Mathematics: Find the direction of vector

Find the direction of vector, №13484919, 07.09.2022 19:45

Physics

Two transverse waves travel along the same taut string. Wave 1 is described by y1(x, t) = A sin(kx - ωt), while wave 2 is described by y2(x, t) = A sin(kx + ωt + φ). The phases (arguments of the sines) are in radians, as usual, and A = 2.1 cm and φ = 0.35π rad. Choose the answer that correctly describes the waves’ directions of travel. MultipleChoice : 1) Both waves travel in the negative x direction. 2) Due to the φ term in the phase of wave 2, it does not travel. Wave 1 travels in the negative x-direction. 3) Due to the φ term in the phase of

Step-by-step answer

P Answered by Specialist

6) Wave 1 travels in the positive x-direction, while wave 2 travels in the negative x-direction.

Explanation:

What matters is the part $kx \pm \omega t$ , the other parts of the equation don't affect time and space variations. We know that when the sign is - the wave propagates to the positive direction while when the sign is + the wave propagates to the negative direction, but here is an explanation of this:

For both cases, + and -, after a certain time $\delta t$ ( $\delta t 0$ ), the displacement y of the wave will be determined by the $kx\pm\omega (t+\delta t)$ term. For simplicity, if we imagine we are looking at the origin (x=0), this will be simply $\pm \omega (t+\delta t)$ .

To know which side, right or left of the origin, would go through the origin after a time $\delta t$ (and thus know the direction of propagation) we have to see how we can achieve that same displacement y not by a time variation but by a space variation $\delta x$ (we would be looking where in space is what we would have in the future in time). The term would be then $k(x+\delta x)\pm\omega t$ , which at the origin is $k \delta x \pm \omega t$ . This would mean that, when the original equation has $kx+\omega t$ , we must have that $\delta x0$ for $k\delta x+\omega t$ to be equal to $kx+\omega\delta t$ , and when the original equation has $kx-\omega t$ , we must have that $\delta x$ for $k\delta x-\omega t$ to be equal to $kx-\omega \delta t$

Note that their values don't matter, although they are a very small variation (we have to be careful since all this is inside a sin function), what matters is if they are positive or negative and as such what is possible or not .

In conclusion, when $kx+\omega t$ , the part of the wave on the positive side ( $\delta x0$ ) is the one that will go through the origin, so the wave is going in the negative direction, and viceversa.

Physics

Two transverse waves travel along the same taut string. Wave 1 is described by y1(x, t) = A sin(kx - ωt), while wave 2 is described by y2(x, t) = A sin(kx + ωt + φ). The phases (arguments of the sines) are in radians, as usual, and A = 2.1 cm and φ = 0.35π rad. Choose the answer that correctly describes the waves’ directions of travel. MultipleChoice : 1) Both waves travel in the negative x direction. 2) Due to the φ term in the phase of wave 2, it does not travel. Wave 1 travels in the negative x-direction. 3) Due to the φ term in the phase of

Step-by-step answer

P Answered by Master

6) Wave 1 travels in the positive x-direction, while wave 2 travels in the negative x-direction.

Explanation:

What matters is the part $kx \pm \omega t$ , the other parts of the equation don't affect time and space variations. We know that when the sign is - the wave propagates to the positive direction while when the sign is + the wave propagates to the negative direction, but here is an explanation of this:

For both cases, + and -, after a certain time $\delta t$ ( $\delta t 0$ ), the displacement y of the wave will be determined by the $kx\pm\omega (t+\delta t)$ term. For simplicity, if we imagine we are looking at the origin (x=0), this will be simply $\pm \omega (t+\delta t)$ .

To know which side, right or left of the origin, would go through the origin after a time $\delta t$ (and thus know the direction of propagation) we have to see how we can achieve that same displacement y not by a time variation but by a space variation $\delta x$ (we would be looking where in space is what we would have in the future in time). The term would be then $k(x+\delta x)\pm\omega t$ , which at the origin is $k \delta x \pm \omega t$ . This would mean that, when the original equation has $kx+\omega t$ , we must have that $\delta x0$ for $k\delta x+\omega t$ to be equal to $kx+\omega\delta t$ , and when the original equation has $kx-\omega t$ , we must have that $\delta x$ for $k\delta x-\omega t$ to be equal to $kx-\omega \delta t$

Note that their values don't matter, although they are a very small variation (we have to be careful since all this is inside a sin function), what matters is if they are positive or negative and as such what is possible or not .

