASA wants a satellite to revolve around Earth 3 times a day. What should be the radius of its orbit if we neglect the presence of the Moon? (G = 6.67 × 10-11 N ∙ m2/kg2, Mearth = 5.97 × 1024 kg)

Answers

Answer:

Approximately [tex]2.03 \times 10^{7}\; {\rm m}[/tex].

Explanation:

Assume that the radius of this orbit is [tex]r[/tex].

Let [tex]m[/tex] denote the mass of this satellite and let [tex]M[/tex] denote the mass of the Earth. At a distance of [tex]r[/tex] from the center of the earth, the magnitude of the gravitational attraction on this satellite would be [tex]G\, m\, M / (r^{2})[/tex].

The question implies that the gravitational pull from the earth is the only significant force on this satellite. Hence, the net force on this satellite would be also [tex]G\, m\, M / (r^{2})[/tex].

The acceleration of this satellite would thus be [tex]a = (\text{net force}) / (\text{mass}) = G\, M / (r^{2})[/tex].

Let [tex]\omega[/tex] denote the angular velocity of this satellite. Since this satellite in in a circular motion, the acceleration on this satellite would need to satisfy [tex]a = \omega^{2} \, r[/tex].

In other words:

[tex]\begin{aligned} \frac{G\, M}{r^{2}} = a = \omega^{2} \, r \end{aligned}[/tex].

[tex]\begin{aligned} r &= \left(\frac{G\, M}{\omega^{2}}\right)^{1/3}\end{aligned}[/tex].

The question asks for a rotation of [tex]3\times (2\, \pi) = 6\, \pi\; {\text{rad}}[/tex] within a day, which is [tex]24 \times 3600\; {\rm s}[/tex]. The angular velocity of this satellite should be:

[tex]\begin{aligned}\omega = \frac{6\, \pi}{24 \times 3600\; {\rm s}} \end{aligned}[/tex].

Substitute this value into the expression for [tex]r[/tex] and evaluate:

[tex]\begin{aligned} r &= \left(\frac{G\, M}{\omega^{2}}\right)^{1/3} \\ &= \left(\frac{(6.67 \times 10^{-11}\; {\rm N \cdot m^{2} \cdot kg^{-2}}) \times (5.97 \times 10^{24}\; {\rm kg})}{((6\, \pi) / (24 \times 3600\; {\rm s}))^{2}}\right)^{1/3} \\ &\approx 2.03 \times 10^{7}\; {\rm m}\end{aligned}[/tex].

(Note that [tex]1\; {\rm N} = 1\; {\rm kg \cdot m \cdot s^{-2}}[/tex].)

The focal length of a diverging lens is negative. If f = −26 cm or a particular diverging lens, where will the image be formed of an object located 63 cm to the left of the lens on the optical axis?

.... cm to the left of the lens

....What is the magnification of the image?

Answers

The distance of the image formed is 44.27 cm and the magnification of the image is 0.7.

Location of the image

The image distance formed by the lens is calculated as follows;

1/v + 1/u = -1/f

where;

v is the image distanceu is the object distancef is the focal length of the lens

1/v = -1/f - 1/u

1/v = -(-1/26) - 1/63

1/v = 1/26 - 1/63

1/v = 0.022588

v = 1/0.022588

v = 44.27 cm

What is magnification of lens?

The magnification of a lens is defined as the ratio of the height of an image to the height of an object.

Magnification of the image formed

The magnification of the image is calculated as follows;

Magnification = image distance/object distance

M = 44.27/63

M = 0.7

Thus, the distance of the image formed is 44.27 cm and the magnification of the image is 0.7.

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Three bees are oriented as shown in the figure. B1 has +17 µC of charge, B2 has −5 µC of charge, and B3 has +26 µC of charge.
B1 to b2 5 cm
B2 to b3 10 com



Bee number 3 is stationed at the observation location for this problem, and we want to find the net electric field at B3. We'll do this in a few steps.
(a) What are the x and y components of the electric field

at the observation location (Bee number 3) due to B1? Remember that the components can be positive or negative depending on their directions along the x or y axis.
E1x =
E1y =

(b) What are the x and y components of the electric field

at the observation location (Bee number 3) due to B2? Again, remember that the components can be positive or negative depending on their directions along the x or y axis.
E2x = N/C
E2y =

(c) What are the magnitude and direction of the net electric field at the observation location where B3 is resting?

magnitude

N/C

direction

° counterclockwise from the +x-axis

(d) What is the magnitude of the force on B3 due to this net electric field?

