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3 years ago

8

Den pushes a desk 400 cm across the floor. He exerts a force of 10 N for 8 s to move the desk. What is his power output? (Power:

P = W/t) 1.25 W 5 W 40 W 500 W

1 answer:

White raven [17]3 years ago

7 0

Answer:

5 W

Explanation:

The formula of the power is:

● P = W/t

W is the work and t is the time needed to do it(in seconds)

Let's calculate first the work that the force exerced:

W = Vector F . Vector d

D is the distance ( here 400 cm wich is 4 m)

Make a representation to see how are the vectors F and V.(picture below)

The vector F and d are colinear since Den is pushing the desk on the ground.

● W = 4 × 10 = 40 J

J is Joule

■■■■■■■■■■■■■■■■■■■■■■■■■■

● P = W / t

● P = 40/ 8

● P = 5 W

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A swimming pool has the shape of a right circular cylinder with radius 21 feet and height 10 feet. Suppose that the pool is full

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Answer:

The water required to pump all the water to a platform 2 feet above the top of the pool is is 6061310.32 foot-pound.

Explanation:

Given that,

Radius = 21 feet

Height = 10 feet

Weighing = 62.5 pounds/cubic

Work = 4329507.37572

Height = 2 feet

Let's look at a horizontal slice of water at a height of h from bottom of pool

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$A=\pi r^2$

Put the value into the formula

$A=\pi\times21^2$

$A=441\pi\ feet^2$

Thickness of slice $t=\Delta h\ ft$

The volume is,

$V=(441\pi\times\Delta h)\ ft^3$

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Using formula of force

$F=W\times V$

Where, W = water weight

V = volume

Put the value into the formula

$F=62.5\times(441\pi\times\Delta h)$

$F=27562.5\pi\times\Delta h\ lbs$

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Using formula of work done

$W=F\times d$

Put the value into the formula

$W=27562.5\pi\times\Delta h\times(10-h)\ ft\ lbs$

We do this by integrating from h = 0 to h = 10

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Using formula of work done

$W=\int_{0}^{h}{W}$

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$W=\int_{0}^{10}{27562.5\pi\\times(10-h)}dh$

$W=27562.5\pi(10h-\dfrac{h^2}{2})_{0}^{10}$

$W=27562.5\pi(10\times10-\dfrac{100}{2}-0)$

$W=4329507.37572$

To pump 2 feet above platform, then each slice has to be lifted an extra 2 feet,

So, the total distance to lift slice is (12-h) instead of of 10-h

We need to calculate the water required to pump all the water to a platform 2 feet above the top of the pool

Using formula of work done

$W=\int_{0}^{h}{W}$

Put the value into the formula

$W=\int_{0}^{10}{27562.5\pi\\times(12-h)}dh$

$W=27562.5\pi(12h-\dfrac{h^2}{2})_{0}^{10}$

$W=27562.5\pi(12\times10-\dfrac{100}{2}-0)$

$W=1929375\pi$

$W=6061310.32\ foot- pound$

Hence, The water required to pump all the water to a platform 2 feet above the top of the pool is is 6061310.32 foot-pound.

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An electron moves at a speed of 1.0 x 104 m/s in a circular path of radius 2 cm inside a solenoid. The magnetic field of the sol

iogann1982 [59]

Answer:

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r = $\frac{mv}{qB}$

⇒ B = $\frac{mv}{qr}$

Where: r is the radius, m is the mass of the electron, v is its velocity, q is the charge on the electron and B is the magnetic field

B = $\frac{9.11*10^{-31*1.0*10^{4} } }{1.6*10^{-19}*0.02 }$

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B = 2.85 × $10^{-6}$ Tesla

b. Given that; N/L = 25 turns per centimetre, then the current, I, can be determined by;

B = μ I N/L

⇒ I = B ÷ μN/L

where B is the magnetic field, μ is the permeability of free space = 4.0 × $10^{-7}$ Tm/A, N/L is the number of turns per length.

I = B ÷ μN/L

= $\frac{2.85*10^{-6} }{4*10^{-7} *25}$

I = 0.285 A

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