What can waves be transmitted by? How do you know?
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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
ANSWER
T₂ = 10.19N
EXPLANATION
Given:
• The mass of the ball, m = 1.8kg
First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,
In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,
The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,
We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,
Solve for T₁,
Now, we use the second equation to find the tension in the horizontal string,
Solve for T₂,
Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.
