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timurjin [86]
4 years ago
12

Photons of wavelength 115 nm are incident on a metal. The most energetic electrons ejected from the metal are bent into a circul

ar arc of radius 1.04 cm by a magnetic field having a magnitude of 6.50 10-4 T. What is the work function of the metal?
Physics
1 answer:
katovenus [111]4 years ago
7 0

Answer:

6.8 eV

Explanation:

As you know, momentum (P) equals the mass of an object (M) times its velocity (V)

p=mv

Also, the momentum of the particle can be expressed in terms of the radius,

charge and field:

p=mv = qrB

K.Emax = p²/2m= q²r²B²/ 2m

K.Emax= (1.6x10^{-19})² x (0.0104)² x (6.50 x10^{-4})² / (2 x 9.1x10^{-31})

K.E max= 1.16 x 10^{-48} / 1.82 x10^{-30

K.Emax= 6.37 10^{-19} J

work function 'Ф' = E- K.E max

since E= hc/ λ

Ф= hc/ λ -  K.E max => \frac{(6.63*10^{-34})(3*10^{8} ) }{115*10^{-9} } - 6.37*10^{-19}

Ф = 1.09 x 10^{-18} J

Ф = 1.09 x 10^{-18} / 1.6x10^{-19}

Ф =6.8 eV

Therefore, the work function of the metal is 6.8 eV

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Three blocks are arranged in a stack on a frictionless horizontal surface. The bottom block has a mass of 37.0 kg. A block of ma
Alex777 [14]

Answer:

N₂ = 503.8 N

Explanation:

given,

mass of bottom block = 37 Kg

mass of middle block = 18 Kg

mass of the top block = 16 Kg

force acting on the top block = 170 N

force on the block at top

N₁ be the normal force from block at middle

now,

N₁ = 170 + m g

N₁ = 170 + 16 x 9.8

now, force on block at middle

N₂ be the normal force exerted by the bottom block

N₂ = N₁ + m₂ g

N₂ = 326.8 + 18 x 9.8

N₂ = 503.8 N

hence, normal force by bottom block is equal to N₂ = 503.8 N

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4 years ago
Substance A has a higher heat capacity than does substance B, and substance B has a higher heat capacity than does substance C.
LekaFEV [45]
BABC that what I Think
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3 years ago
An object with a mass of 1.5 kg changes its velocity +15 m/s during a time interval of 3.5 seconds what impulse was delivered to
Arada [10]
The answer is 10.5 kg m/s

Impulse (I) is the multiplication of force (F) and time interval (Δt): I = F · Δt

Force (F) is the multiplication of mass (m) and acceleration (a): F = m · a

Acceleration (a) can be expressed as change in velocity (v) divided by time interval (Δt): a = Δv/Δt

So: 
a = Δv/Δt         ⇒ F = m · a = m · Δv/Δt
F = m · Δv/Δt   ⇒ I = m · Δv/Δt · Δt
Since Δt can be cancelled out, impulse can be expressed as:
I = m · Δv = m · (v2 - v1)

It is given:
m = 1.5 kg
v1 = 15 m/s
v2 = 22 m/s

I = 1.5 · (22 - 15) = 1.5 · 7 = 10.5 kgm/s.
7 0
4 years ago
A block is released from the top of a frictionless incline plane as pictured above. If the total distance travelled by the block
kotegsom [21]

Complete Question

The diagram for this question is showed on the first uploaded image (reference homework solutions )

Answer:

The  velocity at the bottom is  v  = 11.76  \ m/ s

Explanation:

From the question we are told that

   The  total distance traveled is  d =  1.2  \ m

    The mass of the block is  m_b  =  0.3 \ kg

      The  height of the block from the ground is h =  0.60 m  

According the law of  energy  

   PE  =  KE

Where  PE  is the potential energy which is mathematically represented as

      PE  =  m * g  *  h

substituting values

     PE  =   3 *  9.8  *  0.60

      PE  =  17.64 \  J

So

   KE  is the kinetic energy at the bottom which is mathematically represented as

          KE  =  \frac{1}{2}  *  m v^2

So

      \frac{1}{2}  *  m* v ^2  =  PE

substituting values  

  =>    \frac{1}{2}  *  3 * v ^2  = 17.64

=>       v  = \sqrt{ \frac{ 17.64}{ 0.5 * 3 } }

=>    v  = 11.76  \ m/ s

4 0
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