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Bad White [126]
3 years ago
11

You leave your house at 5:02 PM and run 20 yards down the street. You don't realize that you forgot your Wallet back at home and

turn around. You have run a third of the way back when you look at your watch and see that it change to say 5:03 PM. What is your average velocity in m/s up to this point
Engineering
1 answer:
Sidana [21]3 years ago
7 0

Answer:

The average velocity is 0.203 m/s

Explanation:

Given;

initial displacement, x₁ = 20 yards = 18.288 m

final displacement, x₂ = ¹/₃ x 18.288 = 6.096 m

change in time between 5:02 PM and 5:03 PM, Δt = 3 mins - 2 mins = 1 min = 60 s

The average velocity is given by;

V = change in displacement / change in time

V = (x₂ - x₁) / Δt

V = (18.288 - 6.096) / 60

V = 0.203 m/s

Therefore, the average velocity is 0.203 m/s

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5 0
3 years ago
The velocity profile for a thin film of a Newtonian fluid that is confined between the plate and a fixed surface is defined by u
zimovet [89]

Answer:

F = 0.0022N

Explanation:

Given:

Surface area (A) = 4,000mm² = 0.004m²

Viscosity = µ = 0.55 N.s/m²

u = (5y-0.5y²) mm/s

Assume y = 4

Computation:

F/A = µ(du/dy)

F = µA(du/dy)

F = µA[(d/dy)(5y-0.5y²)]

F = (0.55)(0.004)[(5-1(4))]

F = 0.0022N

8 0
3 years ago
Dust, dirt, or metal chips can pose a potential ____ injury risk in a shop.
Liono4ka [1.6K]

Answer: Eye injury

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7 0
2 years ago
How many 10" diameter circles can be cut from a semicircular shape that has a 20"
N76 [4]

9514 1404 393

Answer:

  1

Explanation:

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3 0
3 years ago
Read 2 more answers
A motorcycle starts from rest with an initial acceleration of 3 m/s^2, and the acceleration then changes with the distance s as
katrin2010 [14]

Answer:

Follows are the solution to this question:

Explanation:

Calculating the area under the curve:  

A = as

   =\frac{1}{2}(3 +6 \frac{m}{s^2})(100 \ m)+ \frac{1}{2}(6+4 \frac{m}{s^2})(100 m) \\\\=\frac{1}{2}(9 \frac{m}{s^2})(100 \ m)+ \frac{1}{2}(10\frac{m}{s^2})(100 m) \\\\=\frac{1}{2}(900 \frac{m^2}{s^2})+ \frac{1}{2}(1,000\frac{m^2}{s^2}) \\\\=(450 \frac{m^2}{s^2})+ (500\frac{m^2}{s^2}) \\\\= 950 \ \frac{m^2}{s^2}

Calculating the kinematics equation:

\to v^2 = v^2_{o} + 2as\\\\

        =0+ \sqrt{2as}\\\\ = \sqrt{2(A)}\\\\= \sqrt{2(950 \frac{m^2}{s^2})}\\\\= 43.59 \frac{m}{s}

Calculating the value of acceleration:  

\to a= \frac{dv}{dt}

=\frac{dv}{ds}(\frac{ds}{dt}) \\\\=v\frac{dv}{ds}\\\\\to \frac{dv}{ds}=\frac{a}{v}

\to \frac{dv}{ds} =\frac{4 \frac{m}{s^2}}{43.59 \frac{m}{s}} \\\\

         =\frac{0.092}{s}

3 0
2 years ago
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