In conclusion, when $kx+\omega t$ , the part of the wave on the positive side ( $\delta x0$ ) is the one that will go through the origin, so the wave is going in the negative direction, and viceversa.

Physics

A particle accelerator fires a proton into a region with a magnetic field that points in the +x-direction (a) If the proton is moving in the ty-direction, what is the direction of the magnetic force on the proton? +x-direction -x-direction +y-direction -y-direction +z-direction -z-direction zero force What is the formula used to find the vector magnetic force? What is the right-hand rule for a cross prod (b) If the proton is moving in the -y-direction, what is the direction of the magnetic force on the proton? +x-direction -x-direction +y-direction

Step-by-step answer

P Answered by PhD

a) -z direction, b) +z direction, c) F=0

Explanation:

The magnetic force is given by the expression

F = q v x B

the bold indicate vectors, this equation can be separated in its module

F = a v B sin θ

and where θ is the angles between the speed and the magnetic field.

The direction of the force can be found with the right-hand rule. For a positive charge, the thumb goes in the direction of speed, the fingers extended in the direction of the magnetic field and the palm points in the direction of the force, if the charge is negative the force is in the opposite direction.

a) Let's apply this to our case

the proton is positively charged

moves in the direction of + x

The magnetized field goes in the direction of y

therefore applying the right hand rule the force must be in the direction of the negative part of the z-axis (-z)

The right-hand rule is used to find this address.

b) in this case it indicates that the proton moves in the recode of -y

again we apply the right hand rule and the force is in the direction of + z

c) The proton moves in the x direction

In this case the force is zero because the angle between the field and the speed is zero and the sine is zero, therefore the force is zero

Physics

A particle accelerator fires a proton into a region with a magnetic field that points in the +x-direction (a) If the proton is moving in the ty-direction, what is the direction of the magnetic force on the proton? +x-direction -x-direction +y-direction -y-direction +z-direction -z-direction zero force What is the formula used to find the vector magnetic force? What is the right-hand rule for a cross prod (b) If the proton is moving in the -y-direction, what is the direction of the magnetic force on the proton? +x-direction -x-direction +y-direction

Step-by-step answer

P Answered by PhD

a) -z direction, b) +z direction, c) F=0

Explanation:

The magnetic force is given by the expression

F = q v x B

the bold indicate vectors, this equation can be separated in its module

F = a v B sin θ

and where θ is the angles between the speed and the magnetic field.

The direction of the force can be found with the right-hand rule. For a positive charge, the thumb goes in the direction of speed, the fingers extended in the direction of the magnetic field and the palm points in the direction of the force, if the charge is negative the force is in the opposite direction.

a) Let's apply this to our case

the proton is positively charged

moves in the direction of + x

The magnetized field goes in the direction of y

therefore applying the right hand rule the force must be in the direction of the negative part of the z-axis (-z)

The right-hand rule is used to find this address.

b) in this case it indicates that the proton moves in the recode of -y

again we apply the right hand rule and the force is in the direction of + z

c) The proton moves in the x direction

In this case the force is zero because the angle between the field and the speed is zero and the sine is zero, therefore the force is zero

Physics

A long line of charge with uniform linear charge density λ1 is located on the x-axis and another long line of charge with uniform linear charge density λ2 is located on the y-axis with their centers crossing at the origin. In what direction is the electric field at point z = a on the positive z-axis if λ1 and λ2 are positive? A-) The positive z-direction B-) All directions are possible parallel to the x-y plane C-) The negative z-direction D-) Halfway between the x-direction and the y-direction C-) Cannot be determined

Step-by-step answer

P Answered by PhD

1

A.The positive z-direction

Explanation:

We are given that

Linear charge density of long line which is located on the x-axis= $\lambda_1$

Linear charge density of another long line which is located on the y-axis= $\lambda_2$

We have to find the direction of electric field at z=a on the positive z-axis if $\lambda_1$ and $\lambda_2$ are positive.

The direction of electric field at z=a on the positive z-axis is positive z-direction .

Because $\lambda_1$ and $\lambda_2$ are positive and the electric field is applied away from the positive charge.

Hence, option A is true.

