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Aloiza [94]
3 years ago
7

If at a particular instant and at a certain point in space the electric field is in the x-direction and has a magnitude of 3.70

V/m, what is the magnitude of the magnetic field of the wave at this same point in space and instant in time
Physics
1 answer:
Charra [1.4K]3 years ago
6 0

Answer:

the magnitude of the magnetic field is 1.23 x 10⁸ T.

Explanation:

Given;

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

The magnitude of the magnetic field is calculated as;

E = cB

where;

B is the magnitude of the magnetic field

c is the speed of light = 3 x 10⁸ m/s

From the above equation, the magnetic field, B, is calculated as;

B = \frac{E}{c} \\\\B = \frac{3.7 }{3\times 10^8 } \\\\B = 1.23 \times 10^{-8 } \ T

Therefore, the magnitude of the magnetic field is 1.23 x 10⁸ T.

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Please help!!! A river has a constant current of 3 km per hour. If a motorboat, capable of maintaining a constant speed of 20km
PilotLPTM [1.2K]
Construct a vector diagram. It will be a right-angled triangle. One vector (the hypotenuse) represents the heading of the boat, one represents the current and one represents the resultant speed of the boat, which I'll call x. Their magnitudes are 20, 3 and x. Let the required angle = theta. We have: 

<span>theta = arcsin(3/20) = approx. 8.63° </span>

<span>The boat should head against the current in a direction approx. 8.63° to the line connecting the dock with the point opposite, or approx. 81.37° to the shore line. </span>

<span>x = sqrt(20^2 - 3^2) </span>
<span>= sqrt(400 - 9) </span>
<span>= sqrt 391 </span>

<span>The boat's crossing time = </span>
<span>0.5 km/(sqrt 391 km/hr) </span>
<span>= (0.5/sqrt 391) hr </span>
<span>= approx. 0.025 hr </span>
<span>= approx. 91 seconds</span>
4 0
3 years ago
Both formal and informal
Tema [17]

Explanation:

is this a question????...

5 0
3 years ago
If the velocity of an object is zero, then that object cannot be accelerating. | True or False
valkas [14]

Answer: False

Explanation:

3 0
3 years ago
the measure of each exterior angle of a regular pentagon is ___ the measure of each exterior angle of a regular nonagon
Cerrena [4.2K]

Answer:

(a) 72°

(b) 40°

Explanation:

PENTAGON

First, we calculate the total angles in a Pentagon using:

180(n - 2)

Where n = number of sides of the polygon, in this case, 5.

Hence, the total angle in a polygon is

180(5 - 2) = 180 * 3 = 540°

Therefore, each angle will be:

540°/5 = 108°

Because the interior angle and exterior angle form a straight line (180°), the exterior angle of a regular pentagon will be:

180 - 108 = 72°

The exterior angle of a regular Pentagon is 72°

NONAGON

First, we calculate the total angles in a Nonagon using:

180(n - 2)

Where n = number of sides of the polygon, in this case, 9.

Hence, the total angle in a polygon is

180(9 - 2) = 180 * 7 = 1260°

Therefore, each angle will be:

1260°/9 = 140°

Because the interior angle and exterior angle form a straight line (180°), the exterior angle of a regular nonagon will be:

180 - 140 = 40°

The exterior angle of a regular Nonagon is 40°

4 0
3 years ago
1. What are the units of area under the line/curve?<br> 2. Does the area have any meaning?
TiliK225 [7]

The area under the velocity time graph is 125 m and the meaning of the area is displacement.

<h3>What is area under velocity - time graph?</h3>

The area under a velocity time graph represents the displacement of the object.

total area of the graph = A1 + A2

total area of the graph = ¹/₂ (base₁)(height₁) + ¹/₂ (base₂)(height₂)

total area of the graph = ¹/₂(4)(40) + ¹/₂(3)(30)

total area of the graph = 125 m

Thus, the area under the velocity time graph is 125 m and the meaning of the area is displacement.

Learn more about velocity time graph here: brainly.com/question/4710544

#SPJ1

6 0
2 years ago
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