here's the first part but for the 2nd one all I know is that the word "compression" goes on the spirals that are closer together.
hope this helps!
Unlike a longitudinal wave, the disturbance in a transverse wave is
2 answers:
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Answer:
The tension in each half of the rope, is approximately 4,908.8 N
Explanation:
The mass of the acrobat, m = 60 kg
The length of the rope, l = 10 m
The extent by which the center dropped = 30 cm = 0.3 m
Let, 'T' represent the tension in each half of the rope
Weight, W = Mass, m × The acceleration due to gravity, g
∴ W = m × g
The acceleration due to gravity, g ≈ 9.8 m/s²
∴ The weight of the acrobat, W = 60 kg × 9.8 m/s² ≈ 588 N
The angle the dropped rope makes with the horizontal, θ is given as follows;
θ = arctan((0.3 m)/(5 m)) = arctan(0.06) ≈ 3.434°
At equilibrium, the sum of vertical forces, = 0
The vertical component of the tension, , in each half of the rope is given as follows;
= T × sin(θ)
∴ = W + T × sin(θ) + T × sin(θ) = W + 2 × T × sin(θ)
Plugging in the values, with θ = arctan(0.06) for accuracy, we get;
588 N + 2 × T × sin(arctan(0.06) = 0
∴ 2 × -T × sin(arctan(0.06) = 588 N
-T= 588 N/(2 × sin(arctan(0.06)) = 4,908.81208 N ≈ 4,908.8 N
The tension in each half of the rope, T ≈ 4,908.8 N.
Answer:
The magnitude of applied force,parallel to the incline is 575.38 N and parallel to the floor is 605 N.
Explanation:
Given:
Mass of the piano = 190 kg
Inclined angle = 18 degree
Considering gravity, = 9.8
And
Using, and
<em>FBD diagram is attached with all the force acting on the floor and and the inclined. </em>
We have to find the magnitude of forces,when the man pushes it parallel to the incline and to the floor.
a.
When the man pushes it parallel to the incline.
Balancing the forces as .
⇒
⇒
⇒ Here it is negative as the force is acting downward.
⇒ Plugging the values of mass and angle
.
⇒
⇒ N
b.
When the force is parallel to the floor.
⇒
⇒
⇒ Plugging the values.
⇒
⇒ N
So,
The magnitude of applied force in inclined direction is 575.38 Newton and parallel to the floor is 605 N.
