Soils with a pH of 6.5 are considered slightly acidic true or false

Consider a monochromatic electromagnetic plane wave propagating in the x direction. At a particular point in space, the magnitud

Scrat [10]

Answer:

a) S = 2.35 10³ J/m²2 ,

b)and the tape recorder must be in the positive Z-axis direction.

the answer is 5

c) the direction of the positive x axis

Explanation:

a) The Poynting vector or intensity of an electromagnetic wave is

S = 1 /μ₀ E x B

if we use that the fields are in phase

B = E / c

we substitute

S = E² /μ₀ c

let's calculate

s = 941 2 / (4π 10⁻⁷ 3 10⁸)

S = 2.35 10³ J/m²2

b) the two fields are perpendicular to each other and in the direction of propagation of the radiation

In this case, the electro field is in the y direction and the wave propagates in the ax direction, so the magnetic cap must be in the y-axis direction, and the tape recorder must be in the positive Z-axis direction.

the answer is 5

C) The poynting electrode has the direction of the electric field, by which or which should be in the direction of the positive x axis

5 0

4 years ago

A ball is projected with an angle o from the top of a tower of height h with velocity vo. The ball strikes the ground after a ce

Lostsunrise [7]

Answer:

dvnuncxtumvd7ojf433f and tv36v54f vs Craig

5 0

3 years ago

Liz puts a 1 kg weight and a 10 kg on identical sleds. She then applies a 10N force to each sled. Which does not explain why the

Mariana [72]

Answer: C

Explanation:
The acceleration does not depend directly on the mass of the object.

Newton's Law is Force = Mass x Acceleration.
Therefore, Acceleration = Force/Mass

The same force is applied in both cases.
Therefore acceleration is inversely proportional to mass.
As mass decreases, acceleration increases.

3 0

4 years ago

Read 2 more answers

A 0.290 cm diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed 30.0 pC charge on it

mojhsa [17]

Answer:

Therefore,

The potential (in V) near its surface is 186.13 Volt.

Explanation:

Given:

Diameter of sphere,

d= 0.29 cm

$radius=\dfrac{d}{2}=\dfrac{0.29}{2}=0.145\ cm$

$r = 0.145\ cm = 0.145\times 10^{-2}\ m$

Charge ,

$Q = 30.0\ pC=30\times 10^{-12}$

To Find:

Electric potential , V = ?

Solution:

Electric Potential at point surface is given as,

$V=\dfrac{1}{4\pi\epsilon_{0}}\times \dfrac{Q}{r}$

Where,

V= Electric potential,

ε0 = permeability free space = 8.85 × 10–12 F/m

Q = Charge

r = Radius

Substituting the values we get

$V=\dfrac{1}{4\times 3.14\times 8.85\times 10^{-12}}\times \dfrac{30\times 10^{-12}}{0.145\times 10^{-2}}$

$V=\dfrac{30}{16.117\times 10^{-2}}=186.13\ Volt$

Therefore,

The potential (in V) near its surface is 186.13 Volt.

3 0

4 years ago

9) If, after viewing a specimen at low power, you switch to high-dry power and, after using fine focus, cannot find the specimen

xeze [42]

Answer:

See the answer below

Explanation:

The best thing one can do in this case would be to return the microscope's objective to low power and then re-center the specimen before switching back to high-dry power.

Most of the time, what makes the specimen under the microscope to be out of focus at higher objective powers after being in focus at low power is because they are not properly centered on the stage. Hence, before calling on the instructor, it would be wise to first return to low power, re-center the specimen and bring it into focus after which the high power objective can be returned to and the fine focus adjusted to bring the image back to focus.

After doing the above and the specimen still does not come into focus, then the instructor can be called upon.

4 0

3 years ago