Answers

The x and y components of the electric field due to B₁ are equal to 6.5 × 10⁶ N/C and -4.5 × 10⁶ N/C respectively.The x and y components of the electric field due to B₂ are equal to -4.5 × 10⁶ N/C and 0 N/C respectively.The magnitude of the net electric field at B₃ is equal to 6.04 × 10⁶ N/C.The direction of the net electric field at the observation location is equal to 49.76°.The magnitude of the force on B₃ due to this net electric field is equal to 157.04 Newton.

Given the following data:

Charge of B₁ = +17 µC to C = 17 × 10⁻⁶ C.Charge of B₂ = -5 µC to C = -5 × 10⁻⁶ C.Charge of B₃ = +26 µC to C = 26 × 10⁻⁶ C.Radius of B₁ to B₂ = 5 cm to m = 0.05 meter.Radius of B₂ to B₃ = 10 cm to m = 0.1 meter.

How to determine the x and y components of the electric field?

First of all, we would calculate the electric field due to B₁ to B₃ and the electric field due to B₂ to B₃ respectively.

The electric field due to B₁ to B₃ is given by:

E₁ = kq₁/r₁²

E₁ = (9 × 10⁹ × 17 × 10⁻⁶)/(0.05² + 0.1²)

E₁ = 153 × 10³/0.0125

Electric field, E₁ = 12.24 × 10⁶ N/C.

The electric field due to B₂ to B₃ is given by:

E₂ = kq₂/r₂²

E₂ = (9 × 10⁹ × 5 × 10⁻⁶)/(0.1²)

E₂ = 45 × 10³/0.01

Electric field, E₂ = 4.5 × 10⁶ N/C.

Also, the magnitude of the angle formed is given by tan trigonometry:

Tanθ = 5/10

Tanθ = 0.5

θ = tan⁻¹(0.5)

θ = 26.56°.

Next, we would calculate the x-component of the electric field at the observation location (Bee number 3) due to B₁:

E₁x = E₁cosθ - E₂

E₁x = 12.24 × 10⁶ × cos(26.56) - 4.5 × 10⁶

E₁x = 10.95 × 10⁶ - 4.5 × 10⁶

E₁x = 6.5 × 10⁶ N/C.

Similarly, we would calculate the y-component of the electric field at the observation location (Bee number 3) due to B₁:

E₁y = -E₁sinθ

E₁y = -12.24 × 10⁶ × sin(26.56)

E₁y = -12.24 × 10⁶ × 0.4471

E₁y = -5.5 × 10⁶ N/C.

Also, the magnitude of the electric field is given by:

Exy = √(E₁x² + E₁y²)

Exy = √(6.5 × 10⁶)² + (-5.5 × 10⁶)²)

Exy = √4.225 × 10¹³ + 3.025 × 10¹³)

Exy = √(7.25 × 10¹³)

Exy = 8.5 × 10⁶ N/C.

Part B.

We would calculate the x-component of the electric field at the observation location (Bee number 3) due to B₂:

E₂x = -E₂sinθ

E₂x = -4.5 × 10⁶ × sin(90)

E₂x = -4.5 × 10⁶ N/C.

Similarly, we would calculate the y-component of the electric field at the observation location (Bee number 3) due to B₁:

E₂y = E₂cosθ

E₂y = -4.5 × 10⁶ × cos(90)

E₂y = 0 N/C.

How to calculate the net electric field at the observation location?

The magnitude of the net electric field at B₃ is given by:

Eₙ = √(E₁x + E₂x)² + E₁y²)

Eₙ = √(6.5 × 10⁶ + (-4.5 × 10⁶))² + (-5.5 × 10⁶)²)

Eₙ = √(2.5 × 10⁶)² + (3.025 × 10¹³)

Eₙ = √(6.25 × 10¹²) + (3.025 × 10¹³)

Eₙ = √(3.65 × 10¹³)

Eₙ = 6.04 × 10⁶ N/C.

For the direction, we have:

Tanθ = x/y

Tanθ = 6.5/-5.5

Tanθ = -1.1818

θ = tan⁻¹(-1.1818)

θ = 49.76°.

Part C.

The magnitude of the force on B₃ due to this net electric field is given by:

F = B₃ × Eₙ

F = 26 × 10⁻⁶ × 6.04 × 10⁶

F = 157.04 Newton.

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a man carries a hand bag by hanging on his hand moves horizantaly wher the bag does not up or down what is the work done on the bag

Answers

Since the displacement is completely perpendicular to the direction of the applied force, the work done on the bag is zero.

