Question 1/5

What is the probability that Tina will NOT wear a white t-shirt on the first day of her trip?

katrin2010 [14]

Answer:

4/5

Explanation:

She is not wearing white t-shirt on the first day so she is wearing the other 4 t-shirt

4 0

3 years ago

To compute the energy used by a motor, multiply the power that it draws by the time of operation. Con- sider a motor that draws

ehidna [41]

Answer:

E=52000Hp.h

E=38724920Wh

E=1.028x10^11 ftlb

Explanation:

To solve this problem you must multiply the engine power by the time factor expressed in h / year, to find this value you must perform the conventional unit conversion procedure.

Finally, when you have the result Hp h / year you convert it to Ftlb and Wh

$E=(12.5hp)(\frac{16h}{day} )(\frac{5 days}{week} )(\frac{52week}{year} )\\$

E=52000Hp.h

$E=52000Hp.h(\frac{744.71Wh}{Hp.h} )\\$

E=38724920Wh

$E=52000Hph(\frac{1977378.4 ft lb}{1Hph}$

E=1.028x10^11 ftlb

3 0

3 years ago

Why does teachers grade things that are not due yet

pav-90 [236]

I think because if you’ve already turned it in they might as well grade asap instead of waiting

4 0

3 years ago

Read 2 more answers

A road has a crest curve, where the PVI station is a 71 35. The road transitions from a 2.1% grade to a -3.4% grade. The highest

sveticcg [70]

Answer:

Stat PVC = Stat(82+98.5)

Stat PVT = Stat(59+71.5)

Explanation

PVI = 71 + 35

Let G1 = Grade 1; G2 = Grade 2

G1 = +2.1% ; G2 = -3.4%

Highest point of curve at station = 74 + 10

General equation of a curve:

$y = ax^{2} +bx+c\\dy/dx=2ax+b\\$

At highest point of the curve $dy/dx=o$

$2ax+b=0\\x=-b/2a\\x=G1L/(G2-G1)\\x=L/2 +(stat 74+10)-(stat 71+35)\\x=L/2 + 275$

$-G1L/(G2-G1) = (L/2 + 275)/100\\L = -2327 ft\\Station PVC = Stat(71+35)+(-2327/2)\\\\Stat PVC = 7135-1163.5\\Stat PVC = Stat(82+98.5)\\$

Station PVT

$Station PVT = Stat PVI + (L/2)\\Station PVT = Stat(71+35)+(-2327/2)\\Station PVT = 7135-1163.5\\Stat PVT = Stat(59+71.5)$

3 0

3 years ago

What mass of LP gas is necessary to heat 1.4 L of water from room temperature (25.0 ∘C) to boiling (100.0 ∘C)? Assume that durin

DochEvi [55]

Answer:

$m_{LP}=0.45\,kg$

Explanation:

Let assume that heating and boiling process occurs under an athmospheric pressure of 101.325 kPa. The heat needed to boil water is:

$Q_{water} = (1.4\,L)\cdot(\frac{1\,m^{3}}{1000\,L} )\cdot (1000\,\frac{kg}{m^{3}} )\cdot [(4.187\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (100^{\textdegree}C-25^{\textdegree}C)+2257\,\frac{kJ}{kg}]$