motorcycle starting from rest cover 200 metre distance in 6 second calculate final velocity and acceleration of the motorcycle

1 answer:

6 0

<h2>Answer:</h2><h2>V=200m</h2><h2><em><u>time=6s</u></em></h2><h2><em><u>time=6s initial velocity=0</u></em></h2><h2><em><u>time=6s initial velocity=0 final velocity=200/6</u></em></h2><h2><em><u>time=6s initial velocity=0 final velocity=200/6=100/3m/s </u></em></h2><h2><em><u>time=6s initial velocity=0 final velocity=200/6=100/3m/s v=u+a</u></em></h2><h2><em><u>time=6s initial velocity=0 final velocity=200/6=100/3m/s v=u+aputting value,</u></em></h2><h2><em><u>time=6s initial velocity=0 final velocity=200/6=100/3m/s v=u+aputting value, 200/3=0+ax6</u></em></h2><h2><em><u>time=6s initial velocity=0 final velocity=200/6=100/3m/s v=u+aputting value, 200/3=0+ax6 200/3-ax6 </u></em></h2><h2><em><u>time=6s initial velocity=0 final velocity=200/6=100/3m/s v=u+aputting value, 200/3=0+ax6 200/3-ax6 200/3/6=a </u></em></h2><h2><em><u>time=6s initial velocity=0 final velocity=200/6=100/3m/s v=u+aputting value, 200/3=0+ax6 200/3-ax6 200/3/6=a 200/18-a </u></em></h2><h2><em><u>time=6s initial velocity=0 final velocity=200/6=100/3m/s v=u+aputting value, 200/3=0+ax6 200/3-ax6 200/3/6=a 200/18-a a 100/9m/s^2</u></em></h2>

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What is the maximum value the string tension can have before the can slips? The coefficient of static friction between the can a

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Answer:

T= 38.38 N

Explanation:

Here

mass of can = m = 3 kg

g= 9.8 m/sec2

angle θ = 40°

From figure we see the vertical and horizontal component of tension force T

If the can is to slip - then horizontal component of tension force should become equal to force of friction.

First we find force of friction

Fs= μ R

where

μ = 0.76

R = weight of can = mg = 3 × 9.8 = 29.4 N

Now horizontal component of tension

Tx= T cos 40 = T× 0.7660 N

==>T× 0.7660 = 29.4

==> T= 38.38 N

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