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faust18 [17]
3 years ago
6

3. A 3.0 g bullet traveling at a speed of 400 m/s enters a tree and exits the other side with a speed of 200 m/s. Where did the

I bullet's lost KE go, and what was the energy transferred? Solution:​
Physics
1 answer:
iragen [17]3 years ago
4 0

Answer:I had two questions: 1) Shouldnt the equation be : Change of internal energy= Change in Ke or KEi +Ui=KEf+Uf .. I cant seem to understand how the change in KE+ Change in Internal Energy equals 0 2) Also, it has a note which says that if the bullet alone was chosen as a system, then work would be done on it and the heat transfer woudl occur. How would one formulate an equation for this?

Explanation:

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The answer to your question is: Iron

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A banked circular highway curve is designed for traffic moving at 58 km/h. The radius of the curve is 201 m. Traffic is moving a
skelet666 [1.2K]

Answer:0.077

Explanation:

Given

banked designed for traffic moving at 58 km/h\approx 16.11 m/s

Radius of the curve 201 m

Actual traffic velocity =37 km/h approx  10.27 m/s

For banking of road

tan\theta =\frac{v^2}{rg}

tan\theta =\frac{16.11^2}{201\times 9.81}

\theta =7.49^{\circ}

Centripetal acceleration is given by

a=\frac{v^2}{r}

Taking component of centripetal acceleration

along and perpendicular to surface

a_{parallel}=\frac{v^2cos\theta }{r}

a_{perpendicular}=\frac{v^2sin\theta }{r}

From FBD

mgsin\theta -f_s=ma_{parallel}

f_s=mgsin\theta -ma_{parallel}----1

where f_s is frictional force

N-mgcos\theta =ma_{perpedicular}

N=mgcos\theta +ma_{perpedicular}----2

and we know coefficient of friction is given by

\mu =\frac{f_s}{N}

\mu =\frac{mgsin\theta -ma_{parallel}}{mgcos\theta +ma_{perpedicular}}

\mu =\frac{gsin\theta -\frac{v^2cos\theta }{r}}{gcos\theta +\frac{v^2sin\theta }{r}}

\mu =\frac{1.2804-0.5202}{9.726+0.068}

\mu =0.077

7 0
3 years ago
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