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3 years ago

Find the derivative of y = sin(ln(5x2 − 2x))

Engineering

1 answer:

pickupchik [31]3 years ago

3 0

Answer:

$y = \cos[\ln x + \ln (5\cdot x - 2)]\cdot \left(\frac{1}{x} + \frac{5}{5\cdot x-2} \right)$

Explanation:

Let $y = \sin[\ln(5\cdot x^{2}-2\cdot x)]$ and we proceed to find the derivative by the following steps:

1) $y = \sin[\ln(5\cdot x^{2}-2\cdot x)]$ Given

2) $y = \sin [\ln[x\cdot (5\cdot x - 2)]]$ Distributive property

3) $y = \sin[\ln x + \ln (5\cdot x - 2 )]$ $\ln (a\cdot b) = \ln a + \ln b$

4) $y = \cos[\ln x + \ln (5\cdot x - 2)]\cdot \left(\frac{1}{x} + \frac{5}{5\cdot x-2} \right)$ $\frac{d}{dx} (\sin x) = \cos x$ / $\frac{d}{dx}(\ln x) = \frac{1}{x}$ / $\frac{d}{dx}(c\cdot x^{n}) = n\cdot c\cdot x^{n-1}$ /Rule of chain/Result

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A 220-V electric heater has two heating coils that can be switched such that either coil can be used independently or the two ca

Nana76 [90]

Answer:

The resistances of both coils are 131.7 Ω and 29.64 Ω.

Explanation:

Since, there are two coils, they can be used independently or in series or parallel. The power is given as:

Power = P = VI

but, from Ohm's Law:

V = IR

I = V/R

therefore,

P = V²/R

R = V²/P

Hence, the resistance (R) and (P) are inversely proportional. Therefore, the maximum value of resistance will give minimum power, that is, 300 W. And the maximum resistance will be in series arrangement, as in series the total resistance gets higher than, any individual resistance.

Therefore,

Rmax = V²/Pmin = R1 + R2

R1 + R2 = (220 V)²/300 W

R1 + R2 = 161.333 Ω ______ en (1)

Similarly, the minimum resistance will give maximum power. And the minimum resistance will occur in parallel combination. Because equivalent resistance of parallel combination is less than any individual resistance.

Therefore,

(R1 R2)/(R1 + R2) = (220 V)²/2000 W

using eqn (1), we get:

(R1 R2) / 161.333 Ω = 24.2 Ω

R1 R2 = 3904.266 Ω²

R1 = 3904.266 Ω²/R2 _____ eqn (2)

Using this value of R1 in eqn (1), we get:

3904.266/R2 +R2 = 161.333

(R2)² - 161.333 R2 +3904.266 = 0

Solving this quadratic eqn we get two values of R2 as:

R2 = 131.7 Ω OR R2 = 29.64 Ω

when ,we substitute these values in eqn (1) to find R1, we get get the same two values as R2, alternatively. This means that the two coils have these resistance, and the order does not matter.

<u>Therefore, the resistance of both coils are found to be 131.7 Ω and 29.64 Ω</u>

7 0

3 years ago

g Three unequal point masses are attached to a vertical shaft with light rods. Two masses, m and 3m, are at a distance a from th

Alexxandr [17]

Answer:

12ma²

Explanation:

The moment of inertia I = ∑mr² where m = mass and r = distance from shaft

Since we have three masses,

So, I = m₁r₁² + m₂r₂² + m₃r₃² where m₁ = first mass = m, m₂ = second mass = 3m and m₃ = third mass = 2m. Also, r₁ = distance of first mass from shaft = a, r₂ = distance of second mass from shaft = a and r₃ = distance of third mass from shaft = 2a.

I = m₁r₁² + m₂r₂² + m₃r₃²

I = ma² + 3ma² + 2m(2a)²

I = ma² + 3ma² + 2m(4a²)

I = ma² + 3ma² + 8ma²

I = 12ma²

4 0

3 years ago

A solid cylinder is concentric with a straight pipe. The cylinder is 0.5 m long and has an outside diameter of 8 cm. The pipe ha

poizon [28]

Answer :

The force needed to move the cylinder is 25.6 N

<h2>Further explanation </h2>

Given that,

Length of the cylinder, l = 0.5 m

Outer diameter of the cylinder, d = 8 cm = 0.08 m

Outer radius of the cylinder, $r=0.04\ m$

Inside diameter of the pipe, d = 8.5 cm = 0.085 m

Inside radius of the pipe, $r=0.0425\ m$

Specific gravity of the oil, $\rho=0.92$

Density of oil, $d=\rho\times \rho_w$

Kinematic viscosity of the oil, $v=5.57\times 10^{-4}\ m^2/s$

Velocity of the cylinder, u = 1 m/s

We need to find the force needed to move the cylinder. Let the force is F.

Specific gravity is defined as the ratio of the density of the substance to the density of water.

Kinematic viscosity is the acquired resistance of a fluid when there is no external force is acting except gravity. It is denoted by v.

Absolute viscosity is given by :

$v=\dfrac{\mu}{d}$

Where, $d$ = density of oil

And $d=\rho\times \rho_w$ (density of oil = specific gravity × density of water )

$d=0.92\times 10^3\ kg/m^3$

So,

$\mu=v\times d$ ..............(1)

$\mu=5.57\times 10^{-4}\ m^2/s\times 0.92\times 10^3\ kg/m^3$

$\mu=0.512\ Pa-s$

The separation between the cylinder and pipe is given by :

$dy=\dfrac{d_p-d_c}{2}=\dfrac{8.5-8}{2}=0.25\ cm=0.0025\ m$

$d_p\ and\ d_c$ are diameter of pipe and cylinder respectively.

The mathematical expression for the Newton's law of viscosity can be written as:

$\tau\propto\dfrac{du}{dy}$

$\tau=\mu\times \dfrac{du}{dy}$ ..........(2)

Where

$\tau$ = Shear stress, $\tau=\dfrac{F}{A}$ ............(3)

$\mu$ = viscosity

$\dfrac{du}{dy}$ = rate of shear deformation

On rearranging equation (1), (2) and (3) we get :

$\dfrac{F}{A}=v\times \rho\times \dfrac{du}{dy}$ ...............(4)

A is the area of the cylinder, $A=2\pi rl$

Equation (4) becomes :

$F=v\times \rho\times \dfrac{du}{dy}\times 2\pi rl$ ..............(5)

$A=\pi d\times l$

$A=\pi \times 0.08\ m\times 0.5\ m$

$A=0.125\ m^2$

Now, equation (5) becomes :

$F=(v\times \rho)\times \dfrac{du}{dy}\times 2\pi rl$

$F=(0.512\ Pa-s)\times (\dfrac{1}{0.0025\ m})\times \times 0.125\ m^2$

F = 25.6 N

<h2>Learn more </h2>

Kinematic viscosity : brainly.com/question/12947932

<h2>Keyword : </h2>

Specific gravity, Kinematic viscosity, Area of cylinder, fluid mass density.

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3 years ago

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Juliette [100K]

Answer:

Explanation:

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3 years ago

Mechanic... Mechanical Engineer... What's the difference?

kumpel [21]

Answer:

Mechanic: a person who <u>repairs and maintains machinery</u>

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Explanation:

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4 years ago