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Gre4nikov [31]
3 years ago
11

An excited hydrogen atom releases an electromagnetic wave to return to its normal state. You use your futuristic dual electric/m

agnetic field tester on the electromagnetic wave to find the directions of the electric field and magnetic field. Your device tells you that the electric field is pointing in the positive y direction and the magnetic field is pointing in the negative x direction. In which direction does the released electromagnetic wave travel?
Physics
1 answer:
Norma-Jean [14]3 years ago
6 0

Answer:

Explanation:

Direction of velocity of electromagnetic wave is given by the formula

E = E j ( vector form )

B = - Bi ( vector form )

Direction of velocity = direction of vector E X B

= E X B

= E j x -Bi

= - EB -k

v = EBk

So the direction of velocity will be along z direction.

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Suzie skydiver, who weighs 500 n, reaches terminal velocity of 90 km/h. the air resistance on suzie is then
masha68 [24]
<span>Weight of the skydiver m = 500 N
 Terminal velocity V = 90 km/h
 Here the weight of the person acts as the force, so based on the Newton's third law the applied is the force what we but in the opposite direction making the resistance. So the air resistance exerted on Suzie will be her weight that is 500N</span>
7 0
3 years ago
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A car with a mass of 600 kg is traveling at a velocity of 10 m/s. How much kinetic energy does it have?
Viefleur [7K]

Answer:

KE = 30,000 J

Explanation:

KE = \frac{1}{2} mv²

KE = \frac{1}{2} (600)(10)²

KE = \frac{1}{2} (600)(100)

KE = \frac{1}{2} (60000)

KE = 30,000 J

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Is it possible to apply the same amount of force and do different amounts of work?
Tamiku [17]
Yes, with simple machines
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A pilot drops a bomb from a plane flying horizontally with constant velocity. When the bomb hits the ground, the horizontal loca
polet [3.4K]

Answer: The horizontal location of the plane will BE OVER THE BOMB

Explanation:

As soon as the bomb was dropped, the bomb will fall under gravity (free fall) and the location of the plane continues to increase horizontally till the bomb reaches the ground which is a falling distance to be travelled by the bomb at 9.8m/s²

8 0
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Two identical small metal spheres with q1 &gt; 0 and |q1| &gt; |q2| attract each other with a force of magnitude 72.1 mN when se
Brrunno [24]

1) +2.19\mu C

The electrostatic force between two charges is given by

F=k\frac{q_1 q_2}{r^2} (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

\frac{Q}{2}

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

F=k\frac{(\frac{Q}{2})^2}{r^2}

And since we know that

r = 1.41 m (distance between the spheres)

F= 21.63 mN = 0.02163 N

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C

So, the final charge on the sphere on the right is

\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C

2) q_1 = +6.70 \mu C

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

F = -72.1 mN = -0.0721 N (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

q_2 = Q-q_1

So we can rewrite eq.(1) as

F=k \frac{q_1 (Q-q_1)}{r^2}

Solving for q1,

Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0

Since Q=4.37\cdot 10^{-6} C, we can substituting all numbers into the equation:

8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0

which gives two solutions:

q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

q_1 = +6.70 \mu C

8 0
3 years ago
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