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lakkis [162]
3 years ago
5

Someone help with this also

Physics
1 answer:
Mumz [18]3 years ago
3 0
I’m pretty sure it’s average speed= total distance and total time which is A.
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When a charged particle such as an electron, proton or ion is in motion, magnetic lines of force rotate around the particle. Since electrical current moving through a wire consists of electrons in motion, there is a magnetic field around the wire.      i hope u have smileing day

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Similar to Hippocrates, modern scientists who study etiology believe that
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In  Medicine, "etiology" means the origin or cause of a disease.

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3 years ago
If jack was traveling at 100 miles in 2 hours what was his speed in miles per minute?
Ludmilka [50]

Speed = (distance) / (time)

Speed = (100 mi/2 hr)

Multiply by a fraction equal to 1 :

Speed = (100 mi/2 hr) x (1 hr/60 min)

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4 0
3 years ago
What is the maximum distance that a 60.-watt motor may vertically lift a 90.-newton?weight in 7.5 seconds?a. 2.3mb. 5.0mc. 140md
Naddik [55]

Answer:

b. 5.0 m

Explanation:

Power: Power can be defined as the rate at which work is done.

The S.I unit of power is watt (W).

P = W/t ............................................... Equation 1

Where p = power, W = work, t = time.

But

W = F×d..................................... Equation 2

Where F = force, d = maximum distance.

Substituting equation 2 into equation 1

P = F×d/t

making d the subject of the equation above,

d = Pt/F......................................................... Equation 3

<em>Given: F = 90 Newton, P = 60 watt, t = 7.5 seconds.</em>

<em>Substituting these values into equation 3,</em>

<em>d = 60×7.5/90</em>

<em>d = 5 m</em>

Thus the maximum distance is = 5 m

The right option is b. 5.0 m

5 0
3 years ago
A town requiring 2.0 m3/s of drinking water has two sources, a local well with 15 g/m3 nitrate (as N) and a distant reservoir wi
kati45 [8]

Explanation:

Drinking water requirement in town 2.0 m^3/s of water per second

nitrate in local well  nitrate per 15 \mathrm{m}^{3} of water

nitrate in distant reservoir =5 \mathrm{~g} / \mathrm{m}^{3}

Let the flow rate of well

flow rate of reservoir =y m^{3} / s

Drinking water requirement is 45 \mathrm{ppm} or 45 \mathrm{~g} / \mathrm{m}^{3}

therefore, the total flow of drinking water

 

8 0
2 years ago
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