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3 years ago

My mom and I walked 3,600 m in 90 minutes. What was our speed in m/min?

Physics

1 answer:

PtichkaEL [24]3 years ago

3 0

Fill in the fraction: 3,600/90 = 40; turn it into a unit fraction.

40 mi/min

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While moving in, a new homeowner is pushing a box across the floor at a constant velocity. The coefficient of kinetic friction b

Citrus2011 [14]

Answer:

$\theta = 66.7 degree$

Explanation:

since force is applied downwards at some angle with the horizontal

so here we will have

$F_n = mg + Fsin\theta$

now we know that the box will not move if applied force is balanced by frictional force on it

so we will have

$Fcos\theta = \mu F_n$

$F cos\theta = \mu (mg + F sin\theta)$

$F(cos\theta - \mu sin\theta) = \mu mg$

$F = \frac{\mu mg}{cos\theta - \mu sin\theta}$

so here we can say

$cos\theta - \mu sin\theta > 0$

$tan\theta = \frac{1}{\mu}$

$\theta = tan^{-1}\frac{1}{\mu}$

$\theta = tan^{-1}(\frac{1}{0.43})$

$\theta = 66.7 degree$

6 0

3 years ago

Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl

melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

$Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar$

The boundary layer thickness is equal to:

$\delta=\frac{4.91*1}{Re^{0.5} } =\frac{4.91*1}{246507^{0.5} } =0.0098$ ft

The shear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{1} )(246507)^{0.5} =0.012$

b) If the railing edge is 2 ft, the Reynold number is:

$Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar$

The boundary layer is equal to:

$\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft$

The sear stress is equal to:

$\tau=0.332(\frac{2.359x10^{-5}*3 }{2} )(493015.6^{0.5} )=0.0082$

c) The drag coefficient is equal to:

$C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019$

The friction drag is equal to:

$F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf$

7 0

3 years ago

Two forces, F⃗ 1F→1F_1_vec and F⃗ 2F→2F_2_vec, act at a point,F⃗ 1F→1F_1_vec has a magnitude of 8.80 NN and is directed at an an

castortr0y [4]

Answer:

Fx = -9.15 N
Fy = 1.72 N
F∠γ ≈ 9.31∠-10.6°

Explanation:

You apparently want the sum of forces ...

F = 8.80∠-56° +7.00∠52.8°

Your angle reference is a bit unconventional, so we'll compute the components of the forces as ...

f∠α = (-f·cos(α), -f·sin(α))

This way, the 2nd quadrant angle that has a negative angle measure will have a positive y component.

= -8.80(cos(-56°), sin(-56°)) -7.00(cos(52.8°), sin(52.8°))

≈ (-4.92090, 7.29553) +(-4.23219, -5.57571)

≈ (-9.15309, 1.71982)

The resultant component forces are ...

Fx = -9.15 N
Fy = 1.72 N

Then the magnitude and direction of the resultant are

F∠γ = (√(9.15309² +1.71982²))∠arctan(-1.71982/9.15309)

F∠γ ≈ 9.31∠-10.6°

4 0

3 years ago

A student increased the resister of the voltmeter year not the circuit

Flura [38]

Answer:

Say the full question I can't understand what it is

3 0

3 years ago

Which is a compound A Kr B O 2 c cl2 d CuF2

Volgvan

I think the answer is CuF2

7 0

3 years ago