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sergejj [24]
2 years ago
7

A 60 kg bicyclist going 2 m/s increased his work output by 1,800 J. What was his final velocity?

Physics
2 answers:
Alinara [238K]2 years ago
6 0

<em>8 </em>was the correct answer, thanks

Bas_tet [7]2 years ago
3 0
Bicyclist initial kinetic energy is Ek=(1/2)*m*v² where m is his mass and v is his speed and that is equal to:

Ek=(1/2)*60*2²=120 J.

When we add the increased work output, we get the total kinetic energy:

Ek(total)=Ek+W= 120 J + 1800 J= 1920 J

So Ek(total)=1920 J = (1/2)*m*V² where V is the speed after the bicyclist increased his work output. So lets solve for V:

(1/2)*60*V²=1920

30*V²=1920, we divide by 30,

V²=64, and take the square root of both sides,

V=8 m/s. 

So the speed of the bicyclist after the increased work output is V=8 m/s.
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a film of transparent material 120 nm thick and having refractive index 1.25 is placed on a glass sheet having refractive index
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Answer:

a) 600nm

b) 300nm

Explanation:

the path difference = 2t  

t = thickness of the film

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n = refractive index of glass

(a)

for destructive interference 2t = L'/2 = L/2n

L = 4*t*n

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(b)

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Answer:

0.00001266 m

Explanation:

D = Distance from source to screen

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y_0=0

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y_1=\frac{1\times 633\times 10^{-9}\times 7}{d}\\\Rightarrow d=\frac{1\times 633\times 10^{-9}\times 7}{0.35}\\\Rightarrow d=0.00001266\ m

The slit separation is 0.00001266 m

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Harmonics problem. A square wave of frequency f contains harmonics (sine waves) at f, 3f, 5f, 7f, ... . Suppose a system respond
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Answer:

B

Explanation:

A square of frequency of consists of the infinite sum of sine waves, whose frequencies are the odd multiples of the main frequency f i.e f, 3f,5f, 7f, ... . Given that the range of frequencies, to which the system responds is 20-40 kHz, for a square wave of frequency 10kHz we need to look for the harmonics whose frequencies are in the systems' respond range, which are the  harmonics of 20, 30 and 40 kHz

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<span>an object that appears black absorbs al color. an object that appears white reflects all colors.</span>
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A particle of mass m=5.00 kilograms is at rest at t=0.00 seconds. a varying force f(t)=6.00t2−4.00t+3.00 is acting on the partic
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<span>v =Integral(F/m)dtv=Integral(F/m)dt =â«50(6t2â’4t+3)/5dt=â«05(6t2â’4t+3)/5dt =1/5â—(2t3â’2t2+3t)|(0,5)=1/5â—(2t3â’2t2+3t)|(0,5) =1/5â—(250â’50+15)=1/5â—(250â’50+15) =215/5=215/5 =43m/s.</span>
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Read 2 more answers
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