A wave has a wavelength of 3.0 m, a frequency of 25.0 Hz, and an amplitude of 14.0 cm. The wave travels in the positive x-direct
ion and has a displacement of zero at t = 0 and x = 0. How many complete oscillations has the wave made at t = 20.0 s and x = 4.2 m?
1 answer:
Answer:
The total number of oscillations made by the wave during the time of travel is 1.4 Oscillations. Strictly speaking, the number of complete oscillations is 1.
Explanation:
The required quantity is the number of complete oscillations made by the traveling wave. The amplitude time and frequency are not needed to calculate the number of oscillations as it is the ratio of the distance traveled to the wavelength( minimum distance that must be traveled to complete one oscillation) of the wave. So the total number of oscillations is 1.4 while the number of complete oscillations is 1 (strictly speaking). The detailed solution to this question can be found in the attachment below. Thank you!
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1) The net force is 16 N to the right
2) The net force is 98 N to the left
3) The net force is 0.5 N downward
4) The net force is 170 N to the right
5) The net force is 175 N to the right
Explanation:
1)
To find the net force, we have to analyze all the forces acting on the box.
We have:
- Force to the right:
, the applied force
- Force to the left:
, the force of friction
- Force to the bottom:
, the weight of the box (the weight is always downward vertically)
- Force to the top:
. This is the normal force, which is the reaction force exerted by the table on the box: it points upward and counterbalances the weight of the box, preventing it from falling down)
Therefore, the horizontal net force is
(to the right)
While the vertical force is
So the net force is 16 N to the right.
2)
In this case, we have the following forces:
downward, the weight of the ball
to the left, the force that kicks the ball
to the right, the force of friction
upward, the normal reaction exerted by the field on the ball
Therefore, the horizontal net force is
(to the left)
While the vertical force is
(downward)
And so, the net force is 98 N to the left.
3)
The force acting on the squirrel in this problem are:
downward, the weight of the squirrel
upward, the air resistance, acting upward
Both forces act vertically and there are no other forces acting in other directions, therefore the net force on the squirrel is simply equal to the net force on the vertical direction, which is:
And since the weight is larger than the air resistance, the direction of the net force is downward.
4)
The forces acting on Monkey are:
is the force applied to the right by Bunny
is the force applied by Deer from the left (so, also on the right)
is the weight of Monkey, downward
is the normal reaction exerted by the surface, upward
So, the net force in the horizontal direction is
(to the right)
While the net force in the vertical direction is
And therefore the net force is 170 N to the right
5)
The forces acting on Deer are:
to the right, the combined force applied by Bunny and Monkey
to the left, the force of friction
downward, the weight of the deer
upward, the normal reaction from the surface that balances the weight
So the net horizontal force is
to the right
While the net vertical force is
So the net force is 175 N to the right.
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Answer:
miu = 0.31
Explanation:
The friction force is defined as the product of the normal force by the coefficient of friction.
where:
f = friction force = 3 [N]
miu = friction coefficient
N = normal force [N]
The normal force on a horizontal surface can be calculated by means of the product of mass by gravity.
where:
m = mass = 1 [kg]
g = gravity acceleration = 9.81 [m/s²]
Now replacing in the equation above, we can find the friction coefficient.