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3 years ago

Kalisha is going to cut the remaining 21 1/2 buns into thirds. How many 1/3 size buns will she have?

2 answers:

6 0

Total buns = 21 1/2 = 43/2 buns
1/3 size buns would be: 43/2 / 1/3 = 43*3 / 2 = 129/2

In short, Your Answer would be 129/2 or 64 1/2 or 64.5

Hope this helps!

solmaris [256]3 years ago

4 0

1/2 of 21 buns is 10.5. 1/3 of that is 3.5

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A disk between vertebrae in the spine is subjected to a shearing force of 600 N. Find its shear deformation, taking it to have t

Art [367]

Answer:

3.34×10^-6m

Explanation:

The shear modulus can also be regarded as the rigidity. It is the ratio of shear stress and shear strain

can be expressed as

shear stress/(shear strain)

= (F/A)/(Lo/ . Δx)

Stress=Force/Area

The sheear stress can be expressed below as

F Lo /(A *Δx)

Where A=area of the disk= πd^2/4

F=shearing force force= 600N

Δx= distance

S= shear modulus= 1 x 109 N/m2

Lo= Lenght of the cylinder= 0.700 cm=7×10^-2m

If we make Δx subject of the formula we have

Δx= FLo/(SA)

If we substitute the Area A we have

Δx= FLo/[S(πd^2/4]

Δx=4FLo/(πd^2 *S)

If we input the values we have

(4×600×0.7×10^-2)/10^9 × 3.14 ×(4×10^-2)^2

= 3.35×10^-6m

Therefore, its shear deformation is 3.35×10^-6m

A=area of the disk= πd^2/4

= [3.142×(4×10^-2)^2]/4

7 0

4 years ago

_____ is the passing of a wave through an object.

Vanyuwa [196]

Transmission is the passing of a wave through SOMEthing.

8 0

3 years ago

Read 2 more answers

The 2-Mg truck is traveling at 15 m/s when the brakes on all its wheels are applied, causing it to skid for 10 m before coming t

Scorpion4ik [409]

Answer:

constant horizontal force developed in the coupling C = 11.25KN

the friction force developed between the tires of the truck and the road during this time is 33.75KN

Explanation:

See attached file

8 0

3 years ago

Why does the blue force have a negative sign?

dimaraw [331]

it’s B (because 20 Is smaller than 50)

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3 years ago

The rear window in a car is approximately a rectangle, 1.3 m wide and 0.30 m high. The inside rear-view mirror is 0.50 m from th

maria [59]

Answer:

Height of mirror 0.075 m

width of mirror 0.325 m

Explanation:

given data

wide = 1.3 m

high = 0.30 m

driver’s eyes = 0.50 m

rear window = 1.50 m

solution

we take here height / width of the mirror is

height / width = $\frac{h}{w}$ .................1

and

height /width of the window is

height /width = $\frac{h_w}{w_w}$ .................2

and

distance of eye / window by the mirror is

distance of eye / window = $\frac{x_e}{x_w}$ .................3

so here

θ = θi = θr ....................4

and tanθ for vertical is

tanθ = $\frac{h}{x_e}$

tanθ = $\frac{h_w}{(x_e + x_w)}$ ....................5

h = $h_w \times \frac{x_e}{(x_e + x_e)}$ ....................6

put here value and we get

$h = 0.30 \times \frac{0.50}{(0.50 + 1.50)}$

h = 0.075 m

and

when we take here tanθ for horizontal than it will be

tanθ = $\frac{w}{x_e}$

tanθ = $\frac{w_w}{(x_e + x_w)}$ .......................7

$w = w_w \times \frac{x_e}{(x_e + x_w)}$ ....................8

put here value and we get

$w = 1.3 \times \frac{0.50}{(0.50 + 1.50)}$

w = 0.325 m

7 0

3 years ago