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3 years ago

Is a mirrors surface transparent translucent or opaque how do you know.

1 answer:

4 0

There are actually two different kinds of mirrors, and the answer is different
for each one.

-- Plain old everyday hand mirror, vanity mirror, bathroom mirror, makeup
mirror, etc.

Opaque, reflecting silver coating is on the back of the glass.
Light from your tongue or your teeth flows to the front surface of the glass,
through the glass, out of the back surface of the glass, bounces off of the silver
coating on the back, reverses its direction, enters the back surface of the glass,
comes back through the glass again, leaves the front of the glass, goes into your
eyes, and you can see your teeth or your tongue.

Both surfaces of the glass, as well as the glass in between the surfaces, are
transparent. The silver coating on the back is opaque. I know that, because
when I look at the back of a mirror, I can't see any light coming through it.
The coating on the back is also reflective ... a big part of the reason why
a mirror works.

-- Expensive mirrors used by astronomers and eye-doctors.
Known as "first surface" mirrors.

Opaque, reflecting silver coating is on the front of the glass.
Light from your tongue or your teeth flows toward the front surface of the glass,
but never actually gets there. It bounces off of the silver coating on the front of
the glass, reverses its direction, goes into your eyes, and you can see your teeth
or your tongue.

The glass is transparent, but that doesn't matter, because the light never reaches
the glass. It only goes as far as the opaque silver coating on the front, and is
reflected from there.

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A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in

koban [17]

Answer:

a) The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) The electric force is $2.0\times 10^{-6}$ newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

$E = \frac{F_{e}}{q}$ (1)

Where:

$E$ - Electric field, measured in newtons per coulomb.

$F_{e}$ - Electric force, measured in newtons.

$q$ - Electric charge, measured in coulombs.

If we know that $F_{e} = 4.0\times 10^{-6}\,N$ and $q = 2.0\times 10^{-8}\,C$ , then the electric field at that point is:

$E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}$

$E = 2.0\times 10^{2}\,\frac{N}{C}$

The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) If we know that $E = 2.0\times 10^{2}\,\frac{N}{C}$ and $q = 1.0\times 10^{-8}\,C$ , then the electric force is:

$F_{e} = E\cdot q$

$F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)$

$F_{e} = 2.0\times 10^{-6}\,N$

The electric force is $2.0\times 10^{-6}$ newtons.

7 0

3 years ago

In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the

Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, $u_0= 10 m/s$ .

Angle of launch with the horizontal, $\theta = 53 ^{\circ}$

So, the vertical component of the initial velocity,

$u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i)$ .

The horizontal component of the initial velocity,

$u_0\cos\theta=10 \cos 53 ^{\circ}$

Let, t be the time of flight, to the horizontal distance covered

$D=10 \cos (53 ^{\circ})t\cdots(ii)$ .

Not, applying the equation of motion in the vertical direction.

$s= ut +\frac 1 2 at^2$

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, $u =10 \sin 53 ^{\circ}$ (from equation (i), s=0 (as the final height is same as the launch height) and $a = -9.81 m/s^2$ (negative sign is due to the downward direction).

$\Rightarrow 0 = 10 (\sin 53 ^{\circ})t-\frac 1 2 (9.81)t^2$

$\Rightarrow t= \frac {2\times 10 (\sin 53 ^{\circ})}{9.81}=1.63$ seconds.

So, the total time in the air is 1.63 seconds.

From equation (i),

Total horizontal distance covered is

$D=10 \cos (53 ^{\circ})\times 1.63 = 9.81 m$ .

Now, for the maximum height, H, applying the equation of motion as

$v^2=u^2+2as$

Here, v is the final velocity and v=0 (at the maximum height), and h=H.

So, $0^2=(10 \sin 53 ^{\circ})^2-2(9.81)H$

$\Rightarrow H = \frac {(10 \sin 53 ^{\circ})^2}{2\times 9.81}$

$\Rightarrow H = 3.25 m$ .

Hence, the maximum height covered is 3.25 m.

8 0

3 years ago

A position vector with magnitude 10 m points to the right and up. its x-component is 6.0 m. part a what is the value of its y-co

lidiya [134]

The position vector can be transcribed as:

A = 6 i + y j

i points in the x-direction and j points in the y-direction.

The magnitude of the vector is its dot product with itself:

|A|2 = A·A

102 = (6 i + y j)•(6 i+ y j) Note that i•j = 0, and i•i = j•j = 1

100 = 36 + y2

64 = y2

get the square root of 64 = 8

The vertical component of the vector is 8 cm.

3 0

3 years ago

Which statement best summarizes the central idea of “Applications of Newton’s Laws”?

Shkiper50 [21]

Newton's three forces, normal, tension and friction, are present in a surprising number of physical situations

Newton's Laws, that describe the relationship between an obejct and the forces acting upon it, apply in almost every physical situation, from quantum mechanics to electricity.

The correct answer is:

Newton’s laws can explain the forces that occur between objects every day

3 0

3 years ago

Read 2 more answers

Block B has mass 6.00 kg and sits at rest on a horizontal, frictionless surface. Block A has mass 2.50 kg and sits at rest on to

zzz [600]

Answer:

Explanation:

Block A sits on block B and force is applied on block A . Block A will experience two forces 1) force P and 2 ) friction force in opposite direction of motion . Block B will experience one force that is force of friction in the direction of motion .

Let force on block A be P . friction force on it will be equal to kinetic friction, that is μ mg , where μ is coefficient of friction and m is mass of block A

friction force = .4 x 2.5 x 9.8

= 9.8 N

net force on block A = P - 9.8

acceleration = ( P - 9.8 ) / 2.5

force on block B = 9.8

acceleration = force / mass

= 9.8 / 6

for common acceleration

( P - 9.8 ) / 2.5 = 9.8 / 6

( P - 9.8 ) / 2.5 = 1.63333

P = 13.88 N .

4 0

4 years ago