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3 years ago

Calculate the power needed to raise 32lb. A distance of 12 feet in 4 seconds

Mathematics

2 answers:

Alex3 years ago

8 0

Answer:

130.07 Watts

Step-by-step explanation:

hope this helps

kenny6666 [7]3 years ago

8 0

Answer:

130.1 W

Step-by-step explanation:

Power is work or energy divided by time, so power has the units of joules/second, which is called the watt.

<h3>Formula</h3>

Power = gravitational potential energy / time

Gravitational potential energy = m g h

<em> g = gravity die to acceleration = 9.81 m/s²</em>

<em> h = height</em>

<em />

<h3>Conversion Ib to kg</h3>

1 Ib = 0.453592 kg

32 Ib = 14.5 kg

<h3>Conversion feet to m</h3>

1 feet = 0.3048 m

12 feet = 3.6576 m

Plug value in the formula

(14.5) (9.81) (3.6576)

520.3 J

520.3 / 4

130.1 W

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2 years ago

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mel-nik [20]

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Step-by-step explanation:

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3 years ago

Which Of The Following Rational Functions Is Graphed Below

Blizzard [7]

The given rational functions is graphed in the figure is Option B. f(x) = 1/ ( x-2) (x + 5) is correct.

<h3>What is a function?</h3>

The function is a type of relation, or rule, that maps one input to specific single output.

Here, a graph is given.

We need to find which of the given rational functions is graphed in the figure.

On x-axis, 1 unit = 2 units

Clearly, we can see the graph is not defined at point x = 2 and at x = -5.

Corresponding to x = 2, factor is (x - 2) .

Corresponding to x = -5, factor is (x + 5) .

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Learn more about function here:

brainly.com/question/2253924

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2 years ago

Please please help!!

shepuryov [24]

Answer:

$\large\boxed{x=0\ and\ x=\pi}$

Step-by-step explanation:

$\tan^2x\sec^2x+2\sec^2x-\tan^2x=2\\\\\text{Use}\ \tan x=\dfrac{\sin x}{\cos x},\ \sec x=\dfrac{1}{\cos x}:\\\\\left(\dfrac{\sin x}{\cos x}\right)^2\left(\dfrac{1}{\cos x}\right)^2+2\left(\dfrac{1}{\cos x}\right)^2-\left(\dfrac{\sin x}{\cos x}\right)^2=2\\\\\left(\dfrac{\sin^2x}{\cos^2x}\right)\left(\dfrac{1}{\cos^2x}\right)+\dfrac{2}{\cos^2x}-\dfrac{\sin^2x}{\cos^2x}=2$

$\dfrac{\sin^2x}{(\cos^2x)^2}+\dfrac{2-\sin^2x}{\cos^2x}=2\\\\\text{Use}\ \sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x\\\\\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{2-(1-\cos^2x)}{\cos^2x}=2\\\\\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{2-1+\cos^2x}{\cos^2x}=2\\\\\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{1+\cos^2x}{\cos^2x}=2$

$\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{(1+\cos^2x)(\cos^2x)}{(\cos^2x)^2}=2\qquad\text{Use the distributive property}\\\\\dfrac{1-\cos^2x+\cos^2x+\cos^4x}{\cos^4x}=2\\\\\dfrac{1+\cos^4x}{\cos^4x}=2\qquad\text{multiply both sides by}\ \cos^4x\neq0\\\\1+\cos^4x=2\cos^4x\qquad\text{subtract}\ \cos^4x\ \text{from both sides}\\\\1=\cos^4x\iff \cos x=\pm\sqrt1\to\cos x=\pm1\\\\ x=k\pi\ for\ k\in\mathbb{Z}\\\\\text{On the interval}\ 0\leq x$

4 0

2 years ago