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baherus [9]
3 years ago
12

Question 4 of 20 : Select the best answer for the question. 4.   Which of the following would lower the pitch of a string instru

ment's sound?        A. shortening the effective string length.   B. increasing the thickness of the string used.   C. adding tension within the string.   D. using less force to vibrate the string
Physics
1 answer:
Nady [450]3 years ago
4 0
The correct option is B.
There are four basic factors which affect the properties of stringed instruments, these are frequency, diameter [thickness], tension and density.
Diameter refers to the thickness of the string. Thick strings with large diameter vibrate slower and thus they have lower frequency and lower pitch compared with thin strings.<span />
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A block of ice with mass 2.00 kg slides 0.890m down an inclined plane that slopes downward at an angle of 28.3 degrees below the
Butoxors [25]

Answer:  The final speed, ignoring the effects of friction, will be 2.88 m/s.

Explanation:

The problem can be solved using two different physical principles. If we resort to Newton's second Law, we can say that, neglecting friction (which is reasonable for a ice block) the only force acting on the block  in the direction of the movement along the plane, is the component of the weight that is parallel to the slope, i.e.

Fp = mg sin 28.3º = ma ⇒ a=g sin 28.3º

Now, as we know that g = constant, we can use the following kinematic equation:

vf² - v₀² = 2 a x

if the block starts from rest, this means that v₀ = 0.

Replacing by the values in the equation, and solving for vf, we get;

vf = vf = \sqrt{2. 9.8.0.89. sin 28.3}  = 2.88 m/s

The other approach is using the conservation of energy principle:

When the block starts, it has some potential energy = mgh

This height h, can be expressed in terms of x (the length travelled by the block downward the plane) and the angle that forms with the horizontal, as follows:

h = x sin 28.3 (applying sin definition) ⇒ U = mg x sin 28.3

At the the end of the slide, the potential energy has been converted to kinetic energy, so we can write the following equation:

m. g. x. sin 28.3º = 1/2 m vf²

Simplifying, replacing by the values and solving for vf, we arrive to the same result as above.

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