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3 years ago

a 55.0 kg person having a denity of 1010 kg/m^3 is underwater. What is the net force acting on the person?

Physics

1 answer:

bija089 [108]3 years ago

6 0

Answer:

The net force acting on the person is 5.34 N acting upwards

Explanation:

The information given are;

Mass of the person = 55.0 kg

Density of the person = 1010 kg/m³

Body volume of the person = Mass/Density = 55/1010 = 0.0545 m³

Weight of the person = Mass × Gravity = 55 × 9.81 = 539.55 N↓

Where the density of water = 1000 kg/m³, we have;

Upthrust = weight of the water displaced = Mass of the water displaced × Gravity

Mass of the water displaced = Density of water × Volume of the water

Mass of the water displaced = Density of water × Body volume of the person

= 1000 × 0.0545 = 54.5 kg

The Upthrust = 54.5 × 9.81 = 534.21 N↑

Given that the net force, $F_{Net}$ acting is the sum of the mass of forces acting on the person which are the weight acting downward and the upthrust acting upwards

We have;

F(Net) = 534.21↑ - 539.55↓ = 5.34 N↑

The net force acting on the person = 5.34 N upwards.

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Answer:

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3 years ago

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3 years ago

Read 2 more answers

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

lawyer [7]

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;

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;

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

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

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

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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3 years ago