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Rom4ik [11]
3 years ago
13

Consider an electron traveling horizontally in the positive direction, above, near, and parallel to a current-carrying wire. The

current is traveling in the negative direction.
(a) Will the electron keep on going straight, move towards the wire or away from the wire?


(b) If we replace the electron with a proton and now the proton is coming from above, traveling in the perpendicular direction, moving towards the wire. Will the proton hit the wire, veer to the left? veer to the right? stop? go out of the page? go onto the page?
Physics
1 answer:
allsm [11]3 years ago
6 0

Answer:

(a) The electron will move towards the wire.

The direction of the magnetic fields created by the wire can be found via right-hand rule. If you point your thumb towards the direction of the current, and if you curl your fingers, the direction of your four fingers will give the direction of the magnetic field. In this case, magnetic field is around the wire, and into the page just above the wire, where the electron is located.

\vec{F} = q\vec{v} \times \vec{B}

According to the above formula, the direction of the force the wire applies to the electron can be found by right-hand rule.

Since the electron has a negative charge, the direction of the force is towards the wire.

(b) The proton will veer to the right.

The direction of the magnetic field is the same as the previous part. The proton has a positive charge, and coming from above. The direction of its velocity is downwards. The magnetic field above the wire is pointed into the page. Using the right-hand rule, the magnetic force on the proton is directed to the right, with respect to us.

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Some students set up a circuit and decided to measure the voltage at different points around
Fiesta28 [93]

Answer:

 V₁ = 6 V ,  V₂ = V₃ = 3 V

Explanation:

To solve this circuit we must remember that there are two fundamental types of construction in series and parallel.

* a serial circuit there is only one path for current

in this circuit the constant current in the entire circuit and the voltage is the sum of the voltage of each term

* Parallel circuit in this there are two or more paths for the current

in this circuit the voltage is constant and the east is divided between each branch

with these principles let's analyze the proposed circuit

The DC battery is in parallel with resistor R1 and the equivalent of the other branch,

as in a parallel circuit the voltage is constant

               V₁ = 6 V

in the other branch (23) it forms a series construction, where the current is constant

               6 = iR₂ + iR₃

as they indicate that each resistance has the same value

              6 = 2 iR

              V = V₂ = V₃ = 3 V

3 0
3 years ago
As you ride a bicycle on the sidewalk with a speed of
mixer [17]

Answer:

1.2 seconds

Explanation:

distance = ((final speed + initial speed) * time)/2

Here given:

  • distance: 3.8 meter
  • initial speed: 6.4 m/s
  • final speed: 0 m/s

Solving steps:

3.8 = ((0 + 6.4) * time))/2

3.8 = 3.2(time)

time = 3.8/3.2

time = 1.1875 seconds ≈ 1.2 seconds

5 0
2 years ago
A point charge +2Q is at the origin and a point charge −Q is located along the x axis at x = d as in the figure below. Find a sy
Akimi4 [234]

Answer: A symbolic expression for the net force on a third point charge +Q located along the y axis  

F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}

Explanation:

Let the force on +Q charge y-axis due to +2Q charge be F_1 and force on +Q charge y axis due to -Q charge on x-axis be F_2.

Distance between the +2Q charge and +Q charge = d units

Distance between the -Q charge and +Q charge = \sqrt{2}d units

k_e= Coulomb constant

F_1=k_e\frac{(+2Q)(+Q)}{d^2}=k_e\frac{+2Q^2}{d^2} N

F_2=k_e\frac{(-Q)(+Q)}{(\sqrt{2}d)^2}=k_e\frac{-Q^2}{2d^2} N

Net force on +Q charge on y-axis is:

F_x=F_2sin 45^o=k_e\frac{-Q^2}{2d^2}\times \frac{1}{\sqrt{2}} N

F_y=F_1-F_2cos45^o

F_y=(F_1-F_2cos45^o)=(k_e\frac{+2Q^2}{d^2})-(k_e\frac{-Q^2}{2d^2}\frac{1}{\sqrt{2}})

F_N=\sqrt{F_x^2+F_y^2}

|F_N|=|k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}|

The net froce on the +Q charge on y-axis is

F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}

4 0
3 years ago
Read 2 more answers
The density of a substance is its mass per unit volume. 
Over [174]

Answers:

a) kg/m^{3}

b) derived

Explanation:

Density D is a characteristic property of substances and materials and is defined as the relationship between the mass m and volume V of a specific substance or material, as shown below:

D=\frac{m}{V}

This means density is inversely proportional to volume.

In addition, density is a derived unit, because is based on two basic units of the International Sistem of Units: Mass (kilograms) and Volume (cubic meters)

7 0
3 years ago
Suppose you figured resistance of a color-coded resistor is 100 ± 5 ohms. You did the Ohms law experiment and measured resistanc
aivan3 [116]

Answer:

* I will look at the range of both, such as 100-5 = 95 and 94+4=98. Since the ranges overlap, I can conclude that they are the same within uncertainty.

* I willl draw a dot-and-whisker plot for both. Since they overlap, I can conclude that they are equal within the uncertainty

Explanation:

The value of all physical measurement has an uncertainty, so it is within a range of values, in this case we must compare the range of values ​​of the resistors

The nominal value is 100 ± 5 Ω

therefore its range of possible values ​​is between

             95 <= R <= 105

For the experimentally measured value

          88 <= R <= 96

as the values ​​overlap we must conclude that they are the same.

When reviewing the different options we have two statements that are correct

* I will look at the range of both, such as 100-5 = 95 and 94+4=98. Since the ranges overlap, I can conclude that they are the same within uncertainty.

* I willl draw a dot-and-whisker plot for both. Since they overlap, I can conclude that they are equal within the uncertainty

5 0
3 years ago
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