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arsen [322]
3 years ago
13

A 5.00-g bullet leaves the muzzle of a rifle with a speed of 320 m/s. the expanding gases behind it exert what force on the bull

et while it is traveling down the barrel of the rifle, 0.820 m long? assume constant acceleration and negligible friction.
Physics
1 answer:
muminat3 years ago
3 0

Answer;

= 312 Newtons

Explanation;

The bullet has a mass of 0.005 kg, and a velocity of 320 m/s, so we need to find it's final kinetic energy.  

KE = 1/2*m*v^2

      = 1/2*0.005*320^2

      = 256 Joules.  

Divide this by the distance over which this energy was received and you have the force that provided that energy.  

  = 256/0.820 = 312.195 Newtons  

Rounded off, this is 312 N

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What is the threshold frequency ν0 of cesium? note that 1 ev (electron volt)=1.60×10−19 j. express your answer numerically in he
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The threshold frequency fo of Cesium is 4.83 × 10¹⁴ Hz

<h3>Further explanation</h3>

The term of package of electromagnetic wave radiation energy was first introduced by Max Planck. He termed it with photons with the magnitude is:

\large {\boxed {E = h \times f}}

<em>E = Energi of A Photon ( Joule )</em>

<em>h = Planck's Constant ( 6.63 × 10⁻³⁴ Js )</em>

<em>f = Frequency of Eletromagnetic Wave ( Hz )</em>

The photoelectric effect is an effect in which electrons are released from the metal surface when illuminated by electromagnetic waves with large enough of radiation energy.

\large {\boxed {E = \frac{1}{2}mv^2 + \Phi}}

\large {\boxed {E = qV + \Phi}}

<em>E = Energi of A Photon ( Joule )</em>

<em>m = Mass of an Electron ( kg )</em>

<em>v = Electron Release Speed ( m/s )</em>

<em>Ф = Work Function of Metal ( Joule )</em>

<em>q = Charge of an Electron ( Coulomb )</em>

<em>V = Stopping Potential ( Volt )</em>

Let us now tackle the problem!

<u>Given:</u>

Φ = 2.1 eV = 2 × 1.60 × 10⁻¹⁹ Joule = 3.20 × 10⁻¹⁹ Joule

<u>Unknown:</u>

fo = ?

<u>Solution:</u>

\Phi = h \times fo

3.20 \times 10^{-19} = 6.63 \times 10^{-34} \times fo

fo = ( 3.20 \times 10^{-19} ) \div ( 6.63 \times 10^{-34} )

\large {\boxed{fo = 4.83 \times 10^{14} ~ Hz} }

<h3>Learn more</h3>
  • Photoelectric Effect : brainly.com/question/1408276
  • Statements about the Photoelectric Effect : brainly.com/question/9260704
  • Rutherford model and Photoelecric Effect : brainly.com/question/1458544
  • Photoelectric Threshold Wavelength : brainly.com/question/10015690

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Quantum Physics

Keywords: Quantum , Photoelectric , Effect , Threshold , Frequency , Electronvolt

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How do fungi get food?
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Two carts on a straight track collide head on. The first cart was moving at 3.6 m/s in the positive x direction and the second w
zubka84 [21]

Answer:

vf₁  = -7.2 m/s :  The first car moves in the negative x direction after the collision.

Explanation:

Theory of collisions  

Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:  

p=m*v  

where  

p:Linear momentum  

m: mass  

v:velocity  

There are 3 cases of collisions : elastic, inelastic and plastic.  

For the three cases the total linear momentum quantity is conserved:  

P₀ = Pf   Formula (1)  

P₀ :Initial linear momentum quantity  

Pf : Final linear momentum quantity  

Data

m₁ = m kg : mass of the first car

m₂= 5m kg : mass of the second car

v₀₁ = 3.6 m/s : Initial velocity of m₁  , to the +x axis :

v₀₂= -2.4 m/s m/s : Initial velocity of m₂ ,  to the -x axis

vf₂= -0.24 m/s m/s : Final velocity of m₂ ,  to the -x axis

Problem development

We appy the formula (1):

P₀ = Pf  

m₁*v₀₁ + m₂*v₀₂ = m₁*vf₁ + m₂*vf₂  

We assume that the first car moves in the positive x direction after the collision., so, the sign of the final speeds is positive:

(m)*( 3.6) + (5m)*( -2.4) = (m)*vf₁ +(5m)*(- 0.24)

3.6m  - 12m=  m*vf₁ - 1.2m

We divided by m both sides of the equation

3.6 - 12= vf₁ -1.2

3.6 - 12 +1.2 = vf₁

-7.2  = vf₁

vf₁  = -7.2 m/s  : The first car moves in the negative x direction after the collision.

4 0
2 years ago
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