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Andreyy89
2 years ago
10

(a) A velocity selector consists of electric and magnetic fields described by the expressions E = E and B = B ĵ, with B = 10.0 m

T. Find the value of E (in kV/m) such that a 740 eV electron moving in the negative x-direction is undeflected. kV/m (b) What If? For the value of E found in part (a), what would the kinetic energy of a proton have to be (in MeV) for it to move undeflected in the negative x-direction?
Physics
1 answer:
Allushta [10]2 years ago
8 0

Answer:

(a) 160000 kV/m

(b) 1336 keV

Explanation:

(a) magnetic filed, B = 10 T

energy of electron, E = 740 eV

mass of electron, m = 9.1 x 10^-31 kg

Let v be the velocity of electron.

E = 1/2 mv^2

740 x 1.6 x 10^-19 = 0.5 x 9.1 x 10^-31 x v^2

v = 1.6 x 10^7 m/s

v = E / B

E = v x B = 1.6 x 10^7 x 10 = 16 x 10^7 V/m

E = 160000 kV/m

(b) E = 16 x 10^7 V/m

B = 10 T

Let v be the velocity of protons.

v = E / B = 16 x 10^7 / 10 = 1.6 x 10^7 m/s

Kinetic energy of proton, E = 1/2 mv^2

                                          = 0.5 x 1.67 x 10^-27 x 1.6 x 1.6 x 10^14

                                          = 2.14 x 10^-13 J = 1336000 eV = 1336 keV

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Part B - see attachment

Part C - 4.9 × 10-³J

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Explanation:

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Part D

Three forces act on the spring-monkey system and they do work in different forms: kinetic energy 1/2mv² , elastic potential

energy due to the restoring force in the spring or the tension force 1/2kd², and the gravitational potential energy mgh of the position of the system. So the total energy of the system E = 1/2kd² + 1/2mv² + mgh.

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A cylindrical rod of steel (E = 207 GPa, 30 × 10 6 psi) having a yield strength of 310 MPa (45,000 psi) is to be subjected to a
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Answer:

Diameter of the cylinder will be d=2.998\times 10^4m

Explanation:

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Load F = 11100 KN

Strain is given by strain=\frac{\Delta l}{l}=\frac{0.38}{500}=7.6\times 10^{-4}

We know that young's modulus E=\frac{stress}{strain}

So 207\times 10^9=\frac{stress}{7.6\times 10^{-4}}

stress=1573.2\times 10^{-5}N/m^2

We know that stress =\frac{force}{artea }

So 1573.2\times 10^{-5}=\frac{11100\times 1000}{area}

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6 0
2 years ago
A meter stick balances at the 50.0-cm mark. If a mass of 50.0 g is placed at the 90.0-cm mark, the stick balances at the 61.3-cm
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Answer:

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