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kifflom [539]
3 years ago
11

A wave travels at 1,500 m/s with a frequency of 5 hertz. What is its wavelength?

Physics
1 answer:
Alex777 [14]3 years ago
7 0

Answer:

λ = 300 m

Explanation:

We are given;

Speed; v = 1500 m/s

Frequency; f = 5 Hz.

Formula for wavelength is given by;

λ = v/f

Plugging in the relevant values;

λ = 1500/5

λ = 300 m

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A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex
irina1246 [14]

a)

F_{E_x}=1.19\cdot 10^{-3}N (+x axis)

F_{B_x}=0

F_{B_y}=0

b)

F_{E_x}=1.19\cdot 10^{-3} N (+x axis)

F_{B_x}=0

F_{B_y}=3.21\cdot 10^{-3}N (+z axis)

c)

F_{E_x}=1.19\cdot 10^{-3} N (+x axis)

F_{B_x}=3.21\cdot 10^{-3} N (+y axis)

F_{B_y}=3.21\cdot 10^{-3}N (-x axis)

Explanation:

a)

The electric force exerted on a charged particle is given by

F=qE

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

q=+4.9\mu C=+4.9\cdot 10^{-6}C is the charge

E_x=+242 N/C is the electric field, along the x-direction

So the electric force (along the x-direction) is:

F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N

towards positive x-direction.

The magnetic force instead is given by

F=qvB sin \theta

where

q is the charge

v is the velocity of the charge

B is the magnetic field

\theta is the angle between the directions of v and B

Here the charge is stationary: this means v=0, therefore the magnetic force due to each component of the magnetic field is zero.

b)

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

F_{E_x}=1.19\cdot 10^{-3} N (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- B_x: this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, \theta=0^{\circ}, so the force due to this field is zero.

- B_y: this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, \theta=90^{\circ}. Therefore, \theta=90^{\circ}, so the force due to this field is:

F_{B_y}=qvB_y

where:

q=+4.9\cdot 10^{-6}C is the charge

v=345 m/s is the velocity

B_y = +1.9 T is the magnetic field

Substituting,

F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N

And the direction of this force can be found using the right-hand rule:

- Index finger: direction of the velocity (+x axis)

- Middle finger: direction of the magnetic field (+y axis)

- Thumb: direction of the force (+z axis)

c)

As in part b), the electric force has not change, since it does not depend on the veocity of the particle:

F_{E_x}=1.19\cdot 10^{-3}N (+x axis)

For the field B_x, the velocity (+z axis) is now perpendicular to the magnetic field (+x axis), so the force is

F_{B_x}=qvB_x

And by substituting,

F_{B_x}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+x axis)

- Thumb: force (+y axis)

For the field B_y, the velocity (+z axis) is also perpendicular to the magnetic field (+y axis), so the force is

F_{B_y}=qvB_y

And by substituting,

F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+y axis)

- Thumb: force (-y axis)

3 0
3 years ago
Two light bulbs are installed in two sockets, the sockets are connected in series, and power is then applied to the combination
Arisa [49]

Answer:

False

Explanation:

<em>If one of the bulbs is removed from the series, the other bulb will not come on at all.</em>

This is because the removal of one of the bulbs would interrupt the flow of current though the entire circuit.

Hence, that the other one will get brighter if one of two bulbs in a circuit is removed from its socket is not true.

5 0
3 years ago
The elements least likely to form bonds are found in which group?<br> O 15<br> O 16<br> 017<br> O 18
12345 [234]
The answer is group 18.

Explanation:
Group 18 is known as the Noble Gases. They are the most stable elements because they have a full outer shell of valence electrons, so they have no need to bond with other elements.
7 0
3 years ago
Read 2 more answers
In two or more sentences describe the impact of fossil fuel combustion on the global carbon cycle. In three sentences or less ex
Nuetrik [128]
For many solids<span> dissolved in liquid water, the </span>solubility increases<span> with </span>temperature<span>. The </span>increase<span> in kinetic energy that comes with </span>higher temperatures<span> allows the solvent molecules to more effectively break apart the solute molecules that are held together by intermolecular attractions.</span>
8 0
3 years ago
Read 2 more answers
Un cuerpo se encuentra en reposo sobre una mesa horizontal. Entonces se puede afirmar que:
vazorg [7]

Answer:

C) solo III

Explanation:

Para solucionar este problema debemos analizar cada una de las opciones hasta llegar a la opcion valida.

I) el cuerpo pesa igual que su masa.

Esta opcion no puede ser ya que el peso de un cuerpo se define como el producto de la masa por la aceleracion gravitacion.

w=m*g

donde:

w = peso [N]

m = masa [kg]

g = aceleracion gravitacional = 9.81 [m/s²]

Como podemos ver el peso siempre sera mayar que la masa, ya que el peso es resultado de la multiplicacion de la masa por la gravedad.

II) Por medio de un analisis de fuerzas en el eje-y, la fuerza del peso se dirige hacia abajo mientras que la fuerza normal tiene igual magnitud, pero se dirige hacia arriba. Por esto la segunda opcion no puede ser.

III) El cuerpo se encuentra en equilibrio, es decir las unicas fuerzas que actuan sobre el cuerpo son el peso y la fuerza normal. Pero estas fuerzas son iguales y opuestas en direccion, por la tanto se cancelan y estan en equilibrio.

Esta es la opcion valida, la fuerza neta es nula.

5 0
3 years ago
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