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Masteriza [31]
3 years ago
6

When using two-wire cable to feed a 240-V appliance that does not require a neutral wire, you should A. mark both ends of the wh

ite wire green. B. use the cable as is. C. use only the black wire. D. mark both ends and any exposed area of the white wire black.
Physics
1 answer:
DENIUS [597]3 years ago
3 0

A, you always want to mark wires with a color that is noticeable

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Which of these will be the correct relationship between work input and work output?
dsp73

<u>Answer:</u>

Work input = Work output * Work against friction is your answer so C

<u>Explanation:</u>

I hope this helps you :)

8 0
2 years ago
What is the current in a 160V circuit if the resistance is 2Ω?<br> V=<br> I=<br> R=
Alex787 [66]

Explanation:

v = IR

v= 160 R = 2

<u>160</u> = <u>2I</u>

2 2

I = 80A

4 0
2 years ago
A test car travels in a straight line along the x-axis. The graph in the figure shows the car’s position x as a function of time
svetlana [45]
If it is s-t graph , point is c
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8 0
2 years ago
Read 2 more answers
The isomerization of cyclopropane to form propene is a first-order reaction. At 760 K, 85% of a sample of cyclopropane changes t
Nataliya [291]

Answer:

so rate constant  is 4.00 x 10^-4 s^{-1}

Explanation:

Given data

first-order reactions

85% of a sample

changes to propene t =  79.0 min

to find out

rate constant

solution

we know that

first order reaction are

ln [A]/[A]0 = -kt

here [A]0 = 1 and (85%) = 0.85 has change to propene

so that [A] = 1 - 0.85 = 0.15.

that why

[A] / [A]0= 0.15 / 1

[A] / [A]0 = 0.15

here t = (79) × (60s/min) = 4740 s  

so

k = - {ln[A]/[A]0} / t

k = -ln 0.15 / 4740

k = 4.00 x 10^-4 s^{-1}

so rate constant  is 4.00 x 10^-4 s^{-1}

3 0
2 years ago
Based on the graph, how would you describe the net forces acting on the moving
ladessa [460]

Given the velocity-time graph of an object.

The slope of a velocity-time graph gives the acceleration acting on the object.

From the graph, we can see that the slope of the graph is zero. That is, the velocity of the object is constant and hence the net acceleration acting on the object is zero.

From Newton's second law, the net force acting on an object is given by the product of the mass of the object and its velocity. Therefore when the acceleration of the object is zero, the net force on the object is also zero.

Therefore the net force acting on the given object is zero. Hence, the correct answer is option A.

3 0
1 year ago
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