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matrenka [14]
3 years ago
15

An electron moves through a region with a uniform magnetic field of unknown magnitude directed in the +y direction. At the insta

nt that the electron is moving in the +z direction with a velocity of 2.83e+6 m/s​​, it experiences a magnetic force of 2.87 pN in the +x direction. The magnetic field is 6.34 T.If the electron were replaced with a proton moving with the same velocity, how would the resulting acceleration of the particle change?a) The resulting acceleration of the proton would be weaker in magnitude and in the same direction as the acceleration of the electron.b) The resulting acceleration of the proton would be weaker in magnitude and oppositely directed compared to the acceleration of the electron.c)The resulting acceleration of the proton would be greater in magnitude and in the same direction as the acceleration of the electron.d)The resulting acceleration of the proton would be equal in magnitude and oppositely directed compared to the acceleration of the electron.e) The resulting acceleration of the proton would be greater in magnitude and oppositely directed compared to the acceleration of the electron.f)The resulting acceleration of the proton would be equal in magnitude and in the same direction as the acceleration of the electron.
Physics
1 answer:
suter [353]3 years ago
6 0

Answer:

Since proton is more massive than electron, thus its acceleration will be lesser in magnitude since acceleration is inversely proportional to acceleration.

However, its charge is opposite to that of electron. Thus it will experience the acceleration in the opposite direction.

Thus, the correct option is:

b) The resulting acceleration of the proton would be weaker in magnitude and oppositely directed compared to the acceleration of the electron.

Explanation:

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Answer:

The answer is 80 kN . m (clockwise)

Explanation:

As,

M = P x L

Here, the towline exerts a force is P.

Substituting P for 4000N.

M = -4000N x 20m

   = -80000N.m

   = 80kN.m

Maximum moment about the point O is 80kN.m (Clockwise)

7 0
3 years ago
Find the work done in pumping gasoline that weighs 6600 newtons per cubic meter. A cylindrical gasoline tank 3 meters in diamete
almond37 [142]

Answer:

<em>work done in pumping the entire fuel is 1399761 J</em>

Explanation:

weight per volume of the gasoline = 6600 N/m^3

diameter of the tank = 3 m

length of the tank = 6 m

The height of the tractor tank above the top of the tank = 5 m

The total volume of the fuel is gotten below

we know that the tank is cylindrical.

<em>we assume that the fuel completely fills the tank.</em>

therefore, the volume of a cylinder =  

where r = radius = diameter ÷ 2 = 3/2 = 1.5 m

volume of the cylinder = 3.142 x  x 6 = 42.417 m^3

we then proceed to find the total weight of the fuel in Newton

total weight = (weight per volume) x volume

total weight = 6600 x 42.417 = 279952.2 N

therefore,

the work done to pump the fuel through to the 5 m height = (total weight of the fuel) x (height through which the fuel is pumped)

work done in pumping = 279952.2 x 5 = <em>1399761 J</em>

5 0
3 years ago
The students on the right are applying a force of 100 N to the right. This is shown by the red arrow. The students on the left a
Artyom0805 [142]

Answer:

A

Explanation:

suddenly part of the force pulling to the left is gone.

and now the force pulling to the right is stronger.

so, the rope will move to the right. and that is also the direction of the net force.

8 0
2 years ago
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A student stretches out a rubber band. What primary form of energy does the stretched-out rubber band possess? A. gravitational
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Answer : )b

Explanation:

bc i said its b sznankana

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3 years ago
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Fiber optics are an important part of our modern internet. In these fibers, two different glasses are used to confine the light
professor190 [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

\theta_{max} =18.38^o

b

New  n_{cladding} =1.491

Explanation:

 From the question we are told that

          The refractive index of the core is  n_{core} = 1.497

         The refractive index of the cladding  is   n_{cladding} = 1.421

Generally according to Snell's law

      n_{core} * sin(90- \theta) = n_{cladding} * sin (90)

Where \theta_{max} is the largest angle a largest angle a ray will make with respect to the interface of the fiber and experience total internal reflection

      \theta_{max} = 90 - sin^{-1} [\frac{n_{cladding}}{n_{core}} ]

       \theta_{max} = 90 - sin^{-1} [\frac{1.421}{1.497}} ]

      \theta_{max} =18.38^o

Given from the question the the largest angle is  5°

Generally the refraction index of the cladding is mathematically represented as

           n_{cladding} = n_{core} * sin (90 - 5)

          n_{cladding} =1.491

       

5 0
3 years ago
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