# Posts by Henri

Total # Posts: 56

**math**

t = time d = distance Alana walking rate: 1 meter per second Gilberto walking rate: 2 meters per second Leanne walking rate: 2.5 rating per second Alana Equation: d = t Gilberto Equation: d = 2t Leanne Equation: d = 2.5t I made a table for the first ten seconds of each of the ...

*March 6, 2017*

**physics**

its to the surface

*February 2, 2017*

**physics**

The surface shown is a square and has a side length of 14.0 cm. The electric field acting on this area has a magnitude of 180. N/C at an angle of 27.5°. Calculate the electric flux through the shown surface.

*February 2, 2017*

**Math**

x = red roses y = pink roses y = 3y 2x = 4y + 10 solve the system for your answer

*September 25, 2016*

**math**

divide 16 1/2 by 5 2/5 first create an improper fraction (multiply number by denominator, then add it to the numerator) then divide the fractions (multiply by the reciprocal/opposite of the second number)

*September 25, 2016*

**math**

8s-29=34+3

*December 11, 2013*

**math**

12 cells

*August 31, 2012*

**Math**

give the number whose hundred place is one-third of the greatest one-digit number, the tens place is twice the hundred place, the ones place is four times the digit in the hundreds place divided by 2, and the three digits in the thousands period is the product of the digits in...

*June 19, 2012*

**applied mechanics**

A solid shaft of 65mm outside diameter and a hollow shaft of 85mm outside diameter are connected by 6 bolts with the mean pitch of thread being 155mm. The shear stress due to the torque on the shafts is equal to the shear stress on the bolts. a) Find the diameter of the bolts...

*May 7, 2009*

**applied mechanics**

How do we calculate the "radius of gyration" of a 500g steel rod spinning at 300 RPM and having a lenght of 30cm? We need this to find the moment of Inertia... don't we?

*May 7, 2009*

**French**

7. Pour aller au mariage, Édouard met une chemise et une ---------. Answer: chers Ans. cravate 11. Maurice et Chantal -------- des courses au grand magasin. Answer: taille Ans. font Good job, Bonne Chance, H

*May 6, 2009*

**Physics**

How do we calculate the "radius of gyration" of a 500g steel rod spinning at 300 RPM and having a lenght of 30cm? We need this to find the moment of Inertia... don't we?

*May 6, 2009*

**Physics**

A solid shaft of 65mm outside diameter and a hollow shaft of 85mm outside diameter are connected by 6 bolts with the mean pitch of thread being 155mm. The shear stress due to the torque on the shafts is equal to the shear stress on the bolts. a) Find the diameter of the bolts...

*May 6, 2009*

**french**

*Aller en* Afrique pour aider les éléphants... (sounds better to me and i'm a french Canadian! lol) *à l'Afrique*... is incorrect

*May 5, 2009*

**French**

You could also have said: *je demeure*

*May 4, 2009*

**French**

Oops,missed one:J'ai quatorze ans et (je)*j'*+habite=j'habite

*May 4, 2009*

**French**

Je m'appelle Larry. J'ai quatorze ans et je habite (dans)*à* Vernon, BC au Canada. *Là* où je habite, au printemps il fait beau(.)*et* en été il fait chaud. Mais (en)*à l'automne* (autumne) il fait *plus* frais(.)*et* (...

*May 4, 2009*

**Applied Machanics (Physics)**

What do you mean by: 1/12ml^2 k^2=radius of gyration

*May 3, 2009*

**Applied Machanics (Physics)**

A steel rod of 500g and 30 cm long spins at 300 RPM,the rod pivots around the center. a) Find angular momentum Moment of Inertia(I)=mk^2 I=0.5kgX(0.075m)^2 I=0.0028125kg*m^2 To find angular momentum: w=300rpmX2pi/60sec w=31.42 rads/s Ang. Momentum=Iw A.M.=0.0028125X31.42 A.M.=...

*May 3, 2009*

**French 1**

c)*La* vitrine b)des

*May 3, 2009*

**French (URGENT)**

Bonjour, *Ici*("voici" instead) un petit paragraphe à propos de mon ami James. James est NÉ EN Alberta mais il habite Vernon, BC *de*("au" instead) Canada. Il a quinze ans ET il a un frère cadet nommé Thomas. Thomas a quatorze ...

*April 27, 2009*

**Please check my math**

A turnbuckle is needed to apply an 8kN force on a cable. If the efficiency of the turnbuckle is 40% and the force applied is 159.155N with a lever of 20cm in length. a)Find Mechanical Advantage: M.A.=Load/Effort M.A.=8000/159.155 M.A.=50.265 (ans.) b)Find the velocity ratio: ...

*April 26, 2009*

**French**

....et des tennis("espadrilles" instead) Mon père met... Ma mère... "au mois de juillet" and "au mois de janvier"... at the very end of the sentence, not at the begining.

