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Ghella [55]
3 years ago
6

During a tug-of-war, team A pulls on team B by applying a force of 1390 N to the rope between them. How much work does team A do

if they pull team B toward them a distance of 3.68 m
Physics
1 answer:
butalik [34]3 years ago
7 0

Answer:

5115.2 J

Explanation:

Work: This can be defined as product of force and distance. The S.I unit of work is Joules (J).

From the question,

The expression for work is given as

W = F×d................. Equation 1

Where W = Work done by team A, F = Force applied by team A, d = distance moved by team B.

Given: F = 1390 N, d = 3.68 m

Substitute into equation 1

W = 1390×3.68

W = 5115.2 J.

Hence the work done by team A in pulling team B = 5115.2 J

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Explanation:

if you slide a hockey puck on ice, it will eventually stop, because of friction on the ice

kite when the wind changes can be described by the first law

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Through which medium would sound travel the fastest, water, a steel bar, or nitrogen gas explain
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High-voltage power lines are a familiar sight throughout the country. The aluminum wire used for some of these lines has a cross
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Answer:

Explanation:

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R = ρ L / S

where ρ is specific resistance , L is length and S is cross sectional area

Given L = 14 000 m ,

S = 4.8 x 10⁻⁴ m²

specific resistance of aluminum = 2.8 x 10⁻⁸ ohm-meter

Putting the values in the formula

R = 2.8 x 10⁻⁸ x 14 x 10³ /  (4.8 x 10⁻⁴ )

R = 0.8167 ohm .

= .82 ohm .

5 0
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The following are connected in parallel circuit at home EXCEPT:
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8 0
2 years ago
herical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water
ki77a [65]

Answer: Below is the complete question

A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.0x10-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.0x10-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass transfer coefficient (m/s)

answer:

mass transfer coefficient = 9.56 * 10^-5 m/s

Explanation:

Candy density = 1950 kg/m^3

Candy diameter = 1 cm

Velocity of water = 1 m/s

water density = 1000 kg/m^3

Viscosity of water = 1 * 10^-3 kg/m/s

diffusion coefficient of candy in water = 2 * 10^-9 m^2/s

solubility of candy = 2 kg/m^3

<u>Determine the mass transfer coefficient ( m/s ) </u>

( Sh) mass transfer coefficient ( flow across sphere ) = 2 + 0.6Re^1/2 * SC^1/3

where : Re = vdp / μ ,   Sh = KLd / Deff

attached below is the remaining solution .

mass transfer coefficient =  9.56 * 10^-5 m/s

6 0
2 years ago
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