If the back of the truck is 1.3 m above the ground and the ramp is inclined at 22 ∘ , how much time do the workers have to get t
o the piano before it reaches the bottom of the ramp?
1 answer:
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
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a) The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.
b) The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.
Explanation:
Statement is incomplete. The complete description is now described below:
<em>A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turns on, causing an acceleration of 0.250 m/s2 in the x direction. The acceleration lasts for 45.0 s, at which point the thruster turns off. </em>
<em>(a) What is the magnitude of the satellite's velocity when the thruster turns off</em>
<em>(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis. ° counterclockwise from the +x-axis</em>
Let be x and y-directions orthogonal to each other and the satellite is accelerated uniformly from rest in the +x direction and moves at constant velocity in the +y direction. The velocity vector of the satellite (), measured in meters per second, is:
Where:
- Initial velocity in +x direction, measured in meters per second.
- Acceleration in +x direction, measured in meter per square second.
- Time, measured in seconds.
- Velocity in +y direction, measured in meters per second.
If we know that ,
,
and
, the final velocity of the satellite is:
a) The magnitud of the satellite's velocity can be found by the resource of the Pythagorean Theorem:
The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.
b) The direction of the satellite's velocity when the thruster turns off is determined with the help of trigonometric functions:
The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.