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DochEvi [55]
3 years ago
7

An antenna emits an electromagnetic wave. the electric field lines at a certain instant in time are shown.at point a, what is th

e direction of the magnetic field?
Physics
1 answer:
Dmitry [639]3 years ago
5 0

Answer:

The image is missing, but let's try to answer it.

When we have a plane electromagnetic wave that moves in the K direction, we aways must have that the electric field and the magnetic field are orthogonal.

Where the magnetic field direction can be calculated with the equation:

B = KxE

this is a cross product, to find the direction of B, you can point with your hand in the direction of the first vector, and close your hand in the direction of the second vector, the place at where your thumb is pointing is also the direction of B.

This means that if we have that the wave moves in the X direction, and E is in the Z direction, we will have that B is in the XxZ = Y direction.

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What is the significance of the fact that the gravitational constant, G, is a very small number and that Coulomb’s constant, k,
WITCHER [35]

The comparison indicates that the gravitational force is a much weaker force than the electromagnetic force. This is the significance of the fact that the gravitational constant, G, is a very small number and that Coulomb’s constant, k, is a very large number

Answer: The comparison indicates that the gravitational force is a much weaker force than the electromagnetic force.

<u>Explanation: </u>

We know that the universal formula for the estimation of gravitational force is,

                       F=G \times\left(\frac{m_{1} \times m_{2}}{r^{2}}\right)

Where,

F, Gravitational force, m_{1} \text { and } m_{2}- objects masses r- The relative distance between the two objects  

G, Gravitational Constant 6.67 \times 10^{-11} \frac{N m^{2}}{k g^{2}}

And, the electrostatic force between two scalar charges according to the Coulomb’s law is,

                  F=k \times\left(\frac{q_{1} \times q_{2}}{r^{2}}\right)

F, Electrostatic force between two charges,q_{1} \text { and } q_{2}- Two scalar charges, r - The relative distance between two scalar charges, k- Coulomb’s constant 8.987 \times 10^{9} \frac{N m^{2}}{c^{2}}

Now, on comparing the values of G and k, we can easily evaluate that eventually, the gravitational force will be lesser than the coulomb’s force.  

Besides this, we can also judge this fact through various examples such as, a balloon rubbed with a cloth, easily sticks to the wall for some time. Opposing the gravitational force of the Earth which is not the case with the normal balloon.

It drops without having the electrostatic force between the wall and the balloon.  This shows that the gravitational force draws lesser impact on objects as compared to the electrostatic force.

7 0
3 years ago
Read 2 more answers
If the mass of an object is 176 and Net force is 50, what would be the Acceleration?
pochemuha

Answer:

0.28m/s²

Explanation:

Force = mass•acceleration

F = m•a

50 = 176•a Divide both sides by 176:

50/176 = a ≈ 0.28 m/s²

6 0
3 years ago
Which of the following statements is correct regarding the polarization of a neuronal membrane and the formation of a resting me
natka813 [3]
It seems that you have missed the necessary options for us to answer this question so I had to look for it. Anyway, here is the answer. The statement that is correct regarding the polarization of a neuronal membrane and the formation of a resting membrane potential is this: <span>Sodium/Potassium pumps maintain concentration gradients; sodium and potassium move down their concentration gradients through leakage channels.</span>
3 0
3 years ago
Now determine the precise time that the cars started their trip (hour:minute:second) if Car 1 passed the streetlight at 12:25:00
vlada-n [284]

This question is incomplete, the missing part is in the image below;

Answer:

the precise time that the cars started their trip is 12:00:00

Explanation:

Given that;

from the image; The velocity of car1 V1 = 1/60 mile/seconds

Given the precise time that the car 1 passed the streetlight ( 12:25:00 )

if car 1 passed 25 seconds before car 2

meaning Car 2 passed the the streetlight at 12:50:00

now time delay is the difference between the two times; i.e

12:50:00 - 12:25:00 = 25 sec

so the distance between starting point and streetlight = 25×1 = 25 miles

Now, time required for Car1 to travel 25 miles distance will be;

t = distance/velocity of car1 = 25 / (\frac{1}{60} ) = 1500 sec = 25 min

so the precise time that the cars started their trip will be;

12:25:00 - 25 min = 12:00:00

Therefore, the precise time that the cars started their trip is 12:00:00

6 0
3 years ago
A car traveling 85 km/h slows down at a constant 0.53 m/s^2 just by "letting up on the gas." Calculate the distance it travels d
snow_lady [41]

Answer:

a) d = 23.345 m : distance the car travels in the first second

b) d = 111.425 m : distance the car travels in the fifth second

Explanation:

The distance (d) in uniformly accelerated motion is calculated as follows:

d = v₀*t +(1/2)*a*t² Formula (1)

d: distance (m)

v₀: initial speed (m/s)

a: acceleration (m/s²)

t: time (s)

Equivalences

1 km = 1000m

1 h = 3600 s

Data

v₀ = 85 km/h = 85*1000m/3600s = 23.61 m/s

a = - 0.53 m/s²

Calculation of the distance the car travels in the first second

We replace data in formula (1) at t = 1s

d = 23.61*(1)+ (1/2)*(-0.53)*(1)²

d = 23.345 m

Calculation of the distance the car travels in the fifth second

We replace data in formula (1) at t = 5s

d = 23.61*(5)+ (1/2)*(-0.53)*(5)²

d = 111.425 m

7 0
4 years ago
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