Asked by Kyle
During a neighborhood baseball game in a vacant lot, a particularly wild hit sends a 148-g baseball crashing through the pane of a second-floor window in a nearby building. The ball strikes the glass at 10.1 m/s, shatters the glass as it passes through, and leaves the window at 8.95 m/s with no change of direction. Calculate the magnitude of this impulse.
And,
he ball is in contact with the glass for 10.1 ms as it passes through. Find the magnitude of the average force of the glass on the ball (a positive number).
And,
he ball is in contact with the glass for 10.1 ms as it passes through. Find the magnitude of the average force of the glass on the ball (a positive number).
Answers
Answered by
Elena
Δp=FΔt
Δp= mv₂-(-mv₁)= m(v₂+v₁),
m(v₂+v₁)=FΔt
F= m(v₂+v₁)/Δt=
=0.148•(10.1+8.95)/10.1•10⁻³=132.6 N
Δp= mv₂-(-mv₁)= m(v₂+v₁),
m(v₂+v₁)=FΔt
F= m(v₂+v₁)/Δt=
=0.148•(10.1+8.95)/10.1•10⁻³=132.6 N
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