Asked by Priyank
A hockey puck of mass 0.16 kg, sliding on a nearly frictionless surface of ice with a velocity of 2.0 m/s (E), strikes a second puck at rest with a mass of 0.17 kg. the first puck has a velocity of 1.5 (N 31 E) after the collision. Determine the velocity of the second puck after the collision.
Answers
Answered by
LENA
please help Ms. SUE!!!!
Answered by
anton
a proton flying to the east near the surface of the earth at 1000mls. what is the magnitude of the force the proton feels due to earths magnetic field
Answered by
Elena
X: m₁v₁=m₁u₁cosα+m₂u₂cosβ
Y: 0=m₁u₁sinα - m₂u₂cosβ
m₁v₁-m₁u₁cosα = m₂u₂cosβ …(1)
m₁u₁sinα= m₂u₂cosβ ………...(2)
Divide (2) by (1)
tanβ= m₁u₁sinα/ (m₁v₁-m₁u₁cosα) =
=sinα/ (1- cosα) =3.6
β=74.5º
u₂=m₁u₁sinα/ m₂cosβ= 0.16•2•sin31/0.17•sin 74.5 =1 m/s
Y: 0=m₁u₁sinα - m₂u₂cosβ
m₁v₁-m₁u₁cosα = m₂u₂cosβ …(1)
m₁u₁sinα= m₂u₂cosβ ………...(2)
Divide (2) by (1)
tanβ= m₁u₁sinα/ (m₁v₁-m₁u₁cosα) =
=sinα/ (1- cosα) =3.6
β=74.5º
u₂=m₁u₁sinα/ m₂cosβ= 0.16•2•sin31/0.17•sin 74.5 =1 m/s
Answered by
Elena
to Anton:
F=qvB? where q=e
v: m mile/s -> m/s
B= 5•10⁻⁵ T
F=qvB? where q=e
v: m mile/s -> m/s
B= 5•10⁻⁵ T
Answered by
Anonymous
4.5
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