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steposvetlana [31]
3 years ago
10

A plane electromagnetic wave travels northward. At one instant, its electric field has a magnitude of 3.7 V/m and points eastwar

d. What are the magnitude (in T) and direction of the magnetic field at this instant
Physics
1 answer:
UNO [17]3 years ago
3 0

Answer:

Magnetic field, B = 1.24\times 10^{-8}\ T (eastwards)

Explanation:

Given that,

The magnitude of electric field, E = 3.7 V/m

A plane electromagnetic wave travels northward. We need to find the magnitude and direction of the magnetic field at this instant.

The relation between the magnetic field and the electric field of the plane electromagnetic wave is given by :

B=\dfrac{E}{c}

c is the speed of light

B=\dfrac{3.7}{3\times10^8}\\\\B=1.24\times 10^{-8}\ T

So, the magnetic field is 1.24\times 10^{-8}\ T. The direction of magnetic field is perpendicular to the electric field and the direction of wave propagation. Hence, the direction of magnetic field is eastward.

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Please help <br>problems 2a.,2b.,3a.,and 3b.​
Darina [25.2K]

Answer:

2a) x = 32 [mil/h]; 2b) t = 0.5[h]; 3a) t = 2.5 [h]; 3b) x = 185[mil]

Explanation:

2a)

We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

v=\frac{x}{t} \\v=velocity [\frac{mil}{h} ] = 32 [\frac{mil}{h}] \\t=time = 1 [h]\\x=v*t\\x=32[\frac{mil}{h} ]*1[h]\\x=32[mil}

2b)

We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

v=\frac{x}{t} \\t=\frac{x}{v} \\t=\frac{420}{840}\\ t=0.5[h]

3a)

We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

v=\frac{x}{t} \\t=\frac{x}{v} \\t=\frac{35}{14}\\ t=2.5[h]

3b)

We can solve this problem by using the kinematics equation, which relates speed to time and displacement.

v=\frac{x}{t} \\v=velocity [\frac{mil}{h} ] = 74 [\frac{mil}{h}] \\t=time = 2.5 [h]\\x=v*t\\x=74[\frac{mil}{h} ]*2.5[h]\\x=185[mil}

8 0
3 years ago
A baseball hit just above the ground leaves the bat 27 m/s at 45° above the horizontal. A) How far away does the ball strike the
Sedbober [7]

Answer:

A) The ball hits the ground 74.45 m far from the hitting position.

B) Maximum height of the ball = 18.57 m

Explanation:

There are two types of motion in this horizontal and vertical motion.

We have velocity = 27 m/s at 45° above the horizontal

Horizontal velocity = 27cos45 = 19.09 m/s

Vertical velocity = 27sin45 = 19.09 m/s

Time to reach maximum height,

           v = u + at

           0 = 19.09 - 9.81 t

            t = 1.95 s

So total time of flight = 2 x 1.95 = 3.90 s

A) So the ball travels at 19.09 m/s for 3.90 seconds.

     Horizontal distance traveled = 19.09 x 3.90 = 74.45 m

     So the ball hits the ground 74.45 m far from the hitting position.

B) We have vertical displacement

              S = ut + 0.5 at²

              H = 19.09 x 1.95 - 0.5 x 9.81 x 1.95² = 18.57 m

    Maximum height of the ball = 18.57 m

6 0
3 years ago
Assume the following values: d1 = 0.880 m , d2 = 1.11 m , d3 = 0.560 m , d4 = 2.08 m , F1 = 510 N , F2 = 306 N , F3 = 501 N , F4
dsp73

Answer:

= 2630.6 N.m

Explanation:

(FR)x = ΣFx = -F4 = -407 N

(FR)y = ΣFy =-F1-F2 -F3 = -510 - 306 - 501 = -1317 N

(MR)B =ΣM + Σ(±Fd)

= MA + F1(d1 +d2) + F2d2 - F4d3

= 1504 + 510(0.880+1.11) +306(1.11) - 407(0.560)

= 2630.64 N.m (counterclockwise)

6 0
3 years ago
A force of 20.0 N is applied to a 3.00 kg object for 4.00 seconds. Calculate the impulse experienced by the object.​
GenaCL600 [577]

Answer:

Impulse = 80Ns

Explanation:

Given the following data;

Mass = 3kg

Force = 20N

Time = 4 seconds

To find the impulse experienced by the object;

Impulse = force * time

Impulse = 20*4

Impulse = 80Ns

Therefore, the impulse experienced by the object is 80 Newton-seconds.

7 0
2 years ago
An airplane flies between two points on the ground that are 500 km apart. The destination is directly north of the origination o
Ann [662]

Answer:

   θ = 4.78º

with respect to the vertical or 4.78 to the east - north

Explanation:

This is a velocity compound exercise since it is a vector quantity.

The plane takes a direction, the air blows to the west and the result must be to the north, let's use the Pythagorean theorem to find the speed

                  v_fly² = v_nort² + v_air²

                  v_nort² = v_fly² + - v_air²

Let's use trigonometry to find the direction of the plane

        sin θ = v_air / v_fly

        θ = sin⁻¹ (v_air / v_fly)

         

let's calculate

        θ = sin⁻¹ (10/120)

         θ = 4.78º

with respect to the vertical or 4.78 to the north-east

5 0
3 years ago
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