When is the Work done on an object ?

The work is done on an object when the force applied is multiply by the distance moved by the object in the direction of the force applied.

Given that a man carries a hand bag by hanging on his hand moves horizontally where the bag does not up or down.

What is work if the displacement is not in the direction of force ?

The work done can only be zero if the displacement is perpendicular to the direction of force. otherwise, it will not be equal to zero.

Also, the work done will be zero, if the displacement is zero.

In the question above, the displacement is completely perpendicular to the direction of the applied force.

Therefore, the work done on the bag is zero.

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Summarize the step you might use to Carry out an investigation using scientific method

Answers

The summary of the step you might use to carry out an investigation using scientific method include:

ObservationHypothesisExperimentConclusion

What is Scientific method?

This is defined as a systematic approach which helps to predict how several elements in the universe works.

This usually starts with observing a phenomenon and making a hypothesis. This is usually followed by conducting experiment to ascertain its authenticity before a conclusion is drawn.

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A body of mass 80kg moving with a velocity of 6m/s hits a stationary body
of mass of 40kg. If the two bodies stick after the impact, calculate their
common velocity.

Answers

Answer:

4 m/s

Explanation:

Conservation of momentum ,  mv

Before collision  80 kg * 6 m/s  = 480  kg-m/s  

   after the collision,

         the momentum is the same but the mass is 80+40 kg =120 kg

 mv = 480

120 * v = 480

v = 4 m/s

"The acceleration due to gravity on the surface of the earth is 9.8 m/s ²." What does it mean?​

Answers

Answer:  It simply means that a freely falling object would increase its velocity by 9.8 m/s per second

Answer:

Hello!

"The acceleration due to gravity on the surface of the earth is 9.8 m/s ²." What does it means that

That every second an object is in free fall, gravity will cause the velocity of the object to increase 9.8 m/s. So, after one second, the object is traveling at 9.8 m/s.

A triangular plate with a non-uniform areal density has a mass M=0.500 kg. It is suspended by a pivot at P and can oscillate as indicated below. Its center of mass is a distance d=0.300 m from the pivot axis and its moment of inertia about an axis through the CM and parallel to the pivot axis is ICM=9.00×10-3 kg m2.
The plate is released from an angle θ=15°. Calculate the period of the oscillations.

Answers

The period of the oscillations.T = 1.2042s

Opposition is the process of any quantity or measure fluctuating repeatedly about its equilibrium value throughout time. This process is referred to as oscillation. Oscillation, a periodic fluctuation of a substance, can also be described as alternating between two values or rotating around a central value.

Typically, the mathematical formula for the moment of inertia is

T = 2 π √(I / mgd)

Therefore, a moment of inertia

I = 9.00×10-3 + md^2 ;

I=9.00*10^{-3}+ 0.5 * 0.3^2

I=0.054

T=2[tex]\pi \sqrt{0.5*9.8*0.3}[/tex]

T=1.2042s

The period of the oscillations.T = 1.2042s

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Consider the f(x) = cos(Bx) function shown in the figure in blue color. What is the value of parameter B for this function? This parameter is called angular frequency in physics and it is denoted by ω.
As a reference the g(x) = cos(x) function is shown in red color, and green tick marks are drawn at integer multiples of π.

Answers

The value of parameter B (angular frequency) for this function is 2.7.

What is amplitude of a wave?

The amplitude of a wave is the maximum displacement of the wave. It can also be described at the maximum upward displacement of a wave curve.

f(x) = Acos(Bx)

where;

A is amplitude of the waveB is angular frequency of the wave

What is angular frequency of a wave?

Angular frequency is  the angular displacement of any element of the wave per unit time.

From the blue color; at y = -0.9, x = -7.5 cm

-0.9 = cos(-7.5B)

7.5B = arc cos (0.9)

7.5B = 2.7

B = 7.5/2.7

B = 2.7

Thus, the value of parameter B (angular frequency) for this function is 2.7.

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A 126- kg astronaut (including space suit) acquires a speed of 2.70 m/s by pushing off with her legs from a 1800-kg space capsule. Use the reference frame in which the capsule is at rest before the push.
a)What is the velocity of the space capsule after the push in the reference frame?
Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the velocity is in the direction of the velocity of the astronaut and negative value if the direction of the velocity is in the direction opposite to the velocity of the astronaut.
b)If the push lasts 0.600 s , what is the magnitude of the average force exerted by each on the other?
Express your answer to three significant figures and include the appropriate units.
c)What is the kinetic energy of the astronaut after the push in the reference frame?
Express your answer to three significant figures and include the appropriate units.
d)What is the kinetic energy of the capsule after the push in the reference frame?
Express your answer to two significant figures and include the appropriate units.

