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Sauron [17]
3 years ago
5

If the magnitude of the magnetic force on a proton is F when it is moving at 18.0 ∘ with respect to the field, what is the magni

tude of the force (in terms of F) when this charge is moving at 33.0 ∘ with respect to the field?
Physics
1 answer:
Lelu [443]3 years ago
5 0

Answer:

The force when θ = 33° is 1.7625 times of the force when θ = 18°

Explanation:

The force on a moving charge through a magnetic field is given by

F = qvB sin θ

q = charge of the moving particle

v = Velocity of the moving charge

B = Magnetic field strength

θ = angle between the magnetic field and the velocity (direction of the motion) of the moving charge

Because qvB are all constant, we can call the expression K.

F = K sinθ

when θ = 18°,

F = K sin 18° = 0.309K

when θ = 33°, let the force be F₁

F₁ = K sin 33° = 0.5446K

(F₁/F) = (0.5446K/0.309K) = 1.7625

F₁ = 1.7625 F

Hope this Helps!!!

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A small sphere is at rest at the top of a frictionless semicylindrical surface. The sphere is given a slight nudge to the right
V125BC [204]

Answer:

vi = 4.77 ft/s

Explanation:

Given:

- The radius of the surface R = 1.45 ft

- The Angle at which the the sphere leaves

- Initial velocity vi

- Final velocity vf

Find:

Determine the sphere's initial speed.

Solution:

- Newton's second law of motion in centripetal direction is given as:

                         m*g*cos(θ) - N = m*v^2 / R

Where, m: mass of sphere

             g: Gravitational Acceleration

             θ: Angle with the vertical

             N: Normal contact force.

- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:

                         m*g*cos(θ) - 0 = m*vf^2 / R

                         g*cos(θ) = vf^2 / R    

                         vf^2 = R*g*cos(θ)

                         vf^2 = 1.45*32.2*cos(34)

                        vf^2 = 38.708 ft/s

- Using conservation of energy for initial release point and point where sphere leaves cylinder:

                          ΔK.E = ΔP.E

                          0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

                          ( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

                          vi^2 =  vf^2 - 2*g*R*( 1 - cos(θ))

                          vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

                          vi^2 = 22.744

                           vi = 4.77 ft/s

4 0
3 years ago
There's an electric field in some region of space that doesn't change with position. An electron starts moving with a speed of 2
tangare [24]

Answer:

Explanation:

Given

speed of Electron u=2\times 10^7\ m/s

final speed of Electron v=4\times 10^7\ m/s

distance traveled d=1.2\ cm

using equation of motion

v^2-u^2=2as

where v=Final velocity

u=initial velocity

a=acceleration

s=displacement

(4\times 10^7)^2-(2\times 10^7)^2=2\times a\times 1.2\times 10^{-2}

a=5\times 10^{16}\ m/s^2

acceleration is given by a=\frac{qE}{m}

where q=charge of electron

m=mass of electron

E=electric Field strength

5\times 10^{16}=\frac{1.6\times 10^{-19}\cdot E}{9.1\times 10^{-31}}

E=248.3\ kN/C                

5 0
3 years ago
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zlopas [31]
The ability to reproduce is essential for a species, but not for an individual. There are some people that can't reproduce, but they're still alive!
6 0
3 years ago
A two slit pattern is viewed on a screen 1.00m from the slits if the two third-order minima are 22.0 cm apart what is the width
Bingel [31]

Answer:

4.4 cm

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Given:

Distance of the screen from the slit, D = 1 m

Distance between two third order interference minimas, x = 22 cm

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x_n = (n + \frac{1}{2}) \frac{wL}{d}

We have:

2x_2 = 2(2 + \frac{1}{2}) * \frac{w*22}{d}

Calculating further for the width of the central bright fringe, we have:

\frac{w}{d} = \frac{22}{5}

= 4.4 cm

Note: w in representswavelength

8 0
3 years ago
what is the density to the nearest hundredths, of a metal with a volume of 3.00 cm3 and a mass of 8.13g?
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