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Andre45 [30]
2 years ago
12

Consider the specific example of a positive charge qqq moving in the +x direction with the local magnetic field in the +y direct

ion. In which direction is the magnetic force acting on the particle? Express your answer using unit vectors (e.g., i^i^i_unit- j^j^j_unit). (Recall that i^i^i_unit is written \hat i (or alternatively i_unit can be used.))
Physics
2 answers:
Tems11 [23]2 years ago
8 0

Answer:

The direction of F(B) is in the positive z direction.

Explanation:

The magnetic force acting on a charge particle into a constant magnetic field is given by this equation:

\vec{F}_{B}=q(\vec{v}\times \vec{B})

The particle is moving in the positive x direction, it means that the unit <u>vector of the velocity will \hat{i}</u>.

By the other hand, the direction of the magnetic field is in the positive direction of the y axis, <u>so the unit vector of B will be \hat{j}.</u>

As we see, the magnetic force  is given by the cross product between the velocity and the magnetic field, therefore the direction will be the result of the cross product between \hat{i} and \hat{j}. Using the right hand rule we will have:

\hat{i} \times \hat{j}=\hat{k}

It means that the direction of <u>F(B) is in the positive z direction.</u>

<u />

I hope it helps you!

Hatshy [7]2 years ago
5 0

Answer: The direction of magnetic Force (B) is in the positive z direction.

Explanation:

The magnetic force that is acting on a charge particle into a constant magnetic field is given by this equation:

F_B = q ( v × B)

The particle is moving in the positive x direction, it means that the unit vector of the velocity will be i^.

While on the other hand, the direction of the magnetic field is in the positive direction of the y axis, so the unit vector will be j^ .

As we see, the magnetic force  is given by the cross product between the velocity and the magnetic field, therefore the direction will be the product between i^ and j^. Using the right hand rule (right hand rule states that, to find the direction of the magnetic force on a positive moving charge, the thumb of the right hand point in the direction of v, the fingers in the direction of B, and the force (F) is directed perpendicular to the right hand palm) we will have:

i^ × j^ = k

meaning that the direction of the F(B) is in the positive z direction.

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Describe several uses of plastic, and explain why plastic is a good choice for these products
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Plastic is used in many things like bags, toys, nd certain products. Plastic is very durable nd flexible :) hope i helped a bit c;
5 0
2 years ago
: Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at
ser-zykov [4K]

Complete Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.

Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.

Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

Answer:

a

The  expression is   F_{No} =   A [P_{No} - \frac{P_{sea}}{2}]

b

F_{No}= 7771.125 \ N

c

 F_p = 2.2*10^{6} N

d

From the value obtained we can say the that the force required to open the lid is higher at Denver

Explanation:

          The altitude of container in Denver is  d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m

           The surface area of the container lid is A = 0.0155m^2

           The altitude of container in New Orleans  is sea-level

           The air pressure in Denver is  P_D = 79000 \ Pa

            The air pressure in new Orleans is P_{ro} = 100250 \ Pa

Generally force is mathematically represented as

            F_{No} = \Delta P A

  So we are told the pressure inside is  is half the pressure the at sea level so the  the pressure acting on the container would

   The  pressure at sea level is a constant with a  value of  

               P_{sea} = 101000 Pa

So the \Delta P which is the difference in pressure within and outside the container is  

           \Delta P = P_{No} - \frac{P_{sea}}{2}

Therefore

                F_{No} =   A [P_{No} - \frac{P_{sea}}{2}]

Now substituting values

                F_{No} =   0.0155 [100250 - \frac{101000}{2}]

                       F_{No}= 7771.125 \ N

The force to remove the lid in Denver is  

           F_p = \Delta P_d A

So we are told the pressure inside is  is half the pressure the at sea level so the  the pressure acting on the container would

 The  pressure at sea level is a constant with a  value of  

               P_{sea} = 101000 Pa    

 At  sea level the air pressure in Denver is mathematically represented as

              P_D = \rho g h

     =>     g = \frac{P_D}{\rho h}      

Let height at sea level is h = 1

  The air pressure at height d_D

             P_d__{D}} = \rho gd_D

    =>     g = \frac{P_d_D}{\rho d_D}

  Equating the both

                 \frac{P_D}{\rho h}  = \frac{P_d_D}{\rho d_D}

                 P_d_D =  P_D * d_D

Substituting value  

                   P_d__{D}} = 1828.2 * 79000

                    P_d__{D}} = 1.445*10^{8} Pa

    So

              \Delta P_d  = P_{d} _D - \frac{P_{sea}}{2}

=>          \Delta P_d  = 1.445 *10^{8} - \frac{101000}{2}    

                        \Delta P_d = 1.44*10^{8}Pa

  So

               F_p = \Delta P_d A

                  = 1.44*10^8 * 0.0155

              F_p = 2.2*10^{6} N

               

