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masya89 [10]
3 years ago
14

How to get infinite thanks could you show me please ☺️​

Physics
1 answer:
murzikaleks [220]3 years ago
7 0

Answer: divide by zero, or square root of a negative

Explanation: If your question is how to get infinity as an answer to a problem, that generally means that the answer is undefined or doesn't exist.

A couple of ways to get that...

  1. You try to divide by zero. In other words, a problem that asks you to perform something like this:  5/0=   or 23/(4-4)=   Such a problem will give you an error on a calculator because the answer is infinity or doesn't exist.
  2. Another way is to try to get the square root of a negative number. That answer doesn't exist as a real number, so \sqrt{-4\\} will also give you an error on a calculator.

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Because they are both easy to measure (?)

Explanation:

(I'm not really sure, there are no choices. If there were different options I might be able to better answer this)

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Chanel has some cotton candy that came in a cloudy shape. She wants to make it more dense. Which describes the candy before and
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Internal diameter of a test tube<br>​
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3 years ago
An unknown charged particle passes without deflection throughcrossed electric and magnetic fields of strengths 187,500 V/m and0.
UNO [17]

Explanation:

The given data is as follows.

        Electric field strength (E) = 187,500 V/m

    Magnetic field strength (B) = 0.125 T

       Diameter (d) = 25.05 cm = 0.2505 m    (as 1 m = 100 cm)

    Radius (r) = \frac{d}{2}

                    = \frac{0.2505}{2}

                    = 0.12525 m

Formula to calculate the magnetic force (F_{M}) is as follows.

              F_{M} = Bqv ............ (1)

Electrical force is calculated as follows.

             F_{E} = qE ............ (2)

On both electric and magnetic fields the velocity is perpendicular.

       F_{M} - F_{E} = 0

or,             F_{M} = F_{E}

Hence, from equations (1) and (2)

              Bqv = qE

or,            v = \frac{E}{B} ............. (3)

                  = \frac{187500 V/m}{0.125 T}

                  = 1,500,000 m/s

As the particle is moving in a semi-circular trajectory and motion of charged particle is given by the electric field as follows.

              F_{c} = \frac{mv^{2}}{r} ........... (4)

where,    F_{c} = centripetal force

             F_{M} = F_{c}

Using equation (1) and (4) as follows.

            F_{M} = F_{c}

              Bqv = \frac{mv^{2}}{r}

                   \frac{q}{m} = \frac{v}{Br}

                       = \frac{15 \times 10^{5}}{0.125 \times 0.12525}

                       = 958.08 \times 10^{5} C/kg

Thus, we can conclude that charge-to-mass ratio of the given particle is 958.08 \times 10^{5} C/kg.

8 0
4 years ago
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