1. Two forces F~ 1 and F~ 2 are acting on a block of mass m=1.5 kg. The magnitude of force F~ 1 is 12N and it makes an angle of
θ = 37◦ with the horizontal as shown in figure-1. The block is sliding at a constant velocity over a frictionless floor.
(a) Find the value of the normal force on the block.
(b) Find the magnitude of force F~2 that is acting on the block
(c) Find the magnitude of force F~ 2 if the block accelerates with a magnitude of a = 2.5 m/s2 along the direction of F~ 2 .
(a) Find the value of the normal force on the block.
(b) Find the magnitude of force F~2 that is acting on the block
(c) Find the magnitude of force F~ 2 if the block accelerates with a magnitude of a = 2.5 m/s2 along the direction of F~ 2 .
1 answer:
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Answer:
Explanation:
For third normal mode of vibration
l = , λ is wavelength , l is length of string .
.4 =
λ = .267 m
velocity =
T is tension and m is mass unit length
m = .5 x 10⁻³ / 40 x 10⁻²
= .00125 kg / m
Putting the values
velocity =
= 253 m /s
frequency
= velocity / λ
= 253 / .267
= 947.5 Hz .
Answer: 5 units
Let's begin by stating clear that movement is the change of position of a body at a certain time. So, during this movement, the body will have a trajectory and a displacement, being both different:
The trajectory is the path followed by the body (is a scalar magnitude).
The <u>displacement</u> is the distance in a straight line between the initial and final position (is a vector magnitude).
According to this, in the description of the object (figure attached) placed at 0 on a number line and moving some units to the left and some oter units to the right, we are talking about the path followed by the object, hence its trajectory. So, 13 units is its trajectory.
But, if we talk about displacement, we have to draw a straight line between the initial position of the object (point 0) to its final position (point 5).
Now, being this an unidimensional problem, the displacement vector for this object is 5 units.