Answers

The change in the speed of the space capsule will be -0.189 m/s.

The average force exerted by each on the other will be 567 N.

The kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

Given:

Mass of the astronaut, [tex]m_a[/tex] = 126 kg

Speed he acquires, [tex]v_{a}[/tex]  = 2.70 m/s

Mass of the space capsule, [tex]m_{c}[/tex] = 1800kg

The initial momentum of the astronaut-capsule system is zero due to rest.

[tex]P_f = m_av_a + m_cv_c[/tex]

[tex]P_I[/tex] = 0

[tex]m_av_a + m_cv_c = 0[/tex]

[tex]v_c =\frac{- m_a v_a}{m_c}}\\\\[/tex]

   [tex]= \frac{126* 2.70}{1800}[/tex]

   [tex]= - 0.189[/tex] m/s

Therefore,

According, to the impulse-momentum theorem;

FΔt = ΔP

ΔP = m Δv

ΔP = 126×2.70

    = 340.2 kgm/sec

t is time interval = 0.600s

F = ΔP/Δt

F = 340.2/0.600

  = 567 N

Therefore, the average force exerted by each on the other will be 567 N.

The Kinetic Energy of the astronaut;

K.E = [tex]\frac{1}{2} m v^2[/tex]

     [tex]= \frac{1}{2}[/tex] × 126 × [tex](2.70) ^2[/tex]

     = 459.27 J

The Kinetic Energy of the capsule;

K.E = [tex]\frac{1}{2} m v^2[/tex]

     = [tex]\frac{1}{2}[/tex]×1800×[tex](0.189) ^2[/tex]

     = 32.14 J

Therefore, the kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

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Which one of the following statements concerning weight and energy balance is most accurate?

A. People generally need the same amount of physical activity to maintain weight stability.

B. Regular physical activity doesn’t impact the percentage of body fat in children and adolescents.

C. It’s possible to achieve weight stability by doing the equivalent of 60–120 minutes a week of moderate-intensity walking.

D. The optimal amount of physical activity needed to maintain weight is unclear.

Answer: D. I took the test and got it right

Answers

The correct answer choice concerning weight and energy balance which is most accurate is the optimal amount of physical activity needed to maintain weight is unclear.

What is energy balance?

Energy balance refers to the way in balance is achieved when intake of energy is equal to energy expended.

Energy refers to the impetus behind all motion and all activity. If is also the capacity to do work. Energy is measured in a unit dimensioned in mass × distance²/time² (ML²/T²) or the equivalent.

So therefore, the correct answer choice concerning weight and energy balance which is most accurate is the optimal amount of physical activity needed to maintain weight is unclear.

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Complete the missing information from the circuit

Answers

From the first circuit (see attached), it is correct to state that I₄ = 40mA. See the explanation below.

What is a circuit?

A path that is designed to transport electric current is that is referred to as a circuit.

How do we compute for the missing details?

I = 200mA

= 20mA + 140mA + I₄

I₄ = 200 - 160

I₄ = 40mA

From the above, we can rightly posit:

R₄ = 3/(40x 10⁻³)

= 75 Ω

Hence

V₆ =  V₄ - V₅

= 3v - 2v

= 1v

R₆ therefore, =

V₆/I₆

= 1/ (40 x 10 ⁻³)

= 14.28 Ω

I₃ = I₄ + I₆

= 40 + 140

= 180mA

R₃ = 9/(180 x 10⁻³)

= 50Ω

R₂ = 12/20 x (10⁻³)

= 60 Ω

Then,

Vbattery =

200 x 10⁻³ x90

= 18 volts

The above indices is used to complete the image. See Circuit Image II.

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The distance between a carbon atom ( m = 12 u ) and an oxygen atom ( m = 16 u ) in the CO molecule is 1.13 × 10-10 m .
How far from the carbon atom is the center of mass of the molecule?
Express your answer to two significant figures and include the appropriate units.

Answers

6.45 ×[tex]10^{-} ^{11}[/tex] m is far from the carbon atom and is the center of mass of the molecule.

The sum of the products of the masses of the two particles and their respective position vectors equals the product of the total mass of the system and the position vector of the center of mass.

