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3 years ago

Gummy gummy bears gummy gummy bears

Engineering

2 answers:

ivanzaharov [21]3 years ago

8 0

I love that song lol

Dima020 [189]3 years ago

8 0

Answer:

Oh, I'm a gummy bear

Yes, I'm a gummy bear

Oh, I'm a yummy, tummy, funny, lucky gummy bear

I'm a jelly bear

'Cause I'm a gummy bear

Oh, I'm a movin', groovin', jammin', singin' gummy bear

Oh yeaoooh

Gummy Gummy Gummy Gummy Gummibär

Bai ding ba doli party

Bamm bing ba doli party

Breding ba doli party

Party pop

Bai ding ba doli party

Bamm bing ba doli party

Breding ba doli party

Party pop

Oh, I'm a gummy bear

Yes, I'm a gummy bear

Oh, I'm a yummy, tummy, funny, lucky gummy bear

I'm a jelly bear

'Cause I'm a gummy bear

Oh, I'm a movin', groovin', jammin', singin' gummy bear

Oh yeaoooh

Ba Ba Bidubidubi Yum Yum

Three times, you can bite me

Ba Ba Bidubidubi Yum Yum

Three times, you can bite me

Gummy Gummy Gummy Gummy Gummibär

Én vagyok a gumimaci

Gyere már gyere velem

Gyere táncolj már kérlek get get party pop

Én vagyok a gumicummi

Cicci meg a kodile a mambó csoda

Pápé gyere már és táncolj party pop

Oh, I'm a gummy bear

Yes, I'm a gummy bear

Oh, I'm a yummy, tummy, funny, lucky gummy bear

I'm a jelly bear

'Cause I'm a gummy bear

Oh, I'm a movin', groovin', jammin', singin' gummy bear

Oh, I'm a gummy bear

Yes, I'm a gummy bear

Oh, I'm a yummy, tummy, funny, lucky gummy bear

I'm a jelly bear

'Cause I'm a gummy bear

Oh, I'm a movin', groovin', jammin', singin' gummy bear

Oh, I'm a gummy bear

Yes, I'm a gummy bear

Oh, I'm a yummy, tummy, funny, lucky gummy bear

I'm a jelly bear

'Cause I'm a gummy bear

Oh, I'm a movin', groovin', jammin', singin' gummy bear

Party pop

You might be interested in

A high molecular weight hydrocarbon gas A is fed continuously into a heated mixed flow reactor (0.1liter) where it is thermally

dimulka [17.4K]

Answer:

Space velocity = 30 hr⁻¹

Explanation:

Space velocity for reactors express how much reactor volume of feed or reactants can be treated per unit time. For example, a space velocity of 3 hr⁻¹ means the reactor can process 3 times its volume per hour.

It is given mathematically as

Space velocity = (volumetric flow rate of the reactants)/(the reactor volume)

Volumetric flowrate of the reeactants

= (molar flow rate)/(concentration)

Molar flowrate of the reactants = 300 millimol/hr

Concentration of the reactants = 100 millimol/liter

Volumetric flowrate of the reactants = (300/100) = 3 liters/hr

Reactor volume = 0.1 liter

Space velocity = (3/0.1) = 30 /hr = 30 hr⁻¹

Hope this Helps!!!

5 0

3 years ago

Where are the ar manufacturers not fitting the engine in the high end sport cars

fomenos

Answer:

it depends on the but i would recommend check in the front next to the turbo intake.

8 0

2 years ago

Can a 1½ " conduit, with a total area of 2.04 square inches, be filled with wires that total 0.93 square inches if the maximum f

Papessa [141]

Answer:

it is not possible to place the wires in the condui

Explanation:

given data

total area = 2.04 square inches

wires total area = 0.93 square inches

maximum fill conduit = 40%

to find out

Can it is possible place wire in conduit conduit

solution

we know maximum fill is 40%

so here first we get total area of conduit that will be

total area of conduit = 40% × 2.04

total area of conduit = 0.816 square inches

but this area is less than required area of wire that is 0.93 square inches

so we can say it is not possible to place the wires in the conduit

4 0

3 years ago

A hemispherical shell with an external diameter of 500 mm and a thickness of 20 mm is going to be made by casting, located entir

Olenka [21]

Solution :

Given :

External diameter of the hemispherical shell, D = 500 mm

Thickness, t = 20 mm

Internal diameter, d = D - 2t

= 500 - 2(20)

= 460 mm

So, internal radius, r = 230 mm

= 0.23 m

Density of molten metal, ρ = $7.2 \ g/cm^3$

= $7200 \ kg/m^3$

The height of pouring cavity above parting surface is h = 300 mm

= 0.3 m

So, the metallostatic thrust on the upper mold at the end of casting is :

$F=\rho g A h$

Area, A $=2 \pi r^2$

$=2 \pi (0.23)^2$

$=0.3324 \ m^2$

$F=\rho g A h$

$=7200 \times 9.81 \times 0.3324 \times 0.3$

= 7043.42 N

3 0

2 years ago

A light bar AD is suspended from a cable BE and supports a 20-kg block at C. The ends A and D of the bar are in contact with fri

babymother [125]

Answer:

Tension in cable BE= 196.2 N

Reactions A and D both are 73.575 N

Explanation:

The free body diagram is as attached sketch. At equilibrium, sum of forces along y axis will be 0 hence

$T_{BE}-W=0$ hence

$T_{BE}=W=20*9.81=196.2 N$

Therefore, tension in the cable, $T_{BE}=196.2 N$

Taking moments about point A, with clockwise moments as positive while anticlockwise moments as negative then

$196.2\times 0.125- 196.2\times 0.2+ D_x\times 0.2=0$

$24.525-39.24+0.2D_x=0$

$D_x=73.575 N$

Similarly,

$A_x-D_y=0$

$A_x=73.575 N$

Therefore, both reactions at A and D are 73.575 N

7 0

3 years ago