1.5 x 10^3 standard notation
1 answer:
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Refer to the diagram shown below.
Assume that
(a) The piano rolls down on frictionless wheels,
(b) Wind resistance is negligible.
The distance along the ramp is
d = (1.3 m)/sin(22°) = 3.4703 m
The component of the piano's weight along the ramp is
mg sin(22°)
If the acceleration down the ramp is a, then
ma = mg sin(22°)
a = g sin(22°) = (9.8 m/s²) sin(22°) = 3.671 m/s²
The time, t, to travel down the ramp from rest is given by
(3.4703 m) = 0.5*(3.671 m/s²)*(t s)²
t² = 3.4703/1.8355 = 1.8907
t = 1.375 s
Answer: 1.375 s
Assume that
(a) The piano rolls down on frictionless wheels,
(b) Wind resistance is negligible.
The distance along the ramp is
d = (1.3 m)/sin(22°) = 3.4703 m
The component of the piano's weight along the ramp is
mg sin(22°)
If the acceleration down the ramp is a, then
ma = mg sin(22°)
a = g sin(22°) = (9.8 m/s²) sin(22°) = 3.671 m/s²
The time, t, to travel down the ramp from rest is given by
(3.4703 m) = 0.5*(3.671 m/s²)*(t s)²
t² = 3.4703/1.8355 = 1.8907
t = 1.375 s
Answer: 1.375 s