A.The positive z-direction

Physics

A long line of charge with uniform linear charge density λ1 is located on the x-axis and another long line of charge with uniform linear charge density λ2 is located on the y-axis with their centers crossing at the origin. In what direction is the electric field at point z = a on the positive z-axis if λ1 and λ2 are positive? A-) The positive z-direction B-) All directions are possible parallel to the x-y plane C-) The negative z-direction D-) Halfway between the x-direction and the y-direction C-) Cannot be determined

Step-by-step answer

P Answered by PhD

1

A.The positive z-direction

Explanation:

We are given that

Linear charge density of long line which is located on the x-axis= $\lambda_1$

Linear charge density of another long line which is located on the y-axis= $\lambda_2$

We have to find the direction of electric field at z=a on the positive z-axis if $\lambda_1$ and $\lambda_2$ are positive.

The direction of electric field at z=a on the positive z-axis is positive z-direction .

Because $\lambda_1$ and $\lambda_2$ are positive and the electric field is applied away from the positive charge.

Hence, option A is true.

A.The positive z-direction

Physics

A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.50km/s in the +x-direction experiences a force of 2.06�10-16N in the +y-direction, and an electron moving at 4.40km/s in the -z-direction experiences a force of 8.40�10-16N in the +y-direction. Part A What is the magnitude of the magnetic field? B = T Part B What is the direction of the magnetic field? (in the xz-plane) theta = from the -z-direction Part C What is the magnitude of the magnetic force on an electron moving

Step-by-step answer

P Answered by Specialist

Explanation:

velocity of proton v = 1.5 x 10³ i m /s

charge on proton e = 1.6 x 10⁻¹⁹ C

Let the magnetic field be B = Bx i + Bz k

force on charged particle ( proton )

F = e ( v x B )

2.06 x10⁻¹⁶ j = 1.6 x 10⁻¹⁹ [ 1.5 x 10³ i x ( Bx i + Bz k) ]

2.06 x10⁻¹⁶ j = - 1.6 x 10⁻¹⁹ x 1.5 x 10³ Bz j) ]

2.06 x10⁻¹⁶ = - 1.6 x 10⁻¹⁹ x 1.5 x 10³ Bz

Bz = - .8583

force on charged particle ( electron )

F = e ( v x B )

8.40 x10⁻¹⁶ j = -1.6 x 10⁻¹⁹ [ - 4.4 x 10³ k x ( Bx i + Bz k) ]

8.4 x10⁻¹⁶ j = 1.6 x 10⁻¹⁹ x 4.4 x 10³ Bx j ]

- 8.4 x10⁻¹⁶ = 1.6 x 10⁻¹⁹ x 4.4 x 10³ Bx

Bx = - 1.19

Magnetic field = - 1.19 i - .8583 k

magnitude = √ (1.19² + .8583²)

= 1.467 T

If it is making angle θ with x - axis in x -z plane

Tanθ = (.8583 / 1.19 )

36⁰ .

C )

v = - 3.7 x 10³j m /s

e = - 1.6 x 10⁻¹⁶ C

Force = F = e ( v x B )

= -1.6 x 10⁻¹⁹ [ -3.7 x 10³ j x ( Bx i + Bz k) ]

= - 1.6 x 10⁻¹⁹ x 3.7 x 10³ Bx k -1.6 x 10⁻¹⁹ x 3.7 x 10³Bzi ]

= 5.08 i - 7.04 k

Tanθ = 54 ° .

Physics

A vector A has a magnitude of 5 units and points in the −y-direction, while a vector B has triple the magnitude of A and points in the +x-direction. What are the magnitude and direction of the following vectors? Give the directions of each as an angle measured counterclockwise from the +x-direction.

