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valentinak56 [21]
3 years ago
11

Step 1, when solving a two dimensional, multi-charge problem, is to define the vectors. Please identify the next five steps, in

order.
Physics
2 answers:
Lana71 [14]3 years ago
5 0

Answer:

Step 1, when solving a two dimensional, multi-charge problem, is to define the vectors. Please identify the next five steps, in order.

Step 2: calculate a & b mag

Step 3: calculate x,y components

Step 4: sum vector components

Step 5: calculate magnitude of R

Step 6: calculate direction of R

Maksim231197 [3]3 years ago
4 0

Answer: The next 5 steps are.

step 2: Find the magnitude of vectors.

step 3: Calculate 'x' and 'y' components.

step 4: Sum the above obtained vector components.

step 5: Calculate the magnitude of Resultant Vector.

Step 6: Determine the direction of the resultant vector.

Explanation:

Let the two vectors be 'A' and 'B'.

Let A = x_{1} \hat{i}+y_{1}\hat{j}+z_{1}\hat{k} and B = x_{2} \hat{i}+y_{2}\hat{j}+z_{2}\hat{k}

It is given that, when solving a two dimensional, multi-charge problem, is to define the vectors.

So, step 1 is to define the vectors A and B.

In step 2, we find the magnitude of the vectors by,

|A|=\sqrt{x_1^2+y_1^2+z_1^2}

|B|=\sqrt{x_2^2+y_2^2+z_2^2}

In step 3, we find the components 'x' and 'y'

Then in step 4, we sum the components. Now we get the Resultant vector (R).

In step 5, Calculate the magnitude of 'R' using the above formula to find magnitude.

In step 6, we determine the direction of 'R'.

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The momentum of an electron is 1.75 times larger than the value computed non-relativistically. What is the speed of the electron
FrozenT [24]

Answer:

<em>Speed of the electron is 2.46 x 10^8 m/s</em>

<em></em>

Explanation:

momentum of the electron before relativistic effect = M_{0} V

where M_{0} is the rest mass of the electron

V is the velocity of the electron.

under relativistic effect, the mass increases.

under relativistic effect, the new mass M will be

M = M_{0}/ \sqrt{1 - \beta ^{2}  }

where

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The new momentum will therefore be

==> M_{0}V/ \sqrt{1 - \beta ^{2}  }

It is stated that the relativistic momentum is 1.75 times the non-relativistic momentum. Equating, we have

1.75M_{0} V = M_{0}V/ \sqrt{1 - \beta ^{2}  }

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1.75 = 1/ \sqrt{1 - \beta ^{2}  }

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3.0625 = 1/(1 - \beta ^{2} )

3.0625 - 3.0625\beta ^{2} = 1

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V/c = 0.819

V = c x 0.819

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