A skier moving at 5.23 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220
1 answer:
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They were going at a velocity 4.07m/s
<u>Explanation:</u>
Distance s =5 m
initial velocity u= 0.8 m/s
Acceleration a =1.6m/s2
We have to calculate the velocity with which they were going afterwards i.e final velocity.
Use the equation of motion
They were going with a velocity 4.07 m/s afterwards.
Answer:
Explanation:
Let the critical angle be C .
sinC = 1 / μ where μ is index of refraction .
sinC = 1 /1.2
= .833
C = 56°
Then angle of refraction r = 90 - 56 = 34 ( see the image in attached file )
sin i / sinr = 1.2 , i is angle of incidence
sini = 1.2 x sinr = 1.2 x sin 34 = .67
i = 42°.
Answer:
Explanation:
The rod will act as pendulum for small oscillation .
Time period of oscillation
angular frequency ω = 2π / T
=
b )
ω = 20( given )
velocity = ω r = ω l
Let the maximum angular displacement in terms of degree be θ .
1/2 m v ² = mgl ( 1 - cosθ ) ,
[ l-lcosθ is loss of height . we have applied law of conservation of mechanical energy .]
.5 ( ω l )² = gl( 1 - cos θ )
.5 ω² l = g ( 1 - cosθ )
1 - cosθ = .5 ω² l /g
cosθ = 1 - .5 ω² l /g
θ can be calculated , if value of l is given .