*April 24, 2009*

**Physics (is this right?)**

A turnbuckle is needed to apply an 8kN force on a cable. If the efficiency of the turnbuckle is 40% and the force applied is 159.155N with a lever of 20cm in length. a)Find Mechanical Advantage: M.A.=Load/Effort M.A.=8000/159.155 M.A.=50.265 (ans.) b)Find the velocity ratio: ...

*April 24, 2009*

**Physics**

A turnbuckle is needed to apply an 8kN force on a cable. If the efficiency of the turnbuckle is 40% and the force applied is 159.155N with a lever of 20cm in length. a)Find Mechanical Advantage: M.A.=Load/Effort M.A.=8000/159.155 M.A.=50.265 (ans.) b)Find the velocity ratio: ...

*April 23, 2009*

**Physics**

Lubricating oil, with a relative density of 0.79, flows around a 90¨¬ bend. The pipe diameter is 0.45m, and the oil has a pressure head of 7m and the flow is 1.7m3/s. Find the force exerted by the oil on the bend. Oil density = 0.79 X 103 Volumetric mass flow rate = 1....

*April 23, 2009*

**French**

Le printemps est ma saison favorite. Je crois que cette saison a plus d'avantages que d'inconvénients. Il est vrai qu'il pleut beaucoup et quand le temps est beau, c'est plus facile d'être distrait, mais à mon avis, les gens sont plus ...

*April 22, 2009*

**Physics**

A turnbuckle is needed to apply an 8kN force on a cable. If the efficiency of the turnbuckle is 40% and the force applied is 159.155N with a lever of 20cm in length. a)Find Mechanical Advantage: M.A.=Load/Effort M.A.=8000/159.155 M.A.=50.265 (ans.) b)Find the velocity ratio: ...

*April 22, 2009*

**Physics**

Lubricating oil, with a relative density of 0.79, flows around a 90¨¬ bend. The pipe diameter is 0.45m, and the oil has a pressure head of 7m and the flow is 1.7m3/s. Find the force exerted by the oil on the bend. Oil density = 0.79 X 103 Volumetric mass flow rate = 1....

*April 22, 2009*

**Spring stiffness**

Thanks drwls... could you check the next one please

*April 22, 2009*

**Spring stiffness**

A spring has stiffness of 100Nmm, the initial compression of the spring is 15mm, and it is then compressed to 65mm. a)Find potential energy at 15mm: K=100Nmm*1000=100000Nm Ep=Kc^2/2 Ep=(100000*0.015^2)/2 Ep=11.25 joule (ans.) b)Find Ep at 65mm: Ep2=(100000*0.065^2)/2 Ep2=211....

*April 22, 2009*

**Math check please**

A spring has stiffness of 100Nmm, the initial compression of the spring is 15mm, and it is then compressed to 65mm. a)Find potential energy at 15mm: K=100Nmm*1000=100000Nm Ep=Kc^2/2 Ep=(100000*0.015^2)/2 Ep=11.25 joule (ans.) b)Find Ep at 65mm: Ep2=(100000*0.065^2)/2 Ep2=211....

*April 21, 2009*

**Is my math correct(2)**

What load can be lifted by a screw jack given: Toggle1 length:520mm Force applied:230N Toggle2 lenght:445mm Force applied:300.557N Mean pitch of thread:13mm S.J. efficiency:35% My answer: Assuming T2 to be 520mm long, the equivalent force would be: 0.445mX300.557N=0.52m X F2 ...

*April 20, 2009*

**Is my math correct?**

Thank you so much

*April 20, 2009*

**Is my math correct?**

Find the Inside diameter(d) and outside diameter(D) of a hollow shaft given: Power transmitted by shaft=1400kW Shaft speed=45 RPM Angle of twist on shaft=1 degree Shaft lenght(L)=25D Assume d=0.6D Modulus of Rigidity(G)=81.75GPa My answer: Shaft ang. vel.=45*2pi/60s A.V.=4....

*April 20, 2009*

**Applied Mech.2**

Find the Inside diameter(d) and outside diameter(D) of a hollow shaft given: Power transmitted by shaft=1400kW Shaft speed=45 RPM Angle of twist on shaft=1 degree Shaft lenght(L)=25D Assume d=0.6D Modulus of Rigidity(G)=81.75GPa My answer: Shaft ang. vel.=45*2pi/60s A.V.=4....

*April 19, 2009*

**Applied Mech. (with my answer..lol)**

What load can be lifted by a screw jack given: Toggle1 length:520mm Force applied:230N Toggle2 lenght:445mm Force applied:300.557N Mean pitch of thread:13mm S.J. efficiency:35% My answer: Assuming T2 to be 520mm long, the equivalent force would be: 0.445mX300.557N=0.52m X F2 ...

