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lesantik [10]
3 years ago
8

When you turn on the light switch in a room, you have completed the _____. open circuit series circuit parallel circuit switch c

ircuit
Physics
2 answers:
Snowcat [4.5K]3 years ago
7 0
Open circuit...by flipping on the switch you close the circuit and allow electrons to flow to the light bulb
const2013 [10]3 years ago
3 0
A.) <span>When you turn on the light switch in a room, you have completed the "Open Circuit"

Hope this helps!</span>
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HELP PLS i need this right now
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V = 44.4 units.

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V = \sqrt{(V_{x})^{2} +(V_{y})^{2} }

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V=\sqrt{(-43)^{2} +(11.1)^{2} }\\V=44.4 units

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A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.
Mama L [17]

Answer: f_{r} = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

W_{x} = horizontal component of the weight = mgsinФ

W_{y} = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - f_{r} = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

g = acceleration due to gravity = 9.8m/s^{2}

f_{r} = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

v^{2} = u^{2} + 2aS

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u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}

we slot in a into the equation below to get frictional force

mgsinФ - f_{r} = ma

3 * 9.8 * sin 37 - f_{r} = 3* 0.4

17.9633 - f_{r} =  1.2

f_{r} = 17.9633 - 1.2

f_{r} = 16.49N

4 0
3 years ago
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