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2 years ago

_ drives the energy and motion of waves?

Physics

2 answers:

Papessa [141]2 years ago

7 0

Answer:

Wind-driven waves, or surface waves,

Explanation:

____ [38]2 years ago

5 0

Answer:

Kinetic Energy(?)

I hope this helped at all. sorry if it didn't

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Object A has 27 J of kinetic energy. Object B has one-quarter the mass of object A.

andreev551 [17]

Answer:

the final speed of object A changed by a factor of $\frac{1}{\sqrt{3} }$ = 0.58

the final speed of object B changed by a factor of $\sqrt{\frac{5}{3} }$ = 1.29

Explanation:

Given;

kinetic energy of object A, = 27 J

let the mass of object A = $m_A$

then, the mass of object B = $m_B = \frac{m_A}{4}$

work done on object A = -18 J

work done on object B = -18 J

let $v_i$ be the initial speed

let $v_f$ be the final speed

For object A;

$K.E_A = 27\\\\\frac{1}{2} m_A v_i^2 = 27\\\\m_A v_i^2 = 54\\\\m_A = \frac{54}{v_i^2} ----Equation \ (1)\\\\Apply \ work-energy \ theorem;\\\\\delta K.E_A = -18\\\\\frac{1}{2} m_A v_f^2 - \frac{1}{2} m_A v_i^2 = -18\\\\\frac{1}{2} m_A ( v_f^2 \ - v_i^2 )\ =- 18\\\\v_f^2 \ - v_i^2 = -\frac{36}{m_A} ---Equation \ (2)\\\\v_f^2 \ - v_i^2 = -\frac{36v_i^2}{54}\\\\ v_f^2 \ =v_i^2 - \frac{36v_i^2}{54}\\\\ v_f^2 = \frac{54v_i^2 -36v_i^2 }{54} \\\\v_f^2 = \frac{18v_i^2}{54} \\\\v_f^2 = \frac{v_i^2}{3} \\\\$

$v_f = \sqrt{\frac{v_i^2}{3} }\\\\v_f = \frac{1}{\sqrt{3} } \ v_i\\\\$

Thus, the final speed of object A changed by a factor of $\frac{1}{\sqrt{3} }$ = 0.58

To obtain the change in the final speed of object B, apply the following equations.

$K.E_B_i = \frac{1}{2} m_Bv_i^2\\\\m_B = \frac{m_A}{4} \\\\K.E_B_i = \frac{1}{2}(\frac{m_A}{4} )v_i^2\\\\K.E_B_i = \frac{m_Av_i^2}{8} \\\\But, \ m_Av_i^2 = 54 \\\\K.E_B_i = \frac{54}{8} \\\\Apply \ work-energy \ theorem ;\\\\\delta K.E = -18\\\\K.E_f -K.E_i = -18\\\\\frac{1}{2}m_Bv_f^2 - \frac{1}{2} m_Bv_i^2 = -18\\\\Recall \ m_B = \frac{m_A}{4} \\\\\frac{1}{2}(\frac{m_A}{4} )v_f^2 - \frac{1}{2}(\frac{m_A}{4} )v_i^2 = -18\\\\\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\$

$\frac{1}{2}\times \frac{m_A}{4} (v_i^2 -v_f^2) = 18\\\\v_i^2 -v_f^2 = \frac{8}{m_A} \times 18\\\\v_i^2 -v_f^2 =\frac{144}{m_A} \\\\But , m_A = \frac{54}{v_i^2} \\\\v_i^2 -v_f^2 =\frac{144v_i^2}{54} \\\\v_f^2 = v_i^2 - \frac{144v_i^2}{54}\\\\v_f^2 = \frac{54v_i^2-144v_i^2}{54}\\\\ v_f^2 = \frac{-90v_i^2}{54} \\\\v_f^2 = \frac{-5v_i^2}{3} \\\\|v_f| = \sqrt{\frac{5v_i^2}{3}} \\\\|v_f| = \sqrt{\frac{5}{3}} \ v_i$

Thus, the final speed of object B changed by a factor of $\sqrt{\frac{5}{3} }$ = 1.29

3 0

3 years ago

As streams flow through Stone Mountain, layers of sand build up. Over time, the sand particles form a sedimentary rock called sa

Firlakuza [10]

As streams flow through Stone Mountain, layers of sand build up. Over time, the sand particles form a sedimentary rock called sandstone. What causes sandstone to change into metamorphic rock at Stone Mountain? Sandstone experiences intense heat and pressure.

(Correct Answer is above)

Also Mark Brainliest please.

3 0

3 years ago

Read 2 more answers

A coat rack weighs 65.0 lbs when it is filled with winter coats and 40.0 lbs when it is empty. The base of the coat rack has an

Whitepunk [10]

Answer:

0.056 psi more pressure is exerted by filled coat rack than an empty coat rack.

Explanation:

First we find the pressure exerted by the rack without coat. So, for that purpose, we use formula:

P₁ = F/A

where,

P₁ = Pressure exerted by empty rack = ?

F = Force exerted by empty rack = Weight of Empty Rack = 40 lb

A = Base Area = 452.4 in²

Therefore,

P₁ = 40 lb/452.4 in²

P₁ = 0.088 psi

Now, we calculate the pressure exerted by the rack along with the coat.

P₂ = F/A

where,

P₂ = Pressure exerted by rack filled with coats= ?

F = Force exerted by filled rack = Weight of Filled Rack = 65 lb

A = Base Area = 452.4 in²

Therefore,

P₂ = 65 lb/452.4 in²

P₂ = 0.144 psi

Now, the difference between both pressures is:

ΔP = P₂ - P₁

ΔP = 0.144 psi - 0.088 psi

ΔP = 0.056 psi

8 0

4 years ago

PLEASE HELP ME WITH THIS ONE QUESTION

sdas [7]

Answer:

0.34 m

Explanation:

From the question,

v = λf................ Equation 1

Where v = speed of sound, f = frequency, λ = Wave length

Make λ the subject of the equation

λ = v/f............... Equation 2

Given: v = 340 m/s, f = 500 Hz.

Substitute these values into equation 2

λ = 340/500

λ = 0.68 m

But, the distance between a point of rarefaction and the next compression point, in the resulting sound is half wave length

Therefore,

λ/2 = 0.68/2

λ/2 = 0.34 m

Hence, the distance between a point of rarefaction and the next compression point, in the resulting sound is 0.34 m

6 0

3 years ago

1. After a rabbit population reaches the carrying capacity of its habitat, what will most likely happen to

kifflom [539]

C) alternately increase and decrease

6 0

3 years ago

_______ drives the energy and motion of waves? ______

_ drives the energy and motion of waves?