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Novay_Z [31]
4 years ago
11

A proton with a velocity in the positive x-direction enters a region where there is a uniform magnetic field B in the positive y

-direction. You want to balance the magnetic force with an electric force so that the proton will continue along a straight line. The electric field should be in the...which direction?
Physics
1 answer:
Daniel [21]4 years ago
4 0

Answer:

Negative z-direction

Explanation:

The direction of the force exerted by the magnetic field on a moving (positive) charged particle can be found by using the right-hand rule:

- The index finger corresponds to the direction of the velocity of the particle

- The middle finger corresponds to the direction of the magnetic field

- The thumb gives the direction of the force

For the proton in this problem:

- Index finger: positive x-direction (direction of the velocity)

- Middle finger: positive y-direction (direction of magnetic field)

- Thumb: positive z-direction (direction of the force)

So, the magnetic force on the proton acts in the positive z-direction.

The electric force on a positive charged particle, instead, has the same direction as the electric field.

In this problem, in order to balance the magnetic force, we have to apply an electric force in the negative z-direction: therefore, the electric field must be in the negative z-direction as well, so that the proton will continue along a straight line.

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3 years ago
9. During an egg toss, a catcher must cushion the egg by maximizing the time it takes to stop the
Lorico [155]

Answer:

the impulse experienced by the egg is 0.053 kgm/s.

Explanation:

Given;

mass of the egg, m = 60 g = 0.06 kg

initial velocity of the egg, u = 6 m/s

height moved by the egg, h = 50 cm = 0.5 m

Determine the final velocity of the egg as it moves upward;

v² = u² + 2(-g)h

v² = u² - 2gh

where;

v is the final velocity

-g is negative acceleration due gravity as it moves upward

v² = 6² - 2(9.8 x 0.5)

v² = 26.2

v = √26.2

v = 5.12 m/s

The impulse applied to the egg is the change in linear momentum;

J = ΔP

ΔP = mu - mv

ΔP = m(u - v)

ΔP = 0.06(6 - 5.12)

ΔP = 0.053 kgm/s

Therefore, the impulse experienced by the egg is 0.053 kgm/s.

8 0
3 years ago
Vector A has magnitude 8.00 mm and is in the xy-plane at an angle of 127∘ counterclockwise from the +x–axis (37∘ past the +y-axi
Ilia_Sergeevich [38]

Answer:

-75.35°

Explanation:

Let C be the sum of the two vectors A and B. Hence, we can write the following  

A_{x} +B_{x} =C_{x} ......(1)\\A_{y} +B_{y} =C_{y} ......(2)

but since the vector C is in the -y direction, C_{x}  = 0 and C_{y} = —12 m.  

Thus  

B_{x} =-A_{x} =-[-Acos(180-127)]=(8)*cos(53)\\B_{x} =4.81m

similarly, we can determine B_{y} by rearranging equation (1)  

 B_{y} =C_{y} -A_{y} =-12m-[(8)*sin(53)\\B_{y} =-18.4m

so the magnitude of B is

B=\sqrt{B_{x}^2+B_{y}^2  } \\B=19m

Finally, the direction of B can be calculated as follows  

Ф=tan^{-1} (\frac{B_{y} }{B_{x} } )\\=-75.35

hence the vector B makes an angle of 75.35 clockwise with + x axis

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