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Verdich [7]
3 years ago
6

Jill does twice as much work as Jack does and in half the time. Jill's power output is Group of answer choices one-fourth as muc

h as Jack's power output. four times Jack's power output. twice Jack's power output. one-half as much as Jack's power output. the same as Jack's power output.
Physics
1 answer:
Musya8 [376]3 years ago
3 0

Answer:

Second Choice.

Explanation:

Jack's Power = W/t

Jill's Power = 2W/(0.5)*t

2/0.5 = 4

Jill's Power = 4*W/t

Jill's Power is 4 times greater than Jack's

Second Choice

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Calculate the ionization potential for C+5 ( 5 electrons removed for the C atom) and in addition compute the wavelength of the t
sveta [45]

Answer:

Ionization potential of C⁺⁵ is 489.6 eV.

Wavelength of the transition from n=3 to n=2 is 1.83 x 10⁻⁸ m.

Explanation:

The ionization potential of hydrogen like atoms is given by the relation :

E = \frac{13.6Z^{2} }{n^{2} } eV     .....(1)

Here <em>E</em> is ionization potential, <em>Z</em> is atomic number and <em>n</em> is the principal quantum number which represents the state of the atom.

In this problem, the ionization potential of Carbon atom is to determine.

So, substitute 6 for <em>Z</em> and 1 for <em>n</em> in the equation (1).

E = \frac{13.6\times(6)^{2} }{1^{2} }

<em> E = </em>489.6 eV

The wavelength (λ)  of the photon due to the transition of electrons in Hydrogen like atom is given by the relation :

\frac{1}{\lambda} =RZ^{2}[\frac{1}{n_{1} ^{2}}-\frac{1}{n_{2} ^{2} }]     ......(2)

R is Rydberg constant, n₁ and n₂ are the transition states of the atom.

Substitute 6 for Z, 2 for n₁, 3 for n₂ and 1.09 x 10⁷ m⁻¹ for R in equation (2).

\frac{1}{\lambda} =1.09\times10^{7} \times6^{2}[\frac{1}{2 ^{2}}-\frac{1}{3 ^{2} }]

\frac{1}{\lambda}  = 5.45 x 10⁷

λ = 1.83 x 10⁻⁸ m

7 0
4 years ago
An airplane flies 250.0 km at 300 m/s. How long does this take?​
givi [52]
Im not sure if this is physics or mathematics. but if 300 msec then per minute this will equal to 300 X 60 sec =18000 m per minute
3 0
3 years ago
Read 2 more answers
A certain humidifier operates by raising water to the boiling point and then evaporating it. Every minute 30 g of water at 20◦ C
Sveta_85 [38]

Answer:

The value of total energy needed per minute for the humidifier = 77.78 KJ

Explanation:

Total energy per minute the humidifier required = Energy required to heat water to boiling point) + Energy required to convert liquid water into vapor at the boiling point) ----- (1)

Specific heat of water = 4190 \frac{J}{kg k}

The heat of vaporization is =  2256 \frac{KJ}{kg}

Mass = 0.030 kg

Energy needed to heat water to boiling point =  m c ( T_{2} - T_{1} )

Energy needed to heat water to boiling point = 0.030 × 4.19 × (100 - 20)

Energy (E_{1}) = 10.08 KJ

Energy needed to convert liquid water into vapor at the boiling point

E_{2} = 0.030 × 2256 = 67.68 KJ

Thus the total energy needed E =  E_{1} + E_{2}

E = 10.08 + 67.68

E = 77.78 KJ

This is the value of total energy needed per minute for the humidifier.

9 0
4 years ago
A copper ornament has a mass of 0.0693 Kg and changes from a temperature of 18 degrees celcius to 26 degrees Celsius. How much h
zhuklara [117]

Answer: 216.2 J

Explanation: The heat energy of copper is the expressed in Q=mc∆T. First make sure that the mass 0.0693 kg is converted into grams to cancel both units.

Q= 69.3 g x 390 J/g°C x 26 °C-18°C = 216.2J

5 0
4 years ago
Read 2 more answers
The sun is 60° above the horizon. Rays from the sun strike the still surface of a pond and cast a shadow of a stick that is stuc
Kipish [7]

Answer:

shadow length 7.67 cm

Explanation:

given data:

refractive index of water is 1.33

by snell's law we have

n_{air} sin30 =n_{water} sin\theta

1*0.5 = 1.33*sin\theta

solving for\theta

sin\theta = \frac{3}{8}

\theta = sin^{-1}\frac{3}{8}

\theta =  22 degree

from shadow- stick traingle

tan(90-\theta) = cot\theta = \frac{h}{s}

s = \frac{h}{cot\theta} = h tan\theta

s = 19tan22 = 7.67 cm

s = shadow length

5 0
3 years ago
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