A box of mass 60 kg is at rest on a horizontal floor that has a static coefficient of friction of 0.6 and a kinetic coefficient
1 answer:
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Answer:
(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz
(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz
Explanation:
(a) The fundamental, f₁, frequency is given as follows;
Where;
T = The tension in the string
μ = The linear density of the string
L = The length of the string
f₁ = The fundamental frequency = 560 Hz
If the tension in the string is increased by 15%, we will have;
Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz
(b) When the string length is decreased by one-third, we have;
The new length of the string, = 2/3·L
The value of the fundamental frequency will then be given as follows;
When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.
Answer:
Y component = 32.37
Explanation:
Given:
Angle of projection of the rocket is,
Initial velocity of the rocket is,
A vector at an angle with the horizontal can be resolved into mutually perpendicular components; one along the horizontal direction and the other along the vertical direction.
If a vector 'A' makes angle with the horizontal, then the horizontal and vertical components are given as:
Here, as the velocity is a vector quantity and makes an angle of 33.6 with the horizontal, its Y component is given as:
Plug in the given values and solve for . This gives,
Therefore, the Y component of initial velocity is 32.37.