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musickatia [10]
3 years ago
10

You are visiting your friend Fabio's house. You find that, as a joke, he filled his swimming pool with Kool-Aid, which dissolved

perfectly into the water. However, now that you want to swim, you must remove all of the Kool-Aid contaminated water. The swimming pool is round, with a 9 foot radius. It is 8 feet tall and has 7.5 feet of water in it. How much work is required to remove all of the water by pumping it over the side? Use the physical definition of work, and the fact that the density of the Kool-Aid contaminated water is σ=63.5lbs/ft3
Physics
2 answers:
oksian1 [2.3K]3 years ago
3 0

Answer:

Explanation:

The volume of contaminated water

= cross sectional area x height of water level

3.14 x 9 x 9 x 7.5 ft³

= 1907.55 ft³

mass = density x volume

= 1907.55 x 63.5 lbs

m = 121129.425 lbs

This mass has to be raised to the height of 8 ft before evacuation .

There is a rise of centre of mass of

8 - 7.5/2 ft

h = 4.25 ft

Energy required

= mgh

= 121129.425 x 32 x 4.25

= 16473601.8 unit.

skelet666 [1.2K]3 years ago
3 0

Answer:

Workdone= 8.08×10^3Joules

Explanation:

dV= pir^2 dx

dV = pi×9^2 = 81pi

dF= density× pi× r^2 dx

dF= 63.5 × 81pi dx= 5143.5pi dx

W= Integral from7.5ft of water in the pool(8 -×)dx

W= 5143.5pi(8-7.5)

W= 5143.5 × 3.142 × 0.5

W= 8080.44Joules = 8.08×10^3Joules

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A 55-kg block, starting from rest, is pushed a distance of 5.0 m across a floor by a horizontal force Fp whose magnitude is 140
max2010maxim [7]

Answer:

The answer is "151.25 J and -547.64 J".

Explanation:

u = 0\\\\v = 2.35\  \frac{m}{sec}\\\\d = 5.0 \ m\\\\

Using formula:

v^2 = u^2 + 2 \times a \times d\\\\2.35^2 = 0^2 + 2 \times a \times 5\\\\a = \frac{2.35^2}{10} \\\\

   = 0.55 \ \frac{m}{sec^2}\\\\

F_{net} = m \times a\\\\F_{net} = 55 \times 0.55 = 30.25\ N\\\\

Calculating the Work by net force

W = F_{net}\times d\\\\W = 30.25 \times 5 = 151.25 \ J\\\\

The above work is converted into thermal energy.

Now,

F_{net} = F_p - F_f\\\\F_p = 140 \ N\\\\F_f = u_k\times m \times g = u_k \times 55 \times 9.81\\\\F_f = 539.55 \times u_k\\\\30.25 = 140 - u_k \times 55 \times 9.81\\\\u_k = \frac{(140 - 30.25)}{(55\times 9.81)}\\\\uk = 0.203 = \text{Coefficient of friction}\\\\W_f = -F_f \times d\\\\W_f = -0.203 \times 55 \times 9.81 \times 5\\\\Work\ done\  by\ friction = -547.64 \ J

6 0
2 years ago
A driver in a 2290-kg car car traveling at 42.7 m/s slams on the brakes and skids to a stop. If the coefficient of friction betw
8_murik_8 [283]

Answer:

126.56 m

Explanation:

Applying,

-F = ma............. Equation 1

Where F = frictional force, m = mass of the car, a = acceleration.

Note: Frictional force is negative because it act in opposite direction to motion

But,

F = mgμ.......... Equation 2

Where g = acceleration due to gravity, μ = coefficient of friction

Substitute equation 2 in equation 1

-mgμ = ma

a = -gμ.............. Equation 3

From the question,

Given: μ = 0.735

Constant: 9.8 m/s²

Substitute these values in equation 3

a = -9.8×0.735

a = -7.203 m/s²

Finally,

Applying

v² = u²+2as.............. Equation 4

Where v = final velocity, u = initial velocity, s = distance

From the question,

Given: u = 42.7 m/s, v = 0 m/s (to a stop), a = -7.203 m/s²

Substitute these values into equation 4

0² = 42.7²+2(-7.203)s

-1823.29 = -14.406s

s = -1823.29/-14.406

s = 126.56 m

4 0
2 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
VMariaS [17]

At point C because it is at the lowest position.

7 0
3 years ago
If you wish to observe features that are around the size of atoms, say 1 .5 x 100 m, with electromagnetic radiation, the radiati
chubhunter [2.5K]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

If you wish to observe features that around the size of atoms, say 1.5×10⁻¹⁰ m, with electromagnetic radiation, the radiation must have a wavelength about the size of the atom itself.

a) If you had a microscope which was capable of doing this, what would the frequency of electromagnetic radiation be, in hertz that you would have to use?

b) What type of electromagnetic radiation would this be?

Given Information:

Wavelength = λ = 1.5×10⁻¹⁰  m

Required Information:

a) Frequency = f = ?

b) Type of electromagnetic radiation = ?

Answer:

a) Frequency = f = 2×10¹⁸ Hz

b) Type of electromagnetic radiation = X-rays

Explanation:

a) The frequency of the electromagnetic radiation is given by

f = c/ λ

Where λ  is the wavelength of the electromagnetic radiation and c is the speed of light and its value is 3×10⁸ m/s

f = 3×10⁸/1.5×10⁻¹⁰

f = 2×10¹⁸ Hz

Therefore, the frequency of the electromagnetic radiation would be 2×10¹⁸ Hz.

b)

The frequency range of X-rays is 3×10¹⁶ Hz to 3×10¹⁹ Hz

The frequency 2×10¹⁸ lies in that range, therefore, the type of electromagnetic radiation is X-rays

5 0
3 years ago
A very long straight wire has charge per unit length 1.56×10−10 C/m .
77julia77 [94]

Answer:

The magnitude of the electric field equal to 2.40 N/C at 1.1537m from the wire.

Explanation:

using Guass law,

(guessing that a cylinder of radius r and length L around wire such that wire is at centre )

E. A = qin / e0

E ( 2πr L ) = (1.56 x 10^{-10} x L) / (8.85 x 10^{-12})

E = (1.56 x 10^{-10} ) / (2πr x 8.85 x  10^{-12})

so 2.40 = (1.54 x 10^{-10} ) / (2πr x 8.85 x 10^{-12})

2.40 (6.284r) = 0.174 x 10²

15.0816r = 17.4

r = 1.1537m

6 0
2 years ago
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