The center of mass of CO molecule will be on the line joining C and O atoms. Let the CO molecule be along the X- axis, the C atm being at the origin (x=0). The center of mass relative to the C atom is given by

[tex]x_{cm} = \frac{m_{c}x_{c} + m_{o} x_{o} }{m_{c}+ m_{o} }[/tex]

Where, [tex]m_{c}[/tex] and [tex]m_{o}[/tex] are the respective masses of C and O atoms,  [tex]x_{c}[/tex] and  [tex]x_{o}[/tex] their distances relative to C atom.

Here, [tex]m_{c}[/tex] = 12 u , [tex]m_{o}[/tex] = 16 u and [tex]x_{o} =1.13[/tex]×[tex]10^-^{10}[/tex] m

[tex]x_{cm} = \frac{m_{c}x_{c} + m_{o} x_{o} }{m_{c}+ m_{o} }[/tex]

[tex]x_{cm} =[/tex] (12 × 0 + 16 × 1.13 × [tex]10^{-} ^{10}[/tex] ) / (12 + 16 )

[tex]x_{cm}[/tex] = 0.64571 × [tex]10^{-} ^{10}[/tex]

[tex]x_{cm}[/tex] = 6.45 ×[tex]10^{-} ^{11}[/tex] m

Therefore, 6.45 ×[tex]10^{-} ^{11}[/tex] m is far from the carbon atom is the center of mass of the molecule.

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In thicker materials the particles can move more easily, therefore the resistance has to ____________.
A. Increase
B. Decrease
C. Stay the same
D. Not enough info

Answers

In thicker materials the particles can move more easily, therefore the resistance has to decrease.

Hence Option (b) is correct.

Resistivity, which is sometimes represented by the Greek letter rho, is quantitatively equal to a specimen's resistance R times its cross-sectional area A times its length.

ρ = R × A/ l

R = ρ × l/ A

It means the Resistance of the material is inversely proportional to the area of the material.

As the area of the material increases, the resistance decreases and vice versa.

Hence, In thicker particles, there is a greater area available and that's why resistance has to decrease so that particles can move more easily.

Hence, Option (b) is correct.

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A block and tackle pulley system has 3pulley wheels in the lower movable block. Determine the load that can be lifted by an effort of 350N if the efficiency of the system is 80%​

Answers

A block and tackle pulley design has a velocity ratio 3. <br> Draw a labelled diagram of this system. In your diagram, indicate absolutely the points of application and the directions of the load and effort

How to calculate  the load that can be lifted by an effort of 350N if the efficiency of the system is 80%​?

VR =3

VR = n=3

Efficiency of the system = 80%

Thus , Mechanical asvantage [tex]$=V R \times \eta=$[/tex]80/100×3 =2.4

Man can lift load with effort =350N

Thus,

Load= MA× effort = 2.4×350 = 840 N.

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In 1656, the Burgmeister (mayor) of the town of Magdeburg, Germany, Otto Von Guericke, carried out a dramatic demonstration of the effect resulting from evacuating air from a container. It is the basis for this problem. Two steel hemispheres of radius 0.430 m (1.41 feet) with a rubber seal in between are placed together and air pumped out so that the pressure inside is 15.00 millibar. The atmospheric pressure outside is 940 millibar.
1. Calculate the force required to pull the two hemispheres apart. [Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]
2. Two equal teams of horses, are attached to the hemispheres to pull it apart. If each horse can pull with a force of 1450N (i.e., about 326 lbs), what is the minimum number of horses required?

Answers

The force required to pull the two hemispheres is 46622.72N

Calculation and Parameters

( Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]

The contact area between the hemispheres is (pi x 0.400^2) = 0.5024m^2.

Pressure difference = (940 - 12)

= 928 millibars.

(928 x 100)

= 92,800N/m^2.

Therefore, the required force to pull the two hemispheres is

(92800 x 0.5024)

= 46622.72N.

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the earth orbits is oval in shape.explain how the mangnitude of gravitational between the earth and sun changes as the eart moves from position a to b

Answers

As the distance between the Earth and the Sun decreases, the magnitude of gravitational force between the earth and sun increases and vice versa.

What is gravitational force?

Gravitational force is a force that attracts any two objects with mass.

According to Newton's law of universal gravitation, the force of attraction between two objects in the universe is directly proportional to the product of their masses and inversely proportional to the square of the distance between the two objects.

F = Gm₁m₂/R²

where;

m₁ is mass of Earthm₂ is mass of sunR is the distance between the Earth and Sun

Thus, as the distance between the Earth and the Sun decreases, the magnitude of gravitational force between the earth and sun increases and vice versa.

