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shtirl [24]
3 years ago
15

Why you cannot use an elastic measuring tape to measure distance. What problem you may face if you use it

Physics
1 answer:
Stells [14]3 years ago
3 0

The problem you would encounter is measuring the height of two different people, a tall one and a short one, and getting the same answer for both of them.

No matter WHAT we're hearing out of the White House these days, you CAN'T bend and stretch your standard measuring devices, or any other 'facts', to make them fit the thing that you're measuring.  This does not work.  You're always entitled to your own opinions, but you're not entitled to your own facts.

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How is the voltage V across the resistor related to the current I and the resistance R of the resistor? (Use I for current and R
VladimirAG [237]

Answer:

This relationship is explained by Ohm's law

Explanation:

Ohm's law states that the current flowing through a circuit or a resistor is directly proportional to the voltage across the resistor and inversely proportional to the resistance. Where current is i, voltage is v and resistance is r, Ohm's law can be represented mathematically as

V= IR

8 0
4 years ago
People who drink coffee are more likely to develop cancer
krok68 [10]

Answer:

The answer is False

8 0
3 years ago
Read 2 more answers
Why does a black hole have a stronger gravitational pull than the star that collapse to form it?​
Studentka2010 [4]

Answer:

We consider Black Holes as an object that possesses extreme gravitational pull, but wait aren’t they have the same mass(or less) as that of their parent star. And we know that gravitational pull ‘F’ is directly proportional to the mass of an object, so if the mass is same(or less) then why do black holes have stronger gravity than the stars they evolved from.

The above consideration that F is directly proportional to the mass is partially correct, one should also mention that F is also inversely proportional to the square of the distance between the considered objects.

F = G*(M*m)/(r^2)

Where:

· F is the force acting on you due to star

· M is the mass of Parent star / Black Hole

· m is the mass of an observer, here it is you

· r is the radial distance between the star and you

We know that black hole formed, has much smaller size than that of its parent star and all that mass is compressed to a much smaller scale. If you consider a Star as having a size of an earth then the black hole formed will have a size of small city.

Let us say that you are standing at an r distance away from a star (r>R1), where R1 is the radius of the star, of course (R1>R2), where R2 is the radius of Black Hole.

The Force by which the star in case 1 attracts you will be equal(or less) to the force by which black hole in case 2. So, there is nothing increase in gravitational pull, it is same(or less) as that of the parent star.

Wait a minute, then why people say that black holes have massive gravitational pull.

The gravitational pull increases as we move closer to the black hole, and when we are at its surface, it is enormous as compare to its star surface, because of the difference in the size.

We know that gravitational pull not only depends upon the mass but also depends upon the radial distance between the concerned objects here, it is you and the black hole.

Here, the size of the black hole is much smaller than that of its parent star, i.e (R1>>>R2), and thus we get F1<<<F2, and that is why we say that the black hole has enormous gravitational pull, such that nothing can escape, not even light.

8 0
3 years ago
What is the formula for mechanical advantage?
Sladkaya [172]

Answer:

A. MA=force output/force input

Explanation:

mechanical advantage is the ratio of the load to the effort

8 0
3 years ago
Read 2 more answers
Two parallel plates of area 2.34*10-3 M2 have 7.07*10-7C of charge placed on them. A6.62*10-5C charge q1 is placed between the p
ira [324]

Answer:

The force exerted on the q_1 is  F =  2.25*10^{3} \ N

Explanation:

From the question we are told that

     The area is  A =  2.34*10^{-3} \ m^2

     The magnitude of charge placed on them is  q =  7.07 * 10^{-7} C

     The charge placed between the plate is q_1 = 6.62 *10^{-5} C

   

The electric field generated around the plate  is mathematically represented as

           E =  \frac{q}{A \epsilon_o}

Substituting values

          E =  \frac{7.07*10^{-7}}{2.34*10^{-3} * 8.85 *10^{-12}}

         E = 34*10^{6} \ V/m

The force exerted the charge q_1 is  mathematically represented as

        F =  q_1 * E

Substituting values  

        F =  6.62 *10^{-5} * 34*10^{6}

        F =  2.25*10^{3} \ N

8 0
3 years ago
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