Straight Shot
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Rattleware 20-Ounce Latte Art Milk Frothing Pitcher
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Description20 oz. straight side steaming pitcher as seen in David Schomer's Latte Art video. This steaming pitcher is made of 18/10 high quality stainless steel. This is the most popular espresso milk steaming pitcher available and is the very same frothing pitcher many espresso pros use to create contest winning latte art... |
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Rattleware 12-Ounce Latte Art Milk Frothing Pitcher
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DescriptionThis 12 ounce straight sided steaming pitcher is made from 18/10 high quality stainless steel. Ideally milk for espresso should be steamed as needed on a per drink basis. This espresso milk steaming pitcher would be a recommended size for the home or office, or kept on hand to make smaller drinks in the coffee shop... |
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Disposable Plastic Straight Wall Shooter Cups - 5 oz: Pack of 12 Cups
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DescriptionThese plastic shooter glasses are a must-have for your next party! Offer your guests awesome layered drinks, shots or even desserts, right in the shooter cups. Perfect for serving shooters or even desserts |
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Illuminations
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DescriptionGROBAN JOSH ILLUMINATIONS |
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DescriptionCOODER RY PULL UP SOME DUST AND SIT DOWN |
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DescriptionVHS Tape |
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Playboy - The Ultimate Pamela Anderson (Body Shot Cover) [VHS]
Sale Price: $24.99 |
DescriptionShe's one of pop culture's greatest icons, and now Playboy brings you everything Pamela, from her discovery to present day and all in one place. |
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Danco 89171 Shut-Off Shower Valve, Chrome
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A projectile of mass 1.300 kg is shot straight up with an initial speed of 25.0 m/s.(a) How high would it go?
A projectile of mass 1.300 kg is shot straight up with an initial speed of 25.0 m/s.(a) How high would it go if there were no air friction?(b) If the projectile rises to a maximum height of only 27.1 m, determine the magnitude of the average force due to air resistance.
Given: m=1,300 kg, v0=25.0 m/s, vf= 0 m/s,
g= -9.81 m/s
Using equations of motion
vf^2-v0^2=2g(hmax)
since vf= 0 m/s the equation becomes
-(v0)^2=2g(hmax)
hmax= -(v0)^2/(2g)
hmax= -(25.0 m/s)^2/(2*-9.81 m/s^2)
hmax=(-625 m^2/s^2)/(-19.62 m/s^2)
hmax=31.86 m
b) Given: hmax=27.1 m,m=1,300 kg, v0=25.0 m/s,
vf= 0 m/s, g= -9.81 m/s
vf^2-v0^2=2a(hmax)
since vf^2=0 the equation reduces t0
-v0^2=2a(hmax)
a= -v0^2/(2*hmax)
a=(-25.0 m/s)^2/(2*27.1 m)
a=(-625 m^2/s^2)/(54.2 m)
a= -11.53 m/s^2
Since this value accounts for gravity the
acceleration due to friction can be found from
afric=a-g
afric=(-11.53 m/s^2)-(-9.81 m/s^2)
afric=-1.72 m/s
Ffric=ma
Ffric=(1.300 kg)(-1.72 m/s)
Ffric=-2.236 N
Answer: the average force due to air
resistance is -2.236 N
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