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murzikaleks [220]
3 years ago
9

How do electric and magnetic fields interact in an electromagnetic wave?

Physics
1 answer:
kotegsom [21]3 years ago
8 0

Answer:

Electric and magnetic field waves are oriented at 90 degree angles relative to each other.

Explanation:

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Calculate the kinetic energy of a 0.032 kg ball as it leaves a hand to be thrown upwards at 6.2 m/s
AnnZ [28]

Answer:

The ball will have a kinetic energy of 0.615 Joules.

Explanation:

Use the kinetic energy formula

E_k = \frac{1}{2}mv^2 = \frac{1}{2}0.032kg\cdot 6.2^2 \frac{m^2}{s^2}= 0.615J

The kinetic energy at the moment of leaving the hand will be 0.615 Joules. (From there on, as it ball is traveling upwards, this energy will be gradually traded off with potential energy until the ball's velocity becomes zero at the apex of the flight)

3 0
4 years ago
Conduction circulates heat throughout the atmosphere.<br><br><br> True or False<br><br> Help!!
katrin2010 [14]
Energy is transferred between the ground and the atmosphere via conduction.
Since air is a poor conductor, most energy transfer by conduction<span> occurs right at the earth's surface</span>
7 0
3 years ago
If the magnitude of the resultant force is to be 500n, directed along the positive y axis, determine the magnitude of force f an
scoundrel [369]
The answer is 0=45.2 degrees
5 0
3 years ago
The electric flux through a spherical surface is 1.4 ✕ 105 N · m2/C. What is the net charge (in C) enclosed by the surface?
Anit [1.1K]

Answer:

The  value is   Q_{net} =  1.239 *10^{-6} \  C

Explanation:

From the question we are told that

   The electric flux is \Phi =  1.4*10^{5} \  N\cdot m^2/C

     

Generally the net charge is mathematically represented as

    Q_{net} =  \Phi *  \epsilon_o

Here \epsilon_o is the permetivity of free space with value  

       \epsilon_o =  8.85*10^{-12}  \  \  m^{-3} \cdot kg^{-1}\cdot  s^4 \cdot A^2

So

   Q_{net} =  1.4*10^5 *  8.85*10^{-12}

=>   Q_{net} =  1.239 *10^{-6} \  C

8 0
3 years ago
A hiker begins a trip by first walking 25.0 km southeast from her car. She stops and sets up her tent for the night. On the seco
Sunny_sXe [5.5K]

Answer:

(a)

x_1=17.68km\\y_1=-17.68km\\x_2=20km\\y_2=34.64km

(b) r=41.32km

\alpha =24.23^o

Explanation:

Let us take the north direction to be the positive y-axis and the east to be positive x-axis.

First day:

25.0 km southeast, which implies 45^o south of east. The y-component will be negative and the x-component will be positive.

x_1=25cos45^o=17.68km

y_1=-25sin45^o=-17.68km

Second day:

She starts off at the stopping point of last day. This time, both the y- and x-components are positive.

x_2=40cos60^o=20km

y_2=40sin60^o=34.64km

Therefore, total displacements:

x=x_1+x_2=(17.68+20)km=37.68km

y=y_1+y_2=(-17.68+34.64)km=16.96km

Magnitude of displacements,

r=\sqrt{x^2+y^2}=41.32km

Direction,

\alpha =tan^{-1}(\frac{y}{x})=24.23^o

6 0
4 years ago
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