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4. Four objects are situated along the y axis as follows: a 2.00kg object is at +3.00m, a 3.00kg object is at +2.50m, a 2.50kg object is at the origin, and a 4.00kg object is at -0.500m. Where is the center of mass of these objects?​

Answers

Answer: 1.348 meters

Explanation: Although the sign is missing from the location of the 4.00 kg object, it is assumed to be positive. The net moment of all the objects about the center of mass must be zero. Let the center of mass be on the y axis at a point  c . Adding the four moments together, we get:

(2.00)(3.00−c)+(3.00)(2.50−c)+(2.50)(0−c)+(4.00)(0.500−c)=0

6.00−2.00c+7.50−3.00c+0−2.50c+2.00−4.00c=0

11.5c=15.50

c=  1.348 metres

The center of mass is on the y axis at  y  = 1.348 metres.

How does the suns energy affect the climate of an area

Answers

Answer:

the earth receives the suns energy as radiation which is able to travel without the need of matter from point A to point B and this radiation is also in the form of heat.

Explanation:

Sun's energy fall on the earth in form of radiation and heats up the atmosphere and produces unequal heating due to its spherical nature and impacts climate of different areas.

The Sun is the only source of energy for the Earth's climate system. Fundamental to atmospheric composition is the warming of the atmosphere caused by solar radiation, which also causes global wind patterns and aids in the development of clouds, storms, and rainfall.

When the sun is at its maximum (peak in sunspots, prominences, and flares), it produces more ultraviolet (UV) radiation. More ozone is produced as a result of UV light, warming the stratosphere. The lower atmosphere and Earth's surface warm slightly as a result of the increased solar radiation.

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Compute your average velocity in the following two cases: (a) You
walk 73.2 m at a speed of 1.22 m/s and then run 13.2 m at a speed
of 3.02 m/s along a straight track. (b) You walk for 1.00 min at a
speed of 1.22 m/s and then run for 1.00 min at 3.05 m/s along a
straight track. (c) Graph x versus t for both cases and indicate how
the average velocity is found on the graph.

Answers

(a) Walking 73.2 m at 1.22 m/s would take

[tex]\dfrac{73.2\,\rm m}{1.22 \frac{\rm m}{\rm s}} = 60 \,\rm s[/tex]

and running 13.2 m at 3.02 m/s would take

[tex]\dfrac{13.2\,\rm m}{3.02\frac{\rm m}{\rm s}} \approx 4.37\,\rm s[/tex]

You've undergone a total displacement of 73.2 + 13.2 = 86.4 m in a matter of approximatly 64.37 s, so your average velocity is

[tex]\dfrac{\Delta x}{\Delta t} = \dfrac{86.4\,\mathrm m}{64.37\,\rm s} \approx \boxed{1.34\dfrac{\rm m}{\rm s}}[/tex]

(b) In the first 1.00 min = 60 s, you undergo a displacement of

[tex](60\,\mathrm s) \left(1.22 \dfrac{\rm m}{\rm s}\right) = 73.2 \,\rm m[/tex]

and in the second minute, you undergo a displacement of

[tex](60\,\mathrm s) \left(3.05\dfrac{\rm m}{\rm s}\right) = 183 \,\rm m[/tex]

Your total displacement is then 73.2 + 183 = 256.2 m in a matter of 2.00 min = 120 s, so your average velocity is

[tex]\dfrac{\Delta x}{\Delta t} = \dfrac{256.2\,\mathrm m}{120\,\rm s} \approx \boxed{2.14\dfrac{\rm m}{\rm s}}[/tex]

(c) For part (a), your displacement [tex]x(t)[/tex] (in m) at time [tex]t[/tex] (in s) is given by

[tex]x(t) = \begin{cases}1.22t & \text{for } 0 \le t \le 60 \\ 73.2 + 3.02 (t-60) & \text{for } t > 60\end{cases}[/tex]

and for part (b), your displacement is given by the very similar

[tex]x(t) = \begin{cases}1.22 t & \text{for } 0 \le t \le 60 \\ 73.2 + 3.05(t-60) & \text{for } t > 60 \end{cases}[/tex]

See the attached plots. The average velocity for the given situation is the slope of the dotted line.

which element has more than 5 valency​

Answers

Answer:

Sulfur (Has six valence electrons). It has maximum valency due to belonging to VI groups of the Periodic Table.

