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sesenic [268]
3 years ago
13

At a particular instant, an electron moves toward the east in a uniform magnetic field that is directed straight downward. the m

agnetic force that acts on it is
Physics
1 answer:
alexdok [17]3 years ago
4 0

Answer:

towards the south

Explanation:

When the electron enters the region with magnetic field, it experiences a magnetic force perpendicular to both the directions of the electron's velocity and the magnetic field.

The direction of the force exerted on the electron can be found by using the right-hand rule:

- Index finger: direction of the velocity of the electron --> towards the east

- Middle finger: direction of the magnetic field --> downward

- Thumb: direction of the force on a positive particle --> towards the north

However, the electron is a negatively charged particle, so we must reverse the direction of the force: therefore, the force exerted on the electron is towards the south.

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To lower the voltage from a "120 V" outlet, what is used (first, at least) in most situations
nasty-shy [4]

Answer:

B. Transformer

Explanation:

A transformer is a device that is used to either raise or lower voltages and currents in an electrical circuit. In modern electrical distribution systems, transformers are used to boost voltage levels so as to decrease line losses during transmission. It basically trades voltage for current in a circuit, while not affecting the total electrical power. This means it takes high-voltage electricity with a small current and changes it into low-voltage electricity with a large current, or vice versa.

4 0
3 years ago
Atoms of the same element that differ only in the number of neutrons are known as.
True [87]

Answer:

They are known as isotopes

7 0
2 years ago
Positive charge Q is distributed uniformly along the x-axis fromx=0 to x=a. A positive point charge q is located on the positive
dybincka [34]

Answer:

a. b- x= y

dx = -dy

b. F = \frac{-kQqi}{r (a+r)}

c.  F = \frac{-kQqi}{r^{2} }

Explanation:

a. x components:

dE = \frac{kdq}{(a+r-x^{2}) } \\

     = \frac{kQdx}{(a(a+r-x)^2}

Integrating and solving gives:

b- x= y

dx = -dy

b. the force is given by the equation derived from (a.):

F = \frac{-kQqi}{r (a+r)}

c. Given that r>>a, the expression becomes:

F = \frac{-kQqi}{r^{2} }

Explanation:

When the size of the charge distribution is less than the distance to the deviation point of the charge then the charge distribution would produce the same effect such as a linear charge.

6 0
3 years ago
Sayid made a chart listing data of two colliding objects. A 5-column table titled Collision: Two Objects Stick Together with 2 r
Alborosie

Answer:

6 m/s is the missing final velocity

Explanation:

From the data table we extract that there were two objects (X and Y) that underwent an inelastic collision, moving together after the collision as a new object with mass equal the addition of the two original masses, and a new velocity which is the unknown in the problem).

Object X had a mass of 300 kg, while object Y had a mass of 100 kg.

Object's X initial velocity was positive (let's imagine it on a horizontal axis pointing to the right) of 10 m/s. Object Y had a negative velocity (imagine it as pointing to the left on the horizontal axis) of -6 m/s.

We can solve for the unknown, using conservation of momentum in the collision: Initial total momentum = Final total momentum (where momentum is defined as the product of the mass of the object times its velocity.

In numbers, and calling P_{xi} the initial momentum of object X and P_{yi} the initial momentum of object Y, we can derive the total initial momentum of the system: P_{total}_i=P_{xi}+P_{yi}= 300*10 \frac{kg*m}{s} -100*6\frac{kg*m}{s} =\\=(3000-600 )\frac{kg*m}{s} =2400 \frac{kg*m}{s}

Since in the collision there is conservation of the total momentum, this initial quantity should equal the quantity for the final mometum of the stack together system (that has a total mass of 400 kg):

Final momentum of the system: M * v_f=400kg * v_f

We then set the equality of the momenta (total initial equals final) and proceed to solve the equation for the unknown(final velocity of the system):

2400 \frac{kg*m}{s} =400kg*v_f\\\frac{2400}{400} \frac{m}{s} =v_f\\v_f=6 \frac{m}{s}

7 0
3 years ago
Read 2 more answers
Julie blows a bubble. At first, the pressure of the gas in the bubble is 4kPa. The bubble floats into the air and expands. When
Andrew [12]

Answer:

V₁ = 1.75 m³

Explanation:

Assuming the gas to be an ideal gas. At constant temperature, the relationship between the volume and temperature of an ideal gas is given by Boyle's Law as follows:

P_{1}V_{1} = P_{2}V_{2}

where,

P₁ = Initial Pressure of the Gas = 4 KPa

V₁ = Initial Volume of the Gas = ?

P₂ = Final Pressure of the Gas = 2 KPa

V₂ = Final Volume of the Gas = 3.5 m³

Therefore,

(4\ KPa)V_{1} = (2\ KPa)(3.5\ m^{3})\\\\V_{1}=\frac{2\ KPa}{4\ KPa}(3.5\ m^{3})\\\\

<u>V₁ = 1.75 m³</u>

4 0
3 years ago
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