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Luden [163]
3 years ago
10

What is the momentum of 15 kg bicycle moving at 12 m/s

Physics
1 answer:
azamat3 years ago
8 0
And the answer is 5/4 and decimals to 1.25.
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Read 2 more answers
2) Two railway tracks are parallel to west east direction. Along one track, train A moves with a speed of 45 m/s from
OLEGan [10]

Answer:

Relative\ Velocity = 105m/s

Explanation:

Given

V_A = 45m/s

V_B = 60m/s

Required

Determine the speed of B w.r.t A

The question implies that, we determine the relative velocity of B w.r.t A

Because both trains are moving towards one another, the required velocity is a sum\ of\ velocities\ of both trains:

This is shown below:

Relative\ Velocity = V_A + V_B

Relative\ Velocity = 45m/s + 60m/s

Relative\ Velocity = 105m/s

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3 years ago
A chinook salmon needs to jump a waterfall that is 1.5 m high. If the fish starts from a distance of 1.00 m from the base of the
____ [38]
 <span>let the fsh jump with initial velocity (u) in direction (angle p) with horizontal 

it can cross and reach top of trajectory if its top height h = 1.5m 
and horizontal distance d = (1/2) Range 
--------------------------------------... 
let t be top height time 
at top height, vertical component of its velocity =0 
vy = 0 = u sin p - gt 
t = u sin p/g 
h = [u sin p]*t - 0.5 g[t[^2 
1.5 = u^2 sin^2 p/g - u^2 sin^2 p/2g 
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u^2 sin^2 p = 1.5*2*9.8 = 29.4 
u sin p = 5.42 m/s >>>>>>>>>>>>>>> V-component 
===================== 
t = HALF the time of flight 
d = (1/2) Range (R) = (1/2) [2 u^2 sin p cos p/g] 
1 = u^2 sin p cos p/g 
u sin p * u cos p = 9.8 
5.42 * u cos p = 9.8 
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check>> 
u = sqrt[u^2 cos^2 p + u^2 sin^2 p] = 5.71 m/s 
u < less than fish's potential jump speed 6.26 m/s 

so it will able to cross</span>
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