What is the coefficient of kinetic friction between the block and the surface? Express your answer using two significant figures.
Answer:
0.39
Explanation:
Given information
m1=1.7 Kg
m2=0.011 Kg
v2=670 m/s
d=2.4 m
m2v2=(m1+m2)v hence 
and also 
The deceleration due to friction is given by
therefore 
Therefore, 
Answer:
Explanation:
A )
The ball floats with half of it exposed above the water level . So it must have density half that of water . In other words its density must have been 500 kg / m³
B )
Tension in the ball will be equal to net force acting on the ball
Net force on the ball = buoyant force - weight .
4/3 x π x .21³ x 10⁻⁶ x 9.8 ( 1000 - 893 )
= 40.65 x 10⁻⁶ N .
C )Tension in the 3 rd ball will be equal to net force acting on the ball
Net force on the ball = weight - buoyant force
= 4/3 x π x .21³ x 10⁻⁶ x 9.8 ( 1320 - 1000 )
= 121.6 x 10⁻⁶ N .
Answer:
(a) x=ASin(ωt+Ф₀)=±(√3)A/2
(b) x=±(√2)A/2
Explanation:
For part (a)
V=AωCos(ωt+Ф₀)⇒±0.5Aω=AωCos(ωt+Ф₀)
Cos(ωt+Ф₀)=±0.5⇒ωt+Ф₀=π/3,2π/3,4π/3,5π/3
x=ASin(ωt+Ф₀)=±(√3)A/2
For part(b)
U=0.5E and U+K=E→K=0.5E
E=K(Max)
(1/2)mv²=(0.5)(1/2)m(Vmax)²
V=±(√2)Vmax/2→ωt+Ф₀=π/4,3π/4,7π/4
x=±(√2)A/2
3.1 the only reason i know this is cause i got it wrong
