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Len [333]
3 years ago
14

A fire hose is made of six sections, each 15 feet in length. The sections are joined together with Storz couplings, which are fo

ur inches long each. What is the total length of the hose?
Physics
1 answer:
gayaneshka [121]3 years ago
8 0

Answer:

h=740\ in

Explanation:

Given:

length of each pipe section, l=15\ ft=180\ in

length of each coupling, c=4 \ in

Now according to question we have six hoses which will require 5 couplings to join them.

So the total length of the hose:

h=6\times l+5\times c

h=6\times 180+5\times 4

h=740\ in is  the total length of the hose.

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If an object is being pulled by two forces, one 4 N to the left and the other 2 N to the right, what is the net
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Answer:

2N

Explanation:

subtract  rthe two forces to see which is greater

4-2=2

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2 years ago
An airplane travels from st louis to portland, oregon in 4.33 hours. if the distance traveled is 2,742 kilometers, what is the a
Makovka662 [10]

Average speed = (total distance covered) / (total time to cover the distance)

                       =  (2,742 km)  /  (4.33 hours)

                       =  (2,742 / 4.33)    km/hr

                       =      633 km/hr        (rounded)
4 0
3 years ago
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What body part plays a strong role in releasing power​
Eddi Din [679]

Answer:

I think your bones, muscles, and joints your welcome :)

Explanation:

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3 years ago
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Convert <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%280.779mg%29%28min%29%7D%7BL%7D" id="TexFormula1" title="\frac{(0.779mg)(mi
Orlov [11]

The number converted is 0.0467 \frac{(kg)(s)}{m^3}

Explanation:

In order to convert from the original units to the final units, we have to keep in mind the following conversion factors:

1 kg = 1000 g = 10^6 mg

1 min = 60 s

1 m^3 = 1000 L

The original unit that we have is

\frac{mg\cdot min}{L}

Therefore, it can be rewritten as:

=\frac{mg \frac{1}{10^6 mg/kg}\cdot min\cdot  60 s/min}{L\frac{1}{1000L/m^3}}=0.06 \frac{(kg)(s)}{m^3}

Therefore, since the initial number was 0.779, the final value is

0.779\cdot 0.06 \frac{(kg)(s)}{m^3}=0.0467 \frac{(kg)(s)}{m^3}

#LearnwithBrainly

5 0
3 years ago
A CFL bulb has an efficiency of 8.9% and a power of 22 W. How much light energy does the lightbulb produce in 1 second
hjlf
The power that the light is able to utilize out of the supply is only 0.089 of the given.
                           Power utilized = (0.089)(22 W)
                                                  = 1.958 W
                                                  = 1.958 J/s
The energy required in this item is the product of the power utilized and the time. That is,
                           Energy = (1.958 J/s)(1 s) = 1.958 J
Thus, the light energy that the bulb is able to produce is approximately 1.958 J. 
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