When you see yourself in a plane mirror, the image is always:

two bumper cars at an amusement park collide together abd get stuck together. assuming that the system of the two bumper cars is

Fiesta28 [93]

The answer to this problem is M1v1-m2v2=(m1+m2)v

7 0

3 years ago

Light incident on a glass sheet is partly reflected and partly refracted. How is the reflected ray different from the refracted

kkurt [141]

Answer:

E. The refracted ray is vertically polarized whereas the reflected ray is horizontally polarized.

Explanation:

#PLATOLIVESMATTER

7 0

3 years ago

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

Use the law of universal gravitacion to predict what would happen to Earth if the moon were replaced with an object that had the

Zina [86]

Verrrrry interesting !
If the moon were replaced by something with a vastly greater mass
but at the same distance, then ...

-- The period of its revolution around the Earth would be much shorter.
That is, it would orbit the Earth in much less than 27.3 days. We might
see it go through a complete set of phases in 2 weeks, or even 1 week.

-- The ocean tides would be much greater. Low tides would be
much lower, and high tides would be much higher.

-- Sadly, the land tides, and the forces on the Earth's internal structure,
would also be much greater. That means great increases in earthquake
and volcanic activity.

-- The Earth and moon both revolve around their common center of
mass. Under the current arrangement ... with the Earth having 80 times
the mass of the Moon ... that point is inside the Earth, and it looks a lot
like the Moon is orbiting a stationary Earth.
When the new body arrives to replace the lightweight Moon, that point
will be a lot closer to the new companion ... maybe even inside it.
Then, it will look a lot like the monster is the stationary one, and the
Earth is orbiting it.
I actually don't believe that we would SEE that change, or feel it.

8 0

4 years ago

A 10 m long steel beam is accidentally dropped by a construction crane from a height of 4.89 m. The horizontal component of the

Otrada [13]

Answer:

e = 1.21 mV

Explanation:

given,

length of rod = 10 m

height of drop = 4.89 m

Earth’s magnetic field = 12.4 µT

acceleration of gravity = 9.8 m/s²

velocity of the beam

$v = \sqrt{2gh}$

$v = \sqrt{2\times 9.8 \times 4.89}$

v = 9.79 m/s

emf of the beam

e = B l v

e = 12.4 x 10⁻⁶ x 9.79 x 10

e = 1.21 x 10⁻³ V

e = 1.21 mV

4 0

3 years ago