Step-by-step answer

P Answered by Specialist

A+B; 5√5 units, 341.57°

A-B; 5√5 units, 198.43°

B-A; 5√5 units, 18.43°

Explanation:

Given A = 5 units

By vector notation and the axis of A, it is represented as -5j

B = 3 × 5 = 15 units

Using the vector notations and the axis, B is +15i. The following vectors ate taking as the coordinates of A and B

(a) A + B = -5j + 15i

A+B = 15i -5j

|A+B| = √(15)²+(5)²

= 5√5 units

∆ = arctan(5/15) = 18.43°

The angle ∆ is generally used in the diagrams

∆= 18.43°

The direction of A+B is 341.57° based in the condition given (see attachment for diagrams

(b) A - B = -5j -15i

A-B = -15i -5j

|A-B|= √(15)²+(-5)²

|A-B| = √125

|A-B| = 5√5 units

The direction is 180+18.43°= 198.43°

See attachment for diagrams

(c) B-A = 15i -( -5j) = 15i + 5j

|B-A| = 5√5 units

The direction is 18.43°

See attachment for diagram

A vector A has a magnitude of 5 units and points in the −y-direction, while a vector B has triple th

Physics

A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.50km/s in the +x-direction experiences a force of 2.06�10-16N in the +y-direction, and an electron moving at 4.40km/s in the -z-direction experiences a force of 8.40�10-16N in the +y-direction. Part A What is the magnitude of the magnetic field? B = T Part B What is the direction of the magnetic field? (in the xz-plane) theta = from the -z-direction Part C What is the magnitude of the magnetic force on an electron moving

Step-by-step answer

P Answered by Specialist

Explanation:

velocity of proton v = 1.5 x 10³ i m /s

charge on proton e = 1.6 x 10⁻¹⁹ C

Let the magnetic field be B = Bx i + Bz k

force on charged particle ( proton )

F = e ( v x B )

2.06 x10⁻¹⁶ j = 1.6 x 10⁻¹⁹ [ 1.5 x 10³ i x ( Bx i + Bz k) ]

2.06 x10⁻¹⁶ j = - 1.6 x 10⁻¹⁹ x 1.5 x 10³ Bz j) ]

2.06 x10⁻¹⁶ = - 1.6 x 10⁻¹⁹ x 1.5 x 10³ Bz

Bz = - .8583

force on charged particle ( electron )

F = e ( v x B )

8.40 x10⁻¹⁶ j = -1.6 x 10⁻¹⁹ [ - 4.4 x 10³ k x ( Bx i + Bz k) ]

8.4 x10⁻¹⁶ j = 1.6 x 10⁻¹⁹ x 4.4 x 10³ Bx j ]

- 8.4 x10⁻¹⁶ = 1.6 x 10⁻¹⁹ x 4.4 x 10³ Bx

Bx = - 1.19

Magnetic field = - 1.19 i - .8583 k

magnitude = √ (1.19² + .8583²)

= 1.467 T

If it is making angle θ with x - axis in x -z plane

Tanθ = (.8583 / 1.19 )

36⁰ .

C )

v = - 3.7 x 10³j m /s

e = - 1.6 x 10⁻¹⁶ C

Force = F = e ( v x B )

= -1.6 x 10⁻¹⁹ [ -3.7 x 10³ j x ( Bx i + Bz k) ]

= - 1.6 x 10⁻¹⁹ x 3.7 x 10³ Bx k -1.6 x 10⁻¹⁹ x 3.7 x 10³Bzi ]

= 5.08 i - 7.04 k

Tanθ = 54 ° .

Physics

A vector A has a magnitude of 5 units and points in the −y-direction, while a vector B has triple the magnitude of A and points in the +x-direction. What are the magnitude and direction of the following vectors? Give the directions of each as an angle measured counterclockwise from the +x-direction.

Step-by-step answer

P Answered by Specialist

A+B; 5√5 units, 341.57°

A-B; 5√5 units, 198.43°

B-A; 5√5 units, 18.43°

Explanation:

Given A = 5 units

By vector notation and the axis of A, it is represented as -5j

B = 3 × 5 = 15 units

Using the vector notations and the axis, B is +15i. The following vectors ate taking as the coordinates of A and B

(a) A + B = -5j + 15i

A+B = 15i -5j

|A+B| = √(15)²+(5)²

= 5√5 units

∆ = arctan(5/15) = 18.43°

The angle ∆ is generally used in the diagrams

∆= 18.43°

The direction of A+B is 341.57° based in the condition given (see attachment for diagrams

(b) A - B = -5j -15i

A-B = -15i -5j

|A-B|= √(15)²+(-5)²

|A-B| = √125

|A-B| = 5√5 units

The direction is 180+18.43°= 198.43°

See attachment for diagrams

(c) B-A = 15i -( -5j) = 15i + 5j

|B-A| = 5√5 units

The direction is 18.43°

See attachment for diagram

Find the direction of vector <6,5>

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