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finlep [7]
2 years ago
9

Right hand rule exercise

Physics
1 answer:
saveliy_v [14]2 years ago
6 0

Answer:

A yxiigxih5dd8yixc99uf8g

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The main difference between obsessive-compulsive personality disorder (OCPD) and obsessive-compulsive disorder (OCD) is that ___
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B. people with OCD know their disorder is irrational

Explanation:

Got it right

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The tip of the second hand of a clock moves in a circle of 20 cm circumference. In one minute the hand makes a complete revoluti
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Answer:

v_{avg} = 0

Explanation:

As we know that average velocity is defined as the ratio of total displacement of the object and its time interval.

so here we can say

v_{avg} = \frac{displacement}{time}

now we know that in one complete revolution the total displacement of the tip of the seconds hand is zero

because it will have same position after one complete revolution from where it starts

so here we can say that the average velocity will be zero

v_{avg} = 0

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3 years ago
Which location on the map above is a source of North Atlantic deep water?
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Why do we use microwaves to communicate berween earth and satellites
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Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist
Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}}), where imax=\frac{E}{R}

Given that, at i(2)=\frac{imax}{2} =\frac{E}{2R}

⇒\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})

⇒\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}

⇒\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}

Applying logarithm on both sides,

⇒log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}

⇒log(2)=\frac{2}{\frac{L}{R}}

⇒\frac{L}{R}=\frac{2}{log2}

Now substitute i(t)=\frac{3}{4}imax=\frac{3E}{4R}

⇒\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})

⇒\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}

⇒\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}

Applying logarithm on both sides,

⇒log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}

⇒log(4)=\frac{t}{\frac{L}{R}}

⇒t=log4\frac{L}{R}

now subs. \frac{L}{R}=\frac{2}{log2}

⇒t=log4\frac{2}{log2}

also log4=log2^{2}=2log2

⇒t=2log2\frac{2}{log2}

⇒t=4

5 0
3 years ago
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