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amm1812
1 year ago
13

Two motorcycles travel along a straight road heading due north. At t = 0 motorcycle 1 is at x = 50 m and moves with a constant s

peed of 6.50 m/s; motorcycle 2 starts from rest at x = 0 and moves with constant acceleration. Motorcycle 2 passes motorcycle 1 at the time t = 10.08.
Part B What is the position of the two motorcycles when motorcycle 2 passes motorcycle 1? Express your answer to three significant figures and include the appropriate units.

Part C What is the acceleration of motorcycle 2? Express your answer to three significant figures and include the appropriate units.​
Physics
1 answer:
Oduvanchick [21]1 year ago
3 0

Answer:

Vf = 23 m/s

Explanation:

First we need to find the distance covered by the motorcycle 2 when it passes motorcycle 1. Using the uniform speed equation for motorcycle 1:

s₁ = v₁t₁

where,

s₁ = distance covered by motorcycle 1 = ?

v₁ = speed of motorcycle 1 = 6.5 m/s

t₁ = time = 10 s

Therefore,

s₁ = (6.5 m/s)(10 s)

s₁ = 65 m

Now, for the distance covered by motorcycle 2 at the meeting point. Since, the motorcycle started 50 m ahead of motorcycle 2. Therefore,

s₂ = s₁ + 50 m

s₂ = 65 m + 50 m

s₂ = 115 m

Now, using second equation of motion for motorcycle 2:

s₂ = Vi t + (1/2)at²

where,

Vi = initial velocity of motorcycle 2 = 0 m/s

Therefore,

115 m = (0 m/s)(10 s) + (1/2)(a)(10 s)²

a = 230 m/100 s²

a = 2.3 m/s²

Now, using 1st equation of motion:

Vf = Vi + at

Vf = 0 m/s + (2.3 m/s²)(10 s)

Vf = 23 m/s

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The colored lines in the figure represent paths taken by different people walking around in a city. Assume that each city block
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Answer: 592.37m

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a) 1.59(10)^{-19} J

b) 2.34(10)^{12} electrons

Explanation:

The photoelectric effect consists of the emission of electrons (electric current) that occurs when light falls on a metal surface under certain conditions.  

If the light is a stream of photons and each of them has energy, this energy is able to pull an electron out of the crystalline lattice of the metal and communicate, in addition, a kinetic energy.  

<u>This is what Einstein proposed: </u>

Light behaves like a stream of particles called photons with an energy  E:

E=\frac{hc}{\lambda} (1)  

So, the energy E of the incident photon must be equal to the sum of the Work function \Phi of the metal and the kinetic energy K of the photoelectron:  

E=\Phi+K (2)  

Where \Phi=6.94(10)^{-19} J is the minimum amount of energy required to induce the photoemission of electrons from the surface of Titanium metal.

Knowing this, let's begin with the answers:

<h3 /><h3>a)  Maximum possible kinetic energy of the emitted electrons (K)</h3>

From (1) we can know the energy of one photon of 233 nm light:

E=\frac{hc}{\lambda}

Where:

h=6.63(10)^{-34}J.s is the Planck constant  

\lambda=233 (10)^{-9} m is the wavelength

c=3 (10)^{8} m/s is the speed of light

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E=8.53(10)^{-19} J (4) This is the energy of one 233 nm photon

Substituting (4) in (2):

8.53(10)^{-19} J=6.94(10)^{-19} J+K (5)  

Finding K:

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Since one photon of 233 nm is able to free at most one electron from the Titanium metal, we can calculate the following relation:

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Answer:

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