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3 years ago

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How do you solve 2A= 4lw+2Lh+6wH for w

1 answer:

gizmo_the_mogwai [7]3 years ago

6 0

Answer:

$\displaystyle w=\frac{A-Lh}{2l+3H}$

Explanation:

Solving Equations by Isolation

We have this equation:

$2A= 4lw+2Lh+6wH$

Let's solve it for w step by step. First, we flip both sides

$4lw+2Lh+6wH =2A$

Subtracting 2Lh in both sides

$4lw+2Lh+6wH-2Lh =2A-2Lh$

Simplifying

$4lw+6wH =2A-2Lh$

Factoring w

$w(4l+6H) =2A-2Lh$

Dividing by (4l+6H)

$\displaystyle w=\frac{2A-2Lh}{4l+6H}$

Simplifying by 2:

$\displaystyle \boxed{w=\frac{A-Lh}{2l+3H}}$

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A water wave traveling in a straight line on a lake is described by the equation

ivanzaharov [21]

Answer:

A) The wave equation is given as

$y(x,t) = A\cos(kx + \omega t)=(3.3\times 10^{-2})\cos(0.004x + 5.05t)\\$

According to the above equation, k = 0.004 and ω = 5.05.

Using the following identities, we can find the period of the wave.

$\omega = 2\pi f\\ f = 1/ T$

T = 1.25 s.

For the horizontal distance travelled by one period of time, x = λ.

$\lambda = 2\pi / k = 2\pi / 0.004 = 1.57\times 10^3~m$

$y(x = \lambda,t = T) = 0.033\cos(0.004*1.57*\10^3 + 5.05*1.25) = 0.033~m$

B) The wave number, k = 0.004 . The number of waves per second is the frequency, so f = 0.8.

C) The propagation speed of the wave is

$v = \lambda f = 1.57\times 10^3 * 0.8 = 1.256\times 10^3~m/s$

The velocity of the wave is the derivative of the position function.

$v(x,t) = \frac{dy(x,t)}{dt} = -(5.05\times 0.033)\sin(0.004x + 5.05t)$

The maximum velocity is when the derivative of the velocity function is equal to zero.

$\frac{dv_y(x,t)}{dt} = -(5.05)^2(0.033)\cos(0.004*1.57\times 10^3 + 5.05t) = 0$

In order this to be zero, cosine term must be equal to zero.

$0.004*1.57\times 10^3 + 5.05t = 5\pi /2\\t = 0.31~s$

The reason that cosine term is set to be 5π/2 is that time cannot be zero. For π/2 and 3π/2, t<0.

$v(x=\lambda, t = 0.31) = -(5.05\times0.033)\sin(0.004\times 1.57\times 10^3 + 5.05\times 0.31) = -0.166~m/s$

6 0

3 years ago

While a roofer is working on a roof that slants at 37.0 ∘∘ above the horizontal, he accidentally nudges his 92.0 NN toolbox, cau

eduard

Answer:

The speed is $v =8.17 m/s$

Explanation:

From the question we are told that

The angle of slant is $\theta = 37.0^o$

The weight of the toolbox is $W_t = 92.0N$

The mass of the toolbox is $m = \frac{92}{9.8} = 9.286kg$

The start point is $d = 4.25m$ from lower edge of roof

The kinetic frictional force is $F_f = 22.0N$

Generally the net work done on this tool box can be mathematically represented as

$Net \ work done = Workdone \ due \ to \ Weight + Workdone \ due \ to \ Friction$

The workdone due to weigh is = $mgsin \theta * d$

The workdone due to friction is = $F_f \ cos\theta * d$

Substituting this into the equation for net workdone

$W_{net} = mgsin\theta * d + F_f \ cos \theta *d$

Substituting values

$W_{net} = 92 * sin (37) * 4.25 + 22 cos (37) * 4.25$

$= 309.98 J$

According to work energy theorem

$W_{net} = \Delta Kinetic \ Energy$

$W_{net} = \frac{1}{2} m (v - u)^2$

From the question we are told that it started from rest so u = 0 m/s

$W_{net} = \frac{1}{2} * m v^2$

Making v the subject

$v = \sqrt{\frac{2 W_{net}}{m} }$

Substituting value

$v = \sqrt{\frac{2 * 309.98}{9.286} }$

$v =8.17 m/s$

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3 years ago

NEED ANSWER ASAP!!!

xenn [34]

Answer:

d

Explanation:

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Answer:

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<u>Explanation:</u>

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3 years ago

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Gravitational I think would be the answer, Hope this helps!

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