Compare the amplitude of these waves.

Given,

Force = 10,000 N

Distance = 30 m

Time = 40 s

Work done = force × distance

Work = 10,000 × 30

Work = 300,000 J

______

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The work done on the object is equal to the power it experiences which is equal to 735 watts.

Work done is defined as the product of force applied and the distance over which the force is applied on the object. Work is done on an object when a force is applied to an object and the object is moved through a particular distance.

The time rate of doing work is called as power, which is designated by the symbol P. It is measured in the units of watts, where one watt is equal to one joule energy applied per second time.

Work done on an object is equal to the force (F) multiplied by the distance d in the direction of the force. Where, time interval is Δt.

P = W/Δt = F × d/Δt = m × g × h/Δt = 100 × 9.8 × 30 / 40

Power = 735 watts

The power of the object is 735 watts.

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Answer:

Work = Force × Distance

Therefore Force = Work/Distance

= 45 joules/3 metres

= 15 newton

Explanation:

Answer:

15

Explanation:

15

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question

mass = 80 kg

acceleration = 2 m/s²

We have

force = 80 × 2 = 160

We have the final answer as

Hope this helps you

The maximum displacement angle of the bob is 13⁰.

The given parameters;

The maximum displacement of the bob is calculated by applying the principle of conservation of energy;

[tex]P.E = K.E\\\\mgh = \frac{1}{2} mv^2\\\\h = \frac{v^2}{2g} \\\\h = \frac{0.8^2}{2\times 10} \\\\h = 0.032 \ m[/tex]

The maximum displacement angle is calculated as follows;

[tex]cos \theta = \frac{L-h}{L} \\\\cos \theta = \frac{1.25 - 0.032}{1.25} \\\\\cos \theta = \frac{1.218}{1.25} \\\\cos \theta = 0.9744\\\\\theta = cos^{-1}(0.9744)\\\\\theta = 13\ ^0[/tex]

Thus, the maximum displacement angle of the bob is 13⁰.

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Answer:

you are very kind Po lang

Answer:

The orbital speeds of stars far from the galactic center are surprisingly high, suggesting that these stars are feeling gravitational effects from unseen matter in the halo.

Answer:

"rings" - viewed from a distance, Saturn appears to be surrounded by rings

Answer:

Δx=7.35m

Explanation:

This is a free fall problem (g= -9.8 m/s²).

Displacement: ? <--- what you're trying to find (max height)

Initial Velocity: 12 m/s

Final Velocity: 0 m/s <-- since the can will land back in your hands

Time: Missing <-- not needed to solve

Acceleration: -9.8 m/s²

The equation that you will use is vf²=vi²+2aΔx

Now plug the information into the equation!

(0m/s)²=(12m/s)²+2(-9.8m/s²)Δx

Multiply the 2 with the acceleration (-9.8m/s²)

(0m/s)²=(12m/s)²-19.6m/s²Δx

Now you need to square the final and initial velocity

0m/s=144m/s-19.6m/s²Δx

Move 144 to the other side of the equal sign

-144m/s=-19.6m/s²Δx

Divide both sides by -19.6m/s²

7.346m=Δx

You can round it to the tenth or hundreth place, up to you!

For this example, I'll round to the hundredth...

Δx=7.35m

I would double check this answer or review it yourself to see if it's correct.

Good luck with your studies!

Levelized production cost is a big challenge in the production of wind energy.

The main challenge in the production of wind energy is the high production cost. It has to minimize the Levelized Production Cost (LPC) which is the ratio between energy production cost and economic lifetime. This problem can be solved by doing modification in the structure of the turbine as well as we must ensure minimal energy loss.

The modification in the structure can produce more electrical energy and reduce the energy loss also provides more profit to the company so we can conclude that levelized production cost is a big challenge in the production of wind energy.

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Answer:

the bottom one is wollsiegel

The characteristics of the speed of the waves in strings and the resonance allows to find the change in the fundamental frequency when changing the tension is:

The speed of the chord wave is given by the relationship between the tension and the density of the medium.

          [tex]v= \sqrt{\frac{T}{\mu } }[/tex]  

Where v is the velocity of the wave, T the tension of the string and μ the density

In a rope held at the ends, a process of standing waves occurs, two at the point where it is attached we have a node and a anti-node in the center.

             2L = n λ

Where L is the length of the chord and call the wavelength

Wave speeds are related to wavelength and frequency.

        v = λ f

We substitute.

            [tex]\sqrt{\frac{T}{\mu } } = \frac{2L}{n} \ \ f[/tex]  

For the fundamental frequency n = 1

            f₀ = [tex]f_o = \sqrt{\frac{T}{\mu } } \ \ \frac{1}{2L}[/tex]  

They indicate that the tension increases 1.70%

           T = T₀ + 0.17 T₀

           T = 1.17 T₀

We substitute.

         [tex]f = \sqrt{1.17 } \ \sqrt{\frac{T_o}{\mu } } \ \ \frac{1}{2L}[/tex]

         f = ra1.17 f₀

         f = 1.08 f₀

In conclusion, using the characteristics of the velocity of the waves in strings and the resonance we can find the change in the fundamental frequency when changing the tension is:

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