Explanation:

The electrons found in an element's outermost atomic shell are known as valence electrons.

Sulfur, which has an atomic number of 16, has an electrical configuration of 2, 8, 6, meaning it has six electrons in its outermost shell. As a result, its valence electrons will also be six.

However, in its natural condition, sulfur exists as the S8 molecule, which has the classic chair structure where each sulfur atom is covalently connected to two other sulfur atoms. In that sense, there will be 8 valence electrons.

Consequently, the answer will be 6 if you're asking about the "sulphur atom," but 8 if you're talking about sulfur in general.

Thank you ,

Eddie

explain the use of low energy x-ray?​

Answers

Answer:

The low energy X-ray calibration service is intended for thin-window plane-parallel chambers required for the dosimetry of superficial X-ray beams. Low-energy X-ray therapy is also known as superficial radiation therapy. This therapy uses beams of low-energy X-rays (radiation waves) to destroy skin cancer cells while not harming healthy tissue around or under the upper layers of skin.

Complete the following:

When light is incident parallel to the principal axis and then strikes a lens, ___

the light will remain parallel after refracting through the lens
the light will refract through the focal point on the opposite side of the lens
the light will not refract at all

Answers

When light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.

To find the answer, we have to know about the rules followed by drawing ray-diagram.

What are the rules obeyed by light rays?If the incident ray is parallel to the principal axis, the refracted ray will pass through the opposite side's focus.The refracted ray becomes parallel to the major axis if the incident ray passes through the focus.The refracted ray follows the same path if the incident light passes through the center of the curve.

Thus, we can conclude that, when light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.

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If something is a good conductor, what type of insulator is it?

1. also a good insulator
2. a metal
3. it depends
4. a poor insulator

Answers

4. a poor insulator

If rest other things are kept constant or unchanged then a good conductor can be termed as a poor insulator.

Answer:

4 conductors and insulators are opposites of each other.

Explanation:

Consider a concave spherical mirr or that has focal length f = +19.5 cm.
a) What is the distance of an object from the mirror's vertex if the image is real and has the same height as the object? Follow the sign rules.

Answers

The distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

What is concave mirror?

A concave mirror has a reflective surface that is curved inward and away from the light source.

Concave mirrors reflect light inward to one focal point and it usually form real and virtual images.

Object distance of the concave mirror

Apply mirrors formula as shown below;

1/f = 1/v + 1/u

where;

f is the focal length of the mirrorv is the object distanceu is the image distance

when image height = object height, magnification = 1

u/v = 1

v = u

Substitute the given parameters and solve for the distance of the object from the mirror's vertex

1/f = 1/v + 1/v

1/f = 2/v

v = 2f

v = 2(19.5 cm)

v = 39 cm

Thus, the distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.

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A 29-g rifle bullet traveling 200 m/s embeds itself in a 3.4-kg pendulum hanging on a 2.5-m-long string, which makes the pendulum swing upward in an arc. (Figure 1)
a) Determine the vertical component of the pendulum's maximum displacement.
Express your answer to two significant figures and include the appropriate units.
b)Determine the horizontal component of the pendulum's maximum displacement.
Express your answer to two significant figures and include the appropriate units.

Answers

The vertical and horizontal component of the maximum displacement are 0.15m and 0.85m.

What is the velocity of the pendulum just after the collision?

As per conversation of momentum, mass of bullet × velocity= mass of pendulum × velocity

=> 0.029kg × 200 = 3.429 × velocity of pendulum

=> Velocity of pendulum= 5.8/3.429 = 1.7 m/s

What is the amplitude of the pendulum's motion?As per conversation of energy,

1/2 × mass of pendulum × velocity²=1/2 × k × amplitude²

Here, k = mg/L = (3.429×9.8)/2.5 = 13.44 N/mSo, 3.429×1.7²= 13.44× amplitude²

=> 10 = 13.44× amplitude²

=> Amplitude = √(10/13.44)

= 0.86m

What is the angle made by the pendulum with vertical direction at largest displacement?

Angle = tan inverse (amplitude/length of pendulum)

= tan inverse (0.86/2.5)

= 20°

What is the maximum displacement along vertical direction?

Maximum displacement along vertical direction= L - L×cos20°

= 2.5 - 2.5×cos20°

= 0.15 m

What's the maximum displacement along horizontal direction?

Maximum displacement along horizontal direction= Lsin20°

= 2.5 × sin 20°

= 0.85m

Thus, we can conclude that the vertical and horizontal component of the maximum displacement are 0.15m and 0.85m.

