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3 years ago

A Wii remote flung from a hand through a TV, with a kinetic energy of 1.44J and a mass of 4.5kg. Whats the velocity?

Engineering

1 answer:

Eva8 [605]3 years ago

7 0

Answer:

0.8

Explanation:

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Imagine the arc of a football as it flies through the air. How does this motion illustrate classical mechanics?

BigorU [14]

Answer: Maybe A

Explanation:

7 0

1 year ago

Read 2 more answers

A pump is used to deliver water from a lake to an elevated storage tank. The pipe network consists of 1,800 ft (equivalent lengt

Nataly_w [17]

Answer:

h_f = 15 ft, so option A is correct

Explanation:

The formula for head loss is given by;

h_f = [10.44•L•Q^(1.85)]/(C^(1.85))•D^(4.8655))

Where;

h_f is head loss due to friction in ft

L is length of pipe in ft

Q is flow rate of water in gpm

C is hazen Williams constant

D is diameter of pipe in inches

We are given;

L = 1,800 ft

Q = 600 gpm

C = 120

D = 8 inches

So, plugging in these values into the equation, we have;

h_f = [10.44*1800*600^(1.85)]/(120^(1.85))*8^(4.8655))

h_f = 14.896 ft.

So, h_f is approximately 15 ft

7 0

3 years ago

Find the mathematical equation for SF distribution and BM diagram for the beam shown in figure 1.

Novosadov [1.4K]

Answer:

i) SF: $v(x) = \frac{(w_0* x )^2}{2L}$

ii) BM : $= \frac{(w_0*x)^3}{6L}$

Explanation:

Let's take,

$\frac{y}{w_0} = \frac{x}{L}$

Making y the subject of formula, we have :

$y = \frac{x}{L} * w_0$

For shear force (SF), we have:

This is the area of the diagram.

$v(x) = \frac{1}{2} * y = \frac{1}{2} * \frac{x}{L} * w_0$

$= \frac{(w_0* x )^2}{2L}$

The shear force equation =

$v(x) = \frac{(w_0* x )^2}{2L}$

For bending moment (BM):

$BM = v(x) * \frac{x}{3}$

$= \frac{(w_0* x )^2}{2L} * \frac{x}{3}$

$= \frac{(w_0*x)^3}{6L}$

The bending moment equation =

$= \frac{(w_0*x)^3}{6L}$

5 0

3 years ago

What are the success factors for mechanical engineering?

Eva8 [605]

Answer:

-effective technical skills.

-the ability to work under pressure.

-problem-solving skills.

-creativity.

-interpersonal skills.

-verbal and written communication skills.

-commercial awareness.

-teamworking skills.

Explanation:

is this what ur looking for? if so there ya go lol

7 0

2 years ago

Read 2 more answers

The state of plane strain on an element is:

balu736 [363]

Answer:

a. ε₁=-0.000317

ε₂=0.000017

θ₁= -13.28° and θ₂=76.72°

b. maximum in-plane shear strain =3.335 *10^-4

Associated average normal strain ε(avg) =150 *10^-6

θ = 31.71 or -58.29

Explanation:

$\epsilon _{1,2} =\frac{\epsilon_x + \epsilon_y}{2} \pm \sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\epsilon _{1,2} =\frac{-300 \times 10^{-6} + 0}{2} \pm \sqrt{(\frac{-300 \times 10^{-6}+ 0}{2}) ^2 + (\frac{150 \times 10^-6}{2})^2}\\\\\epsilon _{1,2} = -150 \times 10^{-6} \pm 1.67 \times 10^{-4}$

ε₁=-0.000317

ε₂=0.000017

To determine the orientation of ε₁ and ε₂

$tan 2 \theta_p = \frac{\gamma_xy}{\epsilon_x - \epsilon_y} \\\\tan 2 \theta_p = \frac{150 \times 10^{-6}}{-300 \times 10^{-6}-\ 0}\\\\tan 2 \theta_p = -0.5$

θ= -13.28° and 76.72°

To determine the direction of ε₁ and ε₂

$\epsilon _{x' }=\frac{\epsilon_x + \epsilon_y}{2} + \frac{\epsilon_x -\epsilon_y}{2} cos2\theta + \frac{\gamma_xy}{2}sin2\theta \\\\\epsilon _{x'} =\frac{-300 \times 10^{-6}+ \ 0}{2} + \frac{-300 \times 10^{-6} -\ 0}{2} cos(-26.56) + \frac{150 \times 10^{-6}}{2}sin(-26.56)\\\\$

=-0.000284 -0.0000335 = -0.000317 =ε₁

Therefore θ₁= -13.28° and θ₂=76.72°

b. maximum in-plane shear strain

$\gamma_{max \ in \ plane} =2\sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\gamma_{max \ in \ plane} = 2\sqrt{(\frac{-300 *10^{-6} + 0}{2} )^2 + (\frac{150 *10^{-6}}{2})^2}$

=3.335 *10^-4

$\epsilon_{avg} =(\frac{\epsilon_x + \epsilon_y}{2} )$

ε(avg) =150 *10^-6

orientation of γmax

$tan 2 \theta_s = \frac{-(\epsilon_x - \epsilon_y)}{\gamma_xy} \\\\tan 2 \theta_s = \frac{-(-300*10^{-6} - 0)}{150*10^{-6}}$

θ = 31.71 or -58.29

To determine the direction of γmax

$\gamma _{x'y' }= - \frac{\epsilon_x -\epsilon_y}{2} sin2\theta + \frac{\gamma_xy}{2}cos2\theta \\\\\gamma _{x'y' }= - \frac{-300*10^{-6} - \ 0}{2} sin(63.42) + \frac{150*10^{-6}}{2}cos(63.42)$

= 1.67 *10^-4

4 0

3 years ago