PHYSICS HIGH SCHOOL

At a given location in space, the magnetic field in an electromagnetic wave is increasing. How is the electric field changing at that same location? a. The electric field is increasing.
b. The electric field is decreasing.
c. The electric field is not changing.

Answers

Answer 1
Answer:

Answer:

Option a) is correct

Explanation:

In the given question option a) ( i.e the electric field is increasing ) is correct.

The above option is correct because the electric field and the electromagnetic wave both are in the same phase of the EM wave. Thus, the electromagnetic wave and the electric field are directly proportional to each other.

Answer 2
Answer:

Final answer:

In an electromagnetic wave, the electric and magnetic fields are linked, oscillating in phase. Therefore, when the magnetic field is increasing, the electric field is also increasing.

Explanation:

In an electromagnetic wave, the electric field and the magnetic field are perpendicular to each other and they oscillate in phase. This means when the magnetic field is increasing, the electric field is also increasing, and vice versa. So, the correct answer is option a. The electric field is increasing. It's important to keep this correlation in mind while studying electromagnetic waves, as these oscillations and their interplay are fundamental to how these waves propagate through space.

Learn more about Electromagnetic Wave here:

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Answers

Atoms form two types of bonds: ionic bond and covalent bond. In a covalent bond, valence electrons are shared; in an ionic bond, valence electrons are transferred. The answer is letter A.

Answer:

Atoms form two types of bonds: ionic bond and covalent bond. In a covalent bond, valence electrons are shared; in an ionic bond, valence electrons are transferred.

Explanation:

Therefore, the answer is A

A small piece of Styrofoam packing material is dropped from a height of 1.90 m above the ground. Until it reaches terminal speed, the magnitude of its acceleration is given by a = g − Bv. After falling 0.400 m, the Styrofoam effectively reaches terminal speed, and then takes 5.40 s more to reach the ground.(a) What is the value of the constant B?s-1(b) What is the acceleration at t = 0?m/s2 (down)(c) What is the acceleration when the speed is 0.150 m/s?m/s2 (down)

Answers

Answer:

Explanation:

If a small piece of Styrofoam packing material is dropped from a height of 1.90 m above the ground and reaches a terminal speed after falling 0.400m, the Change in distance will be 1.90m - 0.400 = 1.50m

If it takes 5.4secs fo r the Styrofoam to reach the ground, the terminal velocity will be expressed as;

Vt = change in distance/time

Vt = 1.5m/5.4s

Vt = 0.28m/s

Note that the Styrofoam reaches its final velocity when the acceleration is zero.

To get the constant value B from the equation a = g-Bv

a = 0m/s²

g = 9.81m/s²

v = 0.28m/s

Substituting the parameters into the formula.

0 = 9.81-0.28B

-9.81 = -0.28B

Divide both sides by -0.28

B = -9.81/-0.28

B = 35.04

b) at t = 0sec, the initial terminal velocity is also zero.

Substituting v = 0 into the equation to get the acceleration.

a = g-Bv

a = g-B(0)

a = g

Hence the acceleration at t =0s is equal to the acceleration due to gravity which is 9.81m/s²

c) Given speed v = 0.150m/s

g = 9.81m/s²

B = 35.04

Substituting the given data into the equation a = g-Bv

a = 9.81-35.04(0.15)

a = 9.81 - 5.26

a = 4.55m/s²

Answer:

a

    v  =  0.28 \  m/s

b

    a =  9.8  m/s^2

c

a = 4.55 \ m/s^2

Explanation:

From the question we are told that

The fall height is h = 1.90 \ m

The magnitude of the acceleration is a = g -Bv

The height at which terminal speed is attained is h_t = 0.400 \ m

The time taken to reach the ground from the terminal velocity height is t = 5.40 s

Generally the height which object traveled with terminal velocity is

H = h- h_t

=> H = 1.90 - 0.400

=> H = 1.50 \ m

Generally the terminal velocity is mathematically represented as

v = (H)/(t)

=>        v  =  (1.50)/(5.40)

=>       v  =  0.28 \  m/s

Generally at terminal velocity , acceleration is zero  so

       g -Bv = 0

substituting

       9.8 \ m/s^2

=>    9.8 -B (0.28) = 0

=> v = 35 \  m/s

Generally at t =  0 s the velocity v = 0  (That no motion at time zero)

So  from acceleration equation

    a =  9.8  -35(0)

=>   a =  9.8  m/s^2

Generally when the speed v  =  0.150 m/s

The acceleration is

      a =  9.8  -35(0.150)

=>    a =  4.55 \  m/s^2

   

   

Which phrase is an example of kinetic energy? Responses a car sitting motionless a car sitting motionless a girl running in a race a girl running in a race a sled sitting at the top of a hill a sled sitting at the top of a hill a tank containing gasoline

Answers