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3 years ago

12

What is the angle of the input

1 answer:

olga55 [171]3 years ago

3 0

Absolute positions — latitudes and longitudes
Relative positions — azimuths, bearings, and elevation angles
Spherical distances between point locations

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Sam decides to remove his oversized hooded jacket before he works near the pto driveline is it safe or unsafe

Salsk061 [2.6K]

Its safe because it isn't something with electricity

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3 years ago

State three active materials of a lead acid cell

igomit [66]

Answer:

lead dioxide,sulfate and lead acid

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3 years ago

Read 2 more answers

A train starts from rest at station A and accelerates at 0.6 m/s^2 for 60 s. Afterwards it travels with a constant velocity for

Aleks [24]

Answer:

The distance between the station A and B will be:

$x_{A-B}=55.620\: km$

Explanation:

Let's find the distance that the train traveled during 60 seconds.

$x_{1}=x_{0}+v_{0}t+0.5at^{2}$

We know that starts from rest (v(0)=0) and the acceleration is 0.6 m/s², so the distance will be:

$x_{1}=\frac{1}{2}(0.6)(60)^{2}$

$x_{1}=1080\: m$

Now, we need to find the distance after 25 min at a constant speed. To get it, we need to find the speed at the end of the first distance.

$v_{1}=v_{0}+at$

$v_{1}=(0.6)(60)=36\: m/s$

Then the second distance will be:

$x_{2}=v_{1}*1500$

$x_{2}=(36)(1500)=54000\: m$

The final distance is calculated whit the decelerate value:

$v_{f}^{2}=v_{1}^{2}-2ax_{3}$

The final velocity is zero because it rests at station B. The initial velocity will be v(1).

$0=36^{2}-2(1.2)x_{3}$

$x_{3}=\frac{36^{2}}{2(1.2)}$

$x_{3}=540\: m$

Therefore, the distance between the station A and B will be:

$x_{A-B}=x_{1}+x_{2}+x_{3}$

$x_{A-B}=1080+54000+540=55.620\: km$

I hope it helps you!

7 0

3 years ago

A pin connection supports a load. Which type of connection would result in the smallest shear stress in the pin

Viktor [21]

Answer:

A single- shear pin connection.

Explanation:

8 0

2 years ago

A lake is fed by a polluted stream and a sewage outfall. The stream and sewage wastes have a decay rate coefficient (k) of 0.5/d

joja [24]

Solution :

Given :

k = 0.5 per day

$C_s = 10 \ mg/L \ ; \ \ Q_s= 40 \ m^3/s$

$C_{sw} = 100 \ ppm \ ; \ \ Q_{sw}= 0.5 \ m^3/s$

Volume, V $= 200 \ m^3$

Now, input rate = output rate + KCV ------------- (1)

Input rate $= Q_s C_s + Q_{sw}C_{sw}$

$=(40 \times 10) + (0.5\times 100)$

$= 2 \times 10^5 \ mg/s$

The output rate $= Q_m C_{m}$

= ( 40 + 0.5 ) x C x 1000

$=40.5 \times 10^3 \ C \ mg/s$

Decay rate = KCV

∴ $KCV =\frac{0.5/d \times C \ \times 200 \times 1000}{24 \times 3600}$

= 1.16 C mg/s

Substituting all values in (1)

$2 \times 10^5 = 40.5 \times 10^3 \ C+ 1.16 C$

C = 4.93 mg/L

4 0

3 years ago