Trucking A Way

Hey there.

I work out at Golds gym, and I am a amateur bodybuilder. Girls often look at me when I pass by and comment on my physique and ask for workout tips. I have dated a few girls that I met at the gym, but when I picked them up for a date in my Mustang, they seemed horrified by my outfit. I like to wear clothes that are "body conscious". I mean, why hide what I work so hard to build right? I mean, I dress like most of the other muscle heads at Golds (i’e Golds gym logo tank top, baggy weight lifter pants, and a leather back support belt) Sometimes I wear accessories like wearing wrist bands, head band, or a do rag (even though I’m white), I keep my Ipod, wallet, and pen (to write down my phone number) in a fanny pack that I wear in front of me.
I mean If the girl met me in this outfit, why should she be so horrified that I wore it on our date?
So I am trying to get ready for a date this Friday night and I feel really weird saying this, but I don’t know what to wear. These are my possible choices.
1. Lightning bolt covered weight lifter pants, and Hard rock cafe shirt.
2. Camouflage cargo shorts with a "Tapout" t-shirt.
3. Denim shorts and a skin tight black "underarmor" shirt
4. plain blue jeans and a baggy golds gym sweatshirt (sleeveless)

I know it doesn’t sound like I have a lot of clothes but its because if you are big like me, it’s hard to find clothes that fit.

BTW. I saw some other bodybuilders from the gym out at a bar one night. They seemed what you might be called "metro-sexual". They had their hair gelled up, and had stripped button down shirts on with "True Religion" jeans. Come to think of it, almost every guy at the bar was dressed like that. Do you think I should go that route and climb out of my Golds gym stuff and into some True Religion wear?

Thanks

Bro,

This girl you are spending time with will most likely want to get to know you on a more personal level. She may have met you as a warrior in the gym, but show her your a star outside of the gym.

Suggestions:
Hair: clean, no product
Shirt: plain shirt with a button down shirt open on top ( plain shirt may be tight if you would like)
Pants: Jeans are good, but dark blue please
Shoes: sandals or nice flip flops

This kind of outfit promotes being open and casual. Its makes the mood not as tense and will enable her to be more chatty with you, and eventually lead to great conversations and the Development of a good friendship

In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole. With what minimum speed must the athlete leave the ground in order to lift his center of mass 2.22m and cross the bar with a speed of 0.740m/s?

A spherically shaped balloon has a radius of 8.68m, and is filled with helium. How large a mass of cargo can it lift, assuming that the skin and structure of the balloon have a mass of 816kg? Neglect the bouyant force on the cargo volume itself. The density of air is 1.29 kg/m3 and the density of helium is 0.179 kg/m3.

I would love help on either of these thanks!

Hello,

I will try.

Now, I believe that his motion is like the arc you see in projectile motion. The point at which he is moving 0.740 m/s and 2.22 m higher is actually the peak of the arc of projectile motion. He has stopped moving vertically, he is only moving horizontally at 0.740 m/s. But his horizontal motion still creates some small final kinetic energy.

KEi = PEf + KEf
KEi = (1/2) m vi^2
PEf = mgh
KEf = (1/2) m vf^2

(1/2) m vi^2 = mgh + (1/2) m vf^2
(1/2) vi^2 = gh + (1/2) vf^2
vi = sgrt(2(gh + (1/2) vf^2))
vi = sgrt(2gh + vf^2))
vi = sgrt(2*9.81 m/s^2*2.22 m + (0.740 m/s)2))
vi = 6.64 m/s

b) Draw a free body diagram
There is weight of the balloon (Wb)pulling down.
There is the weight of the HELIUM (Wh) pulling down.
There is a buoyant force pushing up (B).
There is the weight of the cargo (Wc) pulling down.

So to be balanced, we must have;

B = Wb + Wh + Wc

Wb = 816 kg * 9.81 m/s^2 = 8005 N
Wh = Density of helium * Volume of sphere * gravity
Wh = 0.179 kg/m^3 * (4/3) pi (8.68 m)^3 * 9.81 m/s^2
Wh = 4810 N
Wc = to be determined
B = weight of the displaced air
B = Density of air * Volume of sphere * g
B = 1.29 kg/m^3 * (4/3) pi (8.68 m)^3 * 9.81 m/s^2
B = 34665 N

B = Wb + Wh + Wc
Wc = B – Wb – Wh
Wc = 34665 N – 8005 N – 4810 N
Wc = 21850 N

Good Luck nicole