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MrRa [10]
1 year ago
14

an electron, a proton and a deuteron move in a magnetic field with same momentum perpendicularly. the ratio of the radii of thei

r circular paths will be -
Physics
1 answer:
wel1 year ago
6 0

If an electron, a proton, and a deuteron move in a magnetic field with the same momentum perpendicularly, the ratio of the radii of their circular paths will be:

  • 1: √2 : 1

<h3>How is the ratio of the perpendicular parts obtained?</h3>

To obtain the ratio of the perpendicular parts, one begins bdy noting that the mass of the proton = 1m, the mass of deuteron = 2m, and the mass of the alpha particle  = 4m.

The ratio of the radii of the parts can be obtained by finding the root of the masses and dividing this by the charge. When the coefficients are substituted into the formula, we will have:

r = √m/e : √2m/e : √4m/2e

When resolved, the resulting ratios will be:

1: √2 : 1

Learn more about the radii of their circular paths here:

brainly.com/question/16816166

#SPJ4

​

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A diode for which the forward voltage drop is 0.7 V at 1.0 mA is operated at 0.5 V. What is the value of the current
Vesnalui [34]

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the current value is 0.335 \mu A

Explanation:

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z_i = I_s e^{\frac{0.7}{ut} }= 10^{-3}\\\\Z_z = I_s e^{\frac{0.5}{ut} }\\\\\frac{Z_z}{Z_i}= \frac{Z_z}{10^{-3}}  = e^{\frac{0.5\times 0.7}{0.025} }\\\\= 0.335 \mu A

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a vector has an x-component of 19.5m and a y-component of 28.4m. Find the magnitude and direction of the vector
Delicious77 [7]

Answer:

magnitude=34.45 m

direction=55.52\°

Explanation:

Assuming the initial point P1 of this vector is at the origin:

P1=(X1,Y1)=(0,0)

And knowing the other point is P2=(X2,Y2)=(19.5,28.4)

We can find the magnitude and direction of this vector, taking into account a vector has a initial and a final point, with an x-component and a y-component.

For the magnitude we will use the formula to calculate the distance d between two points:

d=\sqrt{{(Y2-Y1)}^{2} +{(X2-X1)}^{2}} (1)

d=\sqrt{{(28.4 m - 0 m)}^{2} +{(19.5 m - 0m)}^{2}} (2)

d=\sqrt{1186.81 m^{2}} (3)

d=34.45 m (4) This is the magnitude of the vector

For the direction, which is the measure of the angle the vector makes with a horizontal line, we will use the following formula:

tan \theta=\frac{Y2-Y1}{X2-X1}  (5)

tan \theta=\frac{24.8 m - 0m}{19.5 m - 0m}  (6)

tan \theta=\frac{24.8}{19.5}  (7)

Finding \theta:

\theta= tan^{-1}(\frac{24.8}{19.5})  (8)

\theta= 55.52\°  (9) This is the direction of the vector

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3 years ago
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