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inn [45]
2 years ago
8

Consider an electron near the Earth's equator. In which direction does it tend to deflect if its velocity is (a) directed downwa

rd?
Physics
1 answer:
Elenna [48]2 years ago
4 0

The electron near the Earth's equator doesn't deflect if it's velocity is directed downwards.

Assume that north is directed along upwards, south is directed along downwards, east is directed along rightwards, and west is directed along leftwards.

The magnetic force on the charge q, moving with velocity v in the magnetic field B is given by

F=q(→v×→B)=qvBsinθ ..(1)

Here,

θ is the angle between the magnetic field and velocity.

Inside the Earth, the Earth's magnetic field is along northward (upwards), and the velocity is given to be along southwards (downwards) making an angle of 180°. In equation (1) sin180°=0. So, no magnetic force acts on the electron

Therefore,

The electron didn't deflect if it is directed downwards.

To know more about "Magnetic field"

Refer this link:

brainly.com/question/25573309

#SPJ4

You might be interested in
Which of the following best describes why we say that light is an electromagnetic wave?
Kaylis [27]

Answer: The passage of a light wave can cause electrically charged particles to move up and down.

Explanation:

Electromagnetic waves are transversal waves, they are a combination of oscillating electric and magnetic fields, which propagate through space carrying energy from one place to another.  

This means the oscillation of the wave occurs in the transversal direction to its propagation. In addition, electromagnetic waves are spread thanks to the electromagnetic fields produced by moving electric charges.

8 0
4 years ago
Three capacitors are connected as follows: 2.8 F capacitor and 5.57 F capacitor are connected in series, then that combination i
DedPeter [7]

ANSWER

8.99 F

EXPLANATION

We know that two capacitors of capacitances 2.8 F and 5.57 F are connected in series, while a third capacitor of capacitance 7.13 F is connected in parallel to that combination,

The capacitance works similarly to the resistance, except that when capacitors are connected in parallel, their capacitances add up, while when they are connected in series, the equivalent capacitance is like we were finding the equivalent resistance of resistors connected in parallel,

C_{eq}=\frac{1}{\frac{1}{C_1}+\frac{1}{C_2}}+C_3=\frac{1}{\frac{1}{2.8F}+\frac{1}{5.57F}}+7.13F\approx8.99F

Hence, the total capacitance is 8.99 F, rounded to the nearest hundredth.

6 0
2 years ago
13.00 cm + 2.56 mm + 0.0047= cm? Solve problem with correct number of significant digits
Drupady [299]
The answer is 13.2607 cm
7 0
3 years ago
Read 2 more answers
An object of mass 2kg is on an incline where there is an applied force of 15N. The
seraphim [82]

a) See free-body diagram in attachment

b) The acceleration is 2.46 m/s^2

Explanation:

a)

The free-body diagram of an object is a diagram representing all the forces acting on the object. Each force is represented by a vector of length proportional to the magnitude of the force, pointing in the same direction as the force.

The free-body diagram for this object is shown in the figure in attachment.

There are three forces acting on the object:

  • The weight of the object, labelled as mg (where m is the mass of the object and g is the acceleration of gravity), acting downward
  • The applied force, F_a, acting up along the plane
  • The force of friction, F_f, acting down along the plane

b)

In order to find the acceleration of the object, we need to write the equation of the forces acting along the direction parallel to the incline. We have:

F_a - F_f - mg sin \theta = ma

where:

F_a = 15 N is the applied force, pushing forward

F_f = 5 N is the frictional force, acting backward

mg sin \theta is the component of the weight parallel to the incline, acting backward, where

m = 2 kg is the mass of the object

g=9.8 m/s^2 is the acceleration of gravity

\theta=15^{\circ} is the angle between the horizontal and the incline (it is not given in the problem, so I assumed this value)

a is the acceleration

Solving for a, we find:

a=\frac{F_a - F_f - mg sin \theta}{m}=\frac{15-5-(2)(9.8)(sin 15^{\circ})}{2}=2.46 m/s^2

Learn more about inclined planes:

brainly.com/question/5884009

#LearnwithBrainly

8 0
3 years ago
1. A buoyant force of 790 N lifts a 214 kg sinking boat.What is the boat's<br> net acceleration?
enot [183]

Answer:

The net acceleration of the boat is approximately 6.12 m/s² downwards

Explanation:

The buoyant or lifting force applied to the boat = 790 N

The mass of the boat lifted by the buoyant force = 214 kg

The force applied to a body is defined as the product of the mass and the acceleration of the body. Therefore, the buoyant force, F, acting on the boat can be presented as follows;

Fₐ = F - W

The weight of the boat = 214 × 9.81 = 2099.34 N

Therefore;

Fₐ = 790 - 2099.34  = -1309.34 N

Fₐ = Mass of the boat × The acceleration of the boat

Given that the buoyant force, Fₐ, is the net force acting on the boat, we have;

F = Mass of the boat × The net acceleration of the boat

F = -1309.34 N =  214 kg × The net acceleration of the boat

∴ The net acceleration of the boat = -1309.34 N/(214 kg) ≈ -6.12 m/s²

The net acceleration of the boat ≈ 6.12 m/s² downwards

7 0
3 years ago
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