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2 years ago

an electromagnetic wave propagates in a vacuum in the x-direction. In what direction does the electric field oscilate

Physics

1 answer:

sweet [91]2 years ago

5 0

Answer:

<em>The electric field can either oscillates in the z-direction, or the y-direction, but must oscillate in a direction perpendicular to the direction of propagation, and the direction of oscillation of the magnetic field.</em>

Explanation:

Electromagnetic waves are waves that have an oscillating magnetic and electric field, that oscillates perpendicularly to one another. Electromagnetic waves are propagated in a direction perpendicular to both the electric and the magnetic field. If the wave is propagated in the x-direction, then the electric field can either oscillate in the y-direction, or the z-direction but must oscillate perpendicularly to both the the direction of oscillation of the magnetic field, and the direction of propagation of the wave.

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Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.

Artist 52 [7]

Answer:

a) C.M $=(\bar x, \bar y)=(0.767,0.7)m$

b) $(x_4,y_4)=(-1.917,-1.75)m$

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

$\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}$

$\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}$

Where M represent the sum of all the masses on the system.

And the center of mass C.M $=(\bar x, \bar y)$

Part a

$m_1= 3 kg, m_2=5kg,m_3=7kg$ represent the masses.

$(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5)$ represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m$

C.M $=(\bar x, \bar y)=(0.767,0.7)m$

Part b

For this case we have an additional mass $m_4=6kg$ and we know that the resulting new center of mass it at the origin C.M $=(\bar x, \bar y)=(0,0)m$ and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m$

If we solve for a we got:

$(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0$

$a=-\frac{(5kg*2.3m)}{6kg}=-1.917m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m$

$(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0$

And solving for b we got:

$b=-\frac{(7kg*1.5m)}{6kg}=-1.75m$

So the coordinates for this new particle are:

$(x_4,y_4)=(-1.917,-1.75)m$

5 0

2 years ago

Determine the value of the information in scientific notation. The moon is approximately 3. 63 × 108 meters from Earth. What is

Usimov [2.4K]

Answer:

Explanation:

7 0

1 year ago

Planet X is in a stable circular orbit around a star, as shown in the figure. Which of the following graphs best predicts the an

damaskus [11]

Answer:

Explanation:

Angular momentum is the product of moment of inertia and angular velocity.

L = I × ω

Since the planet follows a stable circular orbit, I and ω are constant and non-zero. Therefore, the angular momentum is constant and non-zero.

5 0

2 years ago

Could you help me with this question?

Scorpion4ik [409]

Answer:

Magnitude of static friction force is 70 sin40° = 44.99 N.

No, it is not necessary that it is maximum static friction.

Normal force is equal to 70 cos40° = 53.62 N.

Explanation:

We apply newton law of moton equation along the plane and perpendicular to plane;

Along the plane,

70 sin 40° = $f_{r}$ ---------------(1)

70 cos 40° = N --------------(2)

$f_{r}_{max}$ = μN -----------------(3)

So, it depends on the value of μ that the friction is maximum or not .

7 0

2 years ago

a 300-kg crate is placed on a adjustable incline plane. as one end of the incline is raised, the crate begins to move downward.

denis-greek [22]

Answer:

A 300-kg crate is placed on a adjustable incline plane. as one end of the incline is raised, the crate begins to move downward. if the crate slides down the plane with the acceleration of .70 m/s2 , when the incline is 25 degrees, what is the coefficient of kinetic friction between ramp and crate?

Explanation:

6 0

2 years ago