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2 years ago

Which statement best describes wavelength?

Physics

2 answers:

xxMikexx [17]2 years ago

5 0

Answer:

Explanation:

Wave length is the height perpendicular verically of a wave

Zarrin [17]2 years ago

5 0

Answer:

Explanation:

makes the most sense to me

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On Mars, the acceleration due to gravity is 3.77 m / s 2 . 3.77 m/s2. How far would a 15 g 15 g rock fall from rest in 2.5 s 2.5

Zinaida [17]

Answer:

s = 11.78 m

Explanation:

given,

acceleration due to gravity, g = 3.77 m/s²

mass of the rock = 15 g

time = 2.5 s

distance traveled = ?

using equation of motion

$s = ut +\dfrac{1}{2}at^2$

initial speed = 0 m/s

$s = \dfrac{1}{2}at^2$

$s = \dfrac{1}{2}\times 3.77 \times 2.5^2$

s = 11.78 m

distance traveled by the rock is equal to 11.78 m.

7 0

3 years ago

Read 2 more answers

Apply the general results obtained in the full analysis of motion under the influence of a constant force in Section 2.5 to answ

zvonat [6]

Answer:

y(i) = h

v(y.i) = 0

Explanation:

See attachment for elaboration

3 0

2 years ago

An electromagnetic wave is propagating in the positive x direction. at a given moment in time, the magnetic field at the origin

antiseptic1488 [7]

Since in an electromagnetic wave the electric and magnetic fields are perpendicular to each other and perpendicular to the direction of motion, the electric field has to point in the z direction.

6 0

3 years ago

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Find electric field at point p which is a distance l away from the both +q and -q

denis-greek [22]

Answer:

$\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

$=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }$

Where

q is the charge
r is the distance
$E$ is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }$

F2 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

⇒

F1+F2= $\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

8 0

3 years ago

What is the forumla for chlorine? anion

prohojiy [21]

Chlorine gas or just chlorine?

6 0

2 years ago

Read 2 more answers