*April 19, 2009*

**FRENCH**

The way it is used here is to show that the "art pieces" were made in either France or Maroc

*March 22, 2009*

**Torque and shear stress**

Thanks... I couldn't figure out why there's is a thread pitch either!!!

*March 22, 2009*

**Torque and shear stress**

A solid shaft of 65mm outside diameter and a hollow shaft of 85mm outside diameter are connected by 6 bolts with the mean pitch of thread being 155mm. The shear stress due to the torque on the shafts is equal to the shear stress on the bolts. a) Find the diameter of the bolts...

*March 22, 2009*

**Question & Work**

Thank you very much...It's probably my "new" word... 2007 student edition.

*March 21, 2009*

**Question & Work**

Can somebody tell me how come my work doesn't get posted with my question.

*March 21, 2009*

**please check my math2**

This was the original post... Two bodies hang on a string 2.1m long. The two bodies have a mass of 3.75kg and 2.4kg respectively. If the 3.75kg mass is raised to a height of 0.545m and released, and if upon impact the mass adhere. Mass1 = 3.75 kg Mass2 = 2.4 kg String length...

*March 21, 2009*

**please check my math2**

You figured it out, thanks Damon. But how come my work didn't get posted with the question... I did it like you up until the 12.2kg m/s part... Now i know where I went wrong.. T-Y Henri

*March 21, 2009*

**please check my math2**

Two bodies hang on a string 2.1m long. The two bodies have a mass of 3.75kg and 2.4kg respectively. If the 3.75kg mass is raised to a height of 0.545m and released, and if upon impact the mass adhere. Mass1 = 3.75 kg Mass2 = 2.4 kg String length = 2.1 m M1 raised = 0.545m a) ...

*March 21, 2009*

**Please check my math**

How come my work doesn't show up with my question??? Is it to long?

*March 21, 2009*

**Please check my math**

A spring has stiffness of 100Nmm, the initial compression of the spring is 15mm, and it is then compressed to 65mm. 1Nmm = 0.001Nm 100Nmm = 100 X 0.001 = 0.1Nm Comp.1 = 15 mm = 0.015m Comp.2 = 65 mm = 0.065m a) i-Find potential energy at 15mm. ii- Find potential energy at 65mm...

*March 21, 2009*

**Please check my math 2**

A spring has stiffness of 100Nmm, the initial compression of the spring is 15mm, and it is then compressed to 65mm. 1Nmm = 0.001Nm 100Nmm = 100 X 0.001 = 0.1Nm Comp.1 = 15 mm = 0.015m Comp.2 = 65 mm = 0.065m a) i-Find potential energy at 15mm. ii- Find potential energy at 65mm...

*March 21, 2009*

**Please check my math**

A turnbuckle is needed to apply an 8kN force on a cable. If the efficiency of the turnbuckle is 40% and the force applied is 159.155N with a lever of 20cm in length. a) Find the mechanical advantage. Mechanical Advantage=Load/Effort M.A.=8000N/159.155N M.A.=50.265 (Ans.) b) ...

*March 21, 2009*

**quad formula help**

2X2+8x=-1 4+8x=-1 8x=-1-4 x=-5/8 x=-0.625

*March 21, 2009*

**Velocity Ratio**

Does my work make sense? A turnbuckle is needed to apply an 8kN force on a cable. If the efficiency of the turnbuckle is 40% and the force applied is 159.155N with a lever of 20cm in length. a) Find the mechanical advantage. Mechanical Advantage=Load/Effort M.A.=8000N/159.155N...

*March 20, 2009*

**Applied Mechanics**

Is my work correct? Two bodies hang on a string 2.1m long. The two bodies have a mass of 3.75kg and 2.4kg respectively. If the 3.75kg mass is raised to a height of 0.545m and released, and if upon impact the mass adhere. Mass1 = 3.75 kg Mass2 = 2.4 kg String length = 2.1 m M1 ...

*March 20, 2009*

**Fluids in motion**

Thank you, makes so much sence now! lol :-)

*March 20, 2009*

**Fluids in motion**

Lubricating oil, with a relative density of 0.79, flows around a 90º bend. The pipe diameter is 0.45m, and the oil has a pressure head of 7m and the flow is 1.7m3/s. Find the force exerted by the oil on the bend.

*March 20, 2009*

**applied mechanics**

Solid shafts and larger diameter hollow shafts are usally connected throught a flange welded over the smaller diameter of the solid shaft. The flange outside diameter is the same as the hollow shaft. Bolts are then used to secure the two shafts together.

*March 20, 2009*

**applied mechanics**

A solid shaft of 65mm outside diameter and a hollow shaft of 85mm outside diameter are connected by 6 bolts with the mean pitch of thread being 155mm. The shear stress due to the torque on the shafts is equal to the shear stress on the bolts. a) Find the diameter of the bolts...

*March 20, 2009*

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