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gizmo_the_mogwai [7]
2 years ago
8

2. A ladder 16cm long weighs 420N and it's centre of gravity is 7m from one end.It is carried horizontally by two boys,each hold

ing it 1m from the ends. Show that the forces excreted by them are about 180N and 240N​
Physics
2 answers:
ipn [44]2 years ago
8 0

Answer:

well the answers are 180 and 240 obviously

Explanation:

Download pdf
MariettaO [177]2 years ago
7 0

Answer:

Explanation:

Givens

I take the ladder to be 16 m not 16 cm

The position of the boys.

Each boy is 1 m from the end

The mass of the ladder is 6m form one of the boys.

The fulcrum is at his shoulders,

The other boy is 14 feet from the fulcrum

Equation

420 *6 = 14 * x The boy nearest the fulcrum contributes nothing because he is 0 distance to the fulcrum.

Solution

420*6 = 14*x                Divide by 14

420*6/14 = 14x/14        Combine

30 *6 = x

x = 180

Answer

So the boy furthest from the fulcrum lifts 180 N of the 420N ladder

The other boy contributes 420 - 180 = 240 N

Both boys are represented by an arrow pointing upward.

The ladder's mass points down

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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
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

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What is the wavelength associated with 0.113kg ball traveling with velocity of 43 m/s?
lesya [120]

Answer:

2.73×10¯³⁴ m.

Explanation:

The following data were obtained from the question:

Mass (m) = 0.113 Kg

Velocity (v) = 43 m/s

Wavelength (λ) =?

Next, we shall determine the energy of the ball. This can be obtained as follow:

Mass (m) = 0.113 Kg

Velocity (v) = 43 m/s

Energy (E) =?

E = ½m²

E = ½ × 0.113 × 43²

E = 0.0565 × 1849

E = 104.4685 J

Next, we shall determine the frequency. This can be obtained as follow:

Energy (E) = 104.4685 J

Planck's constant (h) = 6.63×10¯³⁴ Js

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f = 104.4685 / 6.63×10¯³⁴

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Finally, we shall determine the wavelength of the ball. This can be obtained as follow:

Velocity (v) = 43 m/s

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43 = λ × 15.76×10³⁴

Divide both side by 15.76×10³⁴

λ = 43 / 15.76×10³⁴

λ = 2.73×10¯³⁴ m

Therefore, the wavelength of the ball is 2.73×10¯³⁴ m.

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