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tester [92]
3 years ago
11

An attacker at the base of a castle wall 3.80 m high throws a rock straight up with speed 9.00 m/s from a height of 1.70 m above

the ground. a) will the rock reach the top? b) if so what is its speed at the top? if not, what initial speed must it have to reach the top? c) find the change in speed of a rock thrown straight down from the top of the wall at an initial speed of 7.4 m/s and moving between the same two points.d) does the change in speed of the downward moving rock agree with the magnitude of the speed change of the rock moving upward between the same elevations? e) why does this agree or not agree.
Physics
1 answer:
blagie [28]3 years ago
8 0

Answer:

First of all the attack is gonna die because its 2020 who throws a rock up over a wall to kill someone(cavemen) and

it's gonna be ↓

explanation:

An attacker at the base of a castle wall 3.60 m high throws a rock straight up with speed 8.00 m/s from a height of 1.70 m above the ground.

(a) Will the rock reach the top of the wall?

(b) If so, what is its speed at the top? If not, what initial speed must it have to reach the top?

(c) Find the change in speed of a rock thrown straight down from the top of the wall at an initial speed of 8.00 m/s and moving between the same two points.

(d) Does the change in speed of the downward-moving rock agree with the magnitude of the speed change of the rock moving upward between the same elevations? Explain physically why it does or does not agree.

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A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she
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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;

f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L}

Where;

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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;

f_{(1  \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}}  }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L}  = 1.3225 \times f_1

f_{(1  \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1

f_{(1  \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz  = 740.6 \  Hz

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, L_{new} = 2/3·L

The value of the fundamental frequency will then be given as follows;

f_{(1  \, new)} =  \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \times \dfrac{2 \times L}{3} }  =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz =  840 \ Hz

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

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