All categories

2 years ago

5. A 15-nC point charge is at the origin in free space. Calculate V₁ if point P, is located at

Engineering

1 answer:

Gala2k [10]2 years ago

6 0

I don’t know lm so sorry

You might be interested in

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

A 3 m aluminum pole is kept at a residential site for construction

Aliun [14]

Answer:

I don't know sorry

Explanation:

5 0

3 years ago

Por favor. alguien me comunique con fatimalisethmateual

lions [1.4K]

Wait why do you want me to

7 0

2 years ago

List at least two things that scientists and explorers have in common.

Tresset [83]

• educated
•very curious
•discover new things
•Investigate
•search unknown

8 0

3 years ago

Read 2 more answers

a ten station assembly machine has ideal cycle time of 6 sec. the fraction defect rate at each station 0.005 and defect always j

MAXImum [283]

Answer:

$T_{P}=(2.6667)(10^{-3})h$

Explanation:

Let's write the equation of the production rate for the assembly machine :

$T_{P}=T_{C}+(n).(m).(p).(T_{D})$

Where $T_{P}$ is the production rate for the assembly machine.

Where $T_{C}$ is the ideal cycle time

Where n is the number of stations.

Where m is the number stations that get jam when the defect occurs.

Where p is the defect rate at each station.

And where $T_{D}$ is the average downtime per breakdown

We are looking for the hourly production rate ⇒

$1h=60min\\1min=60s$ ⇒

$1h=3600s$ ⇒

$6s=\frac{(6s)(1h)}{(3600s)}= \frac{1}{600}h$

$60min=1h$ ⇒

$1.2min=\frac{(1.2min)(1h)}{(60min)}=0.02h$

$T_{P}=\frac{1}{600}h+(10)(1.0)(0.005)(0.02h)=\frac{1}{375}h=(2.6667)(10^{-3})h$

m = 1.0 in the equation.

3 0

3 years ago