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2 years ago

9

What is its characteristic wavelength? [Hint: Recall that the kinetic energy of a moving object is E=12mv2, where m is the mass

of the object and v is the speed of the object.]

1 answer:

Alchen [17]2 years ago

4 0

Answer:

λ = $1.06 * 10^{-11} m$

Explanation:

Using the De Broglie equation, the characteristic wavelength is given by:

λ = $\frac{h}{p}$

where

h = Planck's constant = $6.626 * 10^{-34}$ Js.

p = momentum

Momentum, p, can be calculated using:

p = $\sqrt{2Em}$

where

m = mass of the electron = $9.11 * 10^{-31}$ kg

E = Energy of the electron = 13.4 keV = $13.4 * 10^3 * 1.6 * 10^{-19}$ J = $2.144 * 10^{-15}$ J

=> p = $\sqrt{2 * 2.144 * 10^{-15} * 9.11 * 10^{-31}}$

p = $\sqrt{3.906 * 10^{-45}}$

p = $6.250 * 10^{-23}$ kgm/s

Therefore, characteristic wavelength, λ, is:

λ = $\frac{6.626 * 10^{-34}}{6.250 * 10^{-23}}$

λ = $1.06 * 10^{-11} m$

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The wavelength of light is 5000 angstrom. Express it in nm and m.

Ierofanga [76]

Answer:

1 angstrom = 0.1nm

5000 angstrom = 5000/1 × 0.1nm

<h3>= 500nm</h3>

$1 \: angstrom = 1 \times {10}^{ - 10} m$

5000 angstrom = 5000 × 1 × 10^-10

<h3>= 5 × 10^-7 m</h3>

Hope this helps you

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2 years ago

A glass beaker of unknown mass contains of water. The system absorbs of heat and the temperature rises as a result. What is the

kakasveta [241]

From the information provided in the question, the mass of the beaker is 144.4 g.

From the information provided in the complete question;

volume of water = 74 mL

Mass of water = 74 g

specific heat of glass = 0.18 cal/g ∙ °C

specific heat of water = 1.0 cal/g ∙ C°

Mass of glass = x g

Total heat gained by the system = 2000.0cal

Temperature rise = 20.0°C

Heat gained by system = Heat gained by glass + Heat gained by water

Heat gained by glass = x × 0.18 × 20

Heat gained by water = 74 × 1.0 × 20

Hence;

2000 = (x × 0.18 × 20) + ( 74 × 1.0 × 20)

2000 - 1480 = (x × 0.18 × 20)

x = 520/3.6

x = 144.4 g

Missing parts;

A glass beaker of unknown mass contains 74.0 ml of water. The system absorbs 2000.0cal of heat and the temperature rises 20.0°C as a result. What is the mass of the beaker? The specific heat of glass is 0.18

cal/g °C, and that of water is 1.0 cal/g °C.

Learn more: brainly.com/question/1446583

3 0

1 year ago

you are flying a kite that has two strings attached to either side of it you are holding those strings in your left and right ha

Svetach [21]

The kite would move to the right as well

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2 years ago

Read 2 more answers

Hanging by a thread. Two metal spheres hang from nylon threads and attract each other when brought close together. (i) What can

elena-14-01-66 [18.8K]

Answer:

Explained

Explanation:

i)Two spheres hanging from nylon threads attract each other because either the two spheres are charged with opposite sign or only one of the spheres is charge so the other would be charge by induction of the charged sphere and hence attract each other.

ii)However, when they are touched the charges will be rearranged among the two sphere such that the two sphere have exact same magnitude and sign of charge and now they will repel each other or the magnitude of charges on the two spheres become zero and they neither attract or repel each other.

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2 years ago

A silver wire 2.6 mm in diameter transfers a charge of 420 Cin 80 min. Silver contains 5.8 x 10^{28} free electrons per cubic me

never [62]

1) Current in the wire: 0.0875 A

The current in the wire is given by:

$I=\frac{Q}{t}$

where

Q is the charge passing a given point in the conductor

t is the time elapsed

In this problem, we have

Q = 420 C is the total charge passing through a given point in a time of

t = 80 min = 4800 s

So, the current is

$I=\frac{420 C}{4800 s}=0.0875 A$

2) Drift velocity of the electrons: $1.78\cdot 10^{-6} m/s$

The drift velocity of the electrons in the wire is given by:

$u = \frac{I}{nAq}$

where

I = 0.0875 A is the current

$n=5.8\cdot 10^{28}$ is the number of free electrons per cubic meter

A is the cross-sectional area

$q=1.6\cdot 10^{-19} C$ is the charge of one electron

The radius of the wire is

$r=\frac{d}{2}=\frac{2.6 mm}{2}=1.3 mm=0.0013 m$

So the cross-sectional area is

$A=\pi r^2=\pi (0.0013 m)^2=5.31\cdot 10^{-6} m^2$

So, the drift velocity is

$u = \frac{(0.0875 A)}{(5.8\cdot 10^{28})(5.31\cdot 10^{-6})(1.6\cdot 10^{-19}C)}=1.78\cdot 10^{-6} m/s$

4 0

2 years ago