                 

             

             

6 0
3 years ago
If the rocket has an initial mass of 6300 kg and ejects gas at a relative velocity of magnitude 2000 m/s , how much gas must it
Rzqust [24]

Answer:

The amount of gas that is to be released in the first second in other to attain an acceleration of  27.0 m/s2  is

      \frac{\Delta m}{\Delta t}   = 83.92 \ Kg/s

Explanation:

From the question we are told that

   The mass of the rocket is m = 6300 kg

   The velocity at gas is being ejected is  u =  2000 m/s

    The initial acceleration desired is a =  27.0 \  m/s

   The time taken for  the gas to be ejected is  t = 1 s

Generally this desired acceleration is mathematically represented as

        a = \frac{u *  \frac{\Delta m}{\Delta t} }{M -\frac{\Delta m}{\Delta t}* t}

Here \frac{\Delta m}{\Delta  t }  is the rate at which gas is being ejected with respect to time

Substituting values

      27 = \frac{2000 *  \frac{\Delta m}{\Delta t} }{6300 -\frac{\Delta m}{\Delta t}* 1}

=>   170100 -27* \frac{\Delta m}{\Delta t} = 2000 *  \frac{\Delta m}{\Delta t}

=>   170100  = 2027 *  \frac{\Delta m}{\Delta t}

=>   \frac{\Delta m}{\Delta t}   = \frac{170100}{2027}

=>   \frac{\Delta m}{\Delta t}   = 83.92 \ Kg/s

     

3 0
2 years ago
A Porsche challenges a Honda to a 200-m race.Because the Porsche's acceleration of 3.5 m/s2 is larger than the Honda's 3.0 m/s2,
Blizzard [7]

Answer:

Honda won by 0.14 s

Explanation:

We are given that

Distance =S=200 m

Initial velocity of Honda=u=0m/s

Initial velocity of Porsche=u'=0m/s

Acceleration of Honda=3.0m/s^2

Acceleration of Porsche's=3.5m/s^2

Time taken by Honda  to start=1 s

s=ut+\frac{1}{2}at^2

Substitute the values

200=0(t)+\frac{1}{2}(3)t^2

200=\frac{3}{2}t^2

t^2=\frac{200\times 2}{3}=\frac{400}{3}

t=\sqrt{\frac{400}{3}}=11.55s

Time taken by Honda=11.55 s

Now, time taken by  Porsche

200=\frac{1}{2}(3.5)t^2

t^2=\frac{200\times 2}{3.5}

t=\sqrt{\frac{400}{3.5}}=10.69 s

Total time taken by Porsche=10.69+1=11.69 s

Because it start 1 s late

Time taken by Honda is less than Porsche .Therefore, Honda won and

Time =11.69-11.55=0.14 s

Honda won by 0.14 s

3 0
2 years ago
A square conducting loop 8.4 cm on a side is placed in a uniform B-field so that the plane of the loop is perpendicular to the d
arsen [322]

Answer:

Explanation:

area of square loop A = side²

= 8.4² x 10⁻⁴

A = 70.56 x 10⁻⁴ m²

when it is converted into rectangle , length = 14.7  , width = 2.1

area = length x width

= 14.7 x 2.1 x 10⁻⁴

= 30.87 x 10⁻⁴ m²

Let magnetic field be B

Change in flux = magnetic field x change in area

= B x ( 70.56 x 10⁻⁴ - 30.87 x 10⁻⁴ )

= 39.69 x 10⁻⁴ B

rate of change of flux = change in flux / time taken

= 39.69 x 10⁻⁴ B  / 6.5 x 10⁻³

= 6.1 x 10⁻¹ B

emf induced = 6.1 x 10⁻¹ B

6.1 x 10⁻¹ B  = 14.7 ( given )

B = 2.41 x 10

= 24.1 T

B ) magnetic flux is decreasing , so it needs to be increased as per Lenz's law . Hence current induced will be anticlockwise so that additional  magnetic flux is induced out of the page.

4 0
2 years ago
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