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son4ous [18]
3 years ago
11

Ultraviolet light of wavelength 3500 Angstrom falls on a potassium surface. The maximum energy of the emitted photoelectrons is

1.6 eV. Find the work function of potassium.
Physics
1 answer:
slega [8]3 years ago
8 0

Answer:

The work function of potassium is 1.94 eV.

Explanation:

Given that,

Wavelength \lambda= 3500\ \AA

Kinetic energy = 1.6 eV

We need to calculate the energy

Using formula of Energy

E=\dfrac{hc}{\lambda}

Put the value into the formula

E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{3500\times10^{-10}}

E=5.68\times10^{-19}\ J

E=5.68\times10^{-19}\times6.24\times10^{18}\ ev

E=3.54\ ev

We need to calculate the work function of potassium

Using formula of work function

E_{x}=\phi+k_{max}

Put the value into the formula

3.54=\phi+1.6

\phi=3.54-1.6

\phi=1.94\ ev

Hence, The work function of potassium is 1.94 eV.

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Set local ground level to 700 ft, and record the inches of mercury.
Katen [24]

The altimeter reading is 29.17 from the Kollsman window is 29.17 in Hg

<h3>How to determine the altimeter reading?</h3>

The given parameters are:

  • Ground level = 700 ft i.e. the field elevation
  • Pressure altitude = 1450 ft

The pressure altitude is calculated as:

Altitude = (29.92 – Altimeter reading) * 1,000 + Ground level

Substitute the known values

1450 = (29.92 - Altimeter reading) * 1000 + 700

Subtract 700 from both sides

750 = (29.92 - Altimeter reading) * 1000

Divide through by 1000

0.75 = 29.92 - Altimeter reading

Evaluate the like terms

Altimeter reading = 29.17

Hence, the altimeter reading is 29.17 from the Kollsman window is 29.17 in Hg

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4 0
2 years ago
Why aren't iron, cobalt and iron in the same group of elements?
aev [14]
Because iron is a metal and cobalt is a non-metal
3 0
3 years ago
The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by
Eduardwww [97]

Complete Question:

A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.

The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.

Answer:

Power = 54.07 W

Explanation:

Mass of the block = 10 kg

Angle made with the horizontal, θ = 60°

Distance covered, d = 5 m

Tension in the rope, T = 40 N

Coefficient of kinetic friction, \mu = 0.2

Let the Normal reaction = N

The weight of the block acting downwards = mg

The vertical resolution of the 40 N force, f_{y} = 40sin \theta

\sum f(y) = 0

N + 40 sin \theta - mg = 0\\N = -40sin60 + 10*9.81 = 0\\N = 63.46 N

\sum f(x) = 0

40 cos 60 - f_{r} - ma = 0\\ f_{r} = \mu N\\ f_{r} = 0.2 * 63.46\\ f_{r} = 12.69 N\\40cos 60 - 12.69-10a = 0\\7.31 = 10a\\a = 0.731 m/s^{2}

v^{2}  = u^{2} + 2as\\u = 0 m/s\\v^{2}  =  2 * 0.731 * 5\\v^{2}  = 7.31\\v = \sqrt{7.31} \\v = 2.704 m/s

Power, P = Fvcos \theta

P = 40 *2.704 cos60\\P = 54.074 W

7 0
3 years ago
a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance
aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b)  Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

  = \frac{KQ}{r^{2} } for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b)  Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

(c) Electric field at 3.10 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{3.10^{2} }

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{8^{2} }

E = - 0.77 x 10³ V/m

8 0
3 years ago
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Fittoniya [83]

The added weight of the sand puts more downward pressure on the wheels contacting the rails, which would cause the trains speed to decrease.

3 0
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