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AlladinOne [14]

2 years ago

11

Suppose that an electromagnetic wave is traveling toward the east. At one instant at a given point, the electric field vector po

ints upward. What is the direction of the magnetic field at this same given point and instant in time? Is it:
1.) downward
2.) east
3.) south
4.)north
5.)west
6.)upward

1 answer:

Mrrafil [7]2 years ago

4 0

Answer: east

Explanation: A propagating electromagnetic wave in space has it electric and magnetic field perpendicular to each other.

For this question of ours, the wave is moving in the eastward direction with the electric field upward (north), the magnetic field must be perpendicular at all time to the electric field in the direction of the motion of the electromagnetic wave.

Hence, the magnetic field is due east

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Does the collision between the cart and the brick follow the law of momentum conservation? Make a claim (yes or no) and support

murzikaleks [220]

Answer:

yes it does

Explanation:

Going by the law of conservation of momentum, when two objects collides, the momentum of one object is transferred to the other object.

This collision can be elastic or inelastic coliision

let the mass of the brick be m1

the mass of the cart be m2

the velocity (initial and final) of the brick u1 and v1

the velocity (initial and final) of the cart u2 and v2

For elastic collision

m1u1+m2u2= m1v1+m2v2

For inelastic collision the two bodies will move with the same velocity after impact

m1u1+m2u2=(m1+m2)V

7 0

3 years ago

Transverse waves travel with a speed of 20 m/s on a string under a tension 0f 6.00 N. What tension is required for a wave speed

Aleksandr [31]

Answer:

$T_2=13.5\ N$

Explanation:

Given that,

Speed of transverse wave, v₁ = 20 m/s

Tension in the string, T₁ = 6 N

Let T₂ is the tension required for a wave speed of 30 m/s on the same string. The speed of a transverse wave in a string is given by :

$v=\sqrt{\dfrac{T}{\mu}}$ ........(1)

T is the tension in the string

$\mu$ is mass per unit length

It is clear from equation (1) that :

$v\propto\sqrt{T}$

$\dfrac{v_1}{v_2}=\sqrt{\dfrac{T_1}{T_2}}$

$T_2=T_1\times (\dfrac{v_2}{v_1})^2$

$T_2=6\times (\dfrac{30}{20})^2$

$T_2=13.5\ N$

So, the tension of 13.5 N is required for a wave speed of 30 m/s. Hence, this is the required solution.

5 0

3 years ago

In a series circuit with three bulbs,

elena-14-01-66 [18.8K]

In a series circuit, there is only one path for current to take.
If more bulbs are added, then the same current loses more energy,
making heat and light on its way through more bulbs, so the ones that
were there before become dimmer.

8 0

2 years ago

Car A rear ends Car B, which has twice the mass of A, on an icy road at a speed low enough so that the collision is essentially

mixer [17]

Answer:

D. The momentum of Car B is three times as great in magnitude as that of car A.

Explanation:

I majored in Physics

3 0

2 years ago

Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant

irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

$a_g = \frac{GM}{R^2}$

Here

$M = \text{Mass inside the Orbit of the star}$

$R = \text{Orbital radius}$

$G = \text{Universal Gravitational Constant}$

Mass inside the orbit in terms of Volume and Density is

$M =V \rho$

Where,

V = Volume

$\rho =$ Density

Now considering the volume of the star as a Sphere we have

$V = \frac{4}{3} \pi R^3$

Replacing at the previous equation we have,

$M = (\frac{4}{3}\pi R^3)\rho$

Now replacing the mass at the gravitational acceleration formula we have that

$a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho$

$a_g = \frac{4}{3} G\pi R\rho$

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration. So centripetal acceleration of the star is

$a_c = \frac{4}{3} G\pi R\rho$

At the same time the general expression for the centripetal acceleration is

$a_c = \frac{\Theta^2}{R}$

Where $\Theta$ is the orbital velocity

Using this expression in the left hand side of the equation we have that

$\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2$

$\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}$

$\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R$

Considering the constant values we have that

$\Theta = \text{Constant} \times R$

$\Theta \propto R$

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

6 0

3 years ago