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A tree limb is blown loose from a tree during a storm. As it falls, it gains
speed. Which type of energy is the tree limb gaining as it falls?
O A. Kinetic energy
B. Gravitational potential energy
O C. Nuclear energy
OD. Light energy

Answers

Answer:

B Gravitational potential energy

Explanation:

How can i find Air velocity??????????????

Answers

Answer: By dividing airflow by duct cross section.

Explanation:

In short, air velocity in the ducts is calculated by dividing airflow by duct cross-section. Airflow is expressed as a simple number. Example: Air conditioner has a max. airflow of 600 CFM.


You are writing a short report to a familiar audience and plan to use an informal
writing style. What can you do to make your writing informal?
Use contractions such as won't, it's, let's, and we're.
Include second-person pronouns such as you.
Use passive-voice verbs.
Incorporate longer sentences.

Answers

The thing that the presenter can do to make the presentation to appear informal is to use contractions such as won't, it's, let's, and we're.

What is a report?

The term report refers to any formal presentation. It is usually made as part of an academic course or as part of professional development. It could be given while a person is hoping for promotion at a job that he or she is doing.

We have to bear in mind that the audience to whom the report is being presented is a familiar audience hence the presentation can be quite informal and devoid of the formality of presentation of reports as we know it to be.

Now, the focus is to make the presentation to look informal. The thing that the presenter can do to make the presentation to appear informal is to use contractions such as won't, it's, let's, and we're.

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Why is glass classified the way it is?
A. The bulb lights up when the switch is closed.
B. Free electrons can carry charge.
C. Delocalized electrons create a current.
D. Not enough charge carriers.

Answers

The glass is classified the way it is because of option(b)i.e, Free electrons can carry a charge.

An amorphous solid is a glass. Glass shows all the mechanical properties of a solid even though its atomic-scale structure resembles that of a supercooled liquid. Transparency, heat resistance, pressure and breakage resistance, and chemical resistance are among glass' primary properties. 90% of all manufactured glass is soda-lime glass, making it the most prevalent and least priced type.

High electronegativity and tightly bonded electrons are found in the oxygen atoms that make up glass. This is so that glass can conduct electricity, which requires a significant amount of energy to release the electrons from the oxygen atom. As a result, the glass serves as an insulator at room temperature.

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Other Questions
Read the article Skara Brae:Skara Brae is one of the oldest historical sites in Europe, and it stayed hidden for more than 4,000 years. The ancient village of Skara Brae was discovered on the main island of Orkney, in Scotland. A massive storm in 1850 removed the sand that had covered it for thousands of years, exposing the remains of a small village.Skara Brae is a collection of eight round, stone houses that still have their stone furnishings and even some of the belongings of the inhabitants. The contents can tell us a great deal about how people lived in what is called the Neolithic era, or the last period of the Stone Age. Seven of the eight houses appear to have been family homes, with beds, dressers, fireplaces, and even simple toilets. The eighth house may have been a community workshop. It shows evidence of tool-making and the use of a large fire pit.How was Skara Brae built, and what, eventually, happened to it? Let's journey back in time through the evidence archaeologists have put together to understand the story of Skara Brae.The small, stone village was built some time around 3100 BCE. The first villagers dug down into an existing midden heap, or trash heap, to build the foundations of their homes. The garbage helped to insulate and protect the homes. They built their round huts with stones found locally, sticking each piece together with a paste made from ground-up waste and dung.After the houses were built, the villagers furnished their homes with stone beds, one on each side of the door, a stone dresser, and a fireplace, or hearth. The hearth kept the home warm and allowed the inhabitants to cook their food.The people of Skara Brae then farmed the surrounding lands, with evidence suggesting that they grew barley and other grains. They kept cows and pigs, and they fished in the ocean for a good part of their diet.As the number of people grew, the village expanded. Covered tunnels and alleyways were built to connect all the homes. This would have protected people traveling from one house to another from cold winter weather. Because of the connections between the houses and the shared workspace, archaeologists believe that Skara Brae was a close-knit community.The people made many things, including pottery, stone tools, and bone jewelry. Curiously, no evidence of weaponry was ever found, leading archaeologists to believe that Skara Brae was a peaceful place.Skara Brae was inhabited for approximately six hundred years, at which point, it was abandoned. Many theories exist about what happened to the people of Skara Brae. One of the most popular theories is that a terrible storm, similar to the one that would uncover the remains of the village in 1850, struck and drove the people from their homes. 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