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V125BC [204]
2 years ago
10

What is the potential difference across a 15Ω resistor that has a current of 3.0 A?

Physics
2 answers:
ololo11 [35]2 years ago
7 0

Answer:

45 V

Explanation:

sesenic [268]2 years ago
6 0

V = I • R

V = (15 ohms) x (3 A)

V = 45 volts

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What is the area of the bottom of a tank 30.0 cm long and 15.0 cm wide?​
Igoryamba

Answer:

perimeter of the bottom of the tank is 450 cm

Explanation:

If you want to get area you need the Length, width,and height.

5 0
2 years ago
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A person travels by car from one city to another with different constant speeds between pair of cities. She drives for 36 min at
Softa [21]

 Change minutes to hrs, divide by 60:
30 min = .50 hrs
45 min = .75 hrs
12 min = .20 hrs
----------------
total + 1.45 hrs, total travel time
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let a = average speed for the trip
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Write a dist equation, dist = speed * time
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80(.5) + 100(.20) + 40(.75) = 1.45a
40 + 20 + 30 = 1.45a
90 = 1.45a
a =
a = 62.069 km/h, for the entire trip
and
90 km is the total distance 

3 0
2 years ago
Convert 2 days into second​
eduard

Answer:

<em><u>172,000 second </u></em>

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3 0
2 years ago
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What determines how soon one action potential can follow another?
belka [17]
It is through biopsychological feedback. 

A class of chemical called a neurotransmitter is important in the transmission of nerve impulses. Neurotransmitters are packaged by the cell into small, membrane-bound sacs called vesicles. Upon receiving a chemical signal, the vesicles move toward the cell membrane and fuse with it, releasing the enclosed neurotransmitters from the terminal end of the nerve cell. 
5 0
3 years ago
Two cars cover the same distance in a straight line. Car a covers the distance at a constant velocity. Car b starts from rest an
Artyom0805 [142]

a) For the motion of car with uniform velocity we have , s = ut+\frac{1}{2}at^2, where s is the displacement, u is the initial velocity, t is the time taken a is the acceleration.

In this case s = 520 m, t = 223 seconds, a =0 m/s^2

Substituting

       520 = u*223\\ \\u = 2.33 m/s

 The constant velocity of car a = 2.33 m/s

b) We have s = ut+\frac{1}{2} at^2

s = 520 m, t = 223 seconds, u =0 m/s

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Now we have v = u+at, where v is the final velocity

Substituting

        v = 0+0.0209*223 = 4.66 m/s

So final velocity of car b = 4.66 m/s

c) Acceleration = 0.0209 m/s^2

7 0
3 years ago
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