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Elis [28]
3 years ago
5

Help with these questions

Physics
1 answer:
erma4kov [3.2K]3 years ago
8 0
<h2>Explanation:</h2><h3>3. </h3>

When light bounces back, it is <em>reflected</em>. (That's why you see your <em>reflection</em> in a mirror.) When light is bent from the path it is taking, it is <em>refracted</em>. The only answer choice that makes correct use of these terms is the third choice:

  • Part of the ray is <em>refracted</em> into ray B; part of the ray is <em>reflected</em> as ray R.

_____

<h3>4.</h3>

The index of refraction is the ratio of the sine of the angle of incidence to the sine of the angle of refraction. Both angles are measured from the normal to the surface. The angle of refraction here is 12.5° less than the angle of incidence, 44°, so is 31.5°. Then the index of refraction of the medium is ...

  n = sin(44°)/sin(31.5°) = 0.69466/0.52250 = 1.3299 ≈ 1.33

  • none of the offered choices is correct. The closest is 1.34.
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Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote
Arlecino [84]

Answer:

If all these three charges are positive with a magnitude of 1.8 \times 10^{-8}\; \rm C each, the electric potential at the midpoint of segment \rm AB would be approximately 8.3 \times 10^{3}\; \rm V.

Explanation:

Convert the unit of the length of each side of this triangle to meters: 10\; \rm cm = 0.10\; \rm m.

Distance between the midpoint of \rm AB and each of the three charges:

  • d({\rm A}) = 0.050\; \rm m.
  • d({\rm B}) = 0.050\; \rm m.
  • d({\rm C}) = \sqrt{3} \times (0.050\; \rm m).

Let k denote Coulomb's constant (k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}.)

Electric potential due to the charge at \rm A: \displaystyle \frac{k\, q}{d({\rm A})}.

Electric potential due to the charge at \rm B: \displaystyle \frac{k\, q}{d({\rm B})}.

Electric potential due to the charge at \rm A: \displaystyle \frac{k\, q}{d({\rm C})}.

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of \rm AB due to all these three charges  would be:

\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}.

4 0
3 years ago
Briefly answer the following questions. a) A clock is mounted on the wall. As you look at it, what is the direction of the angul
Lelu [443]

Answer:

Explanation:

a ) The direction of angular velocity vector of second hand will be along the line going into the plane of dial perpendicular to it.

b )  If the angular acceleration of a rigid body is zero, the angular velocity will remain constant.

c )  If another planet the same size as Earth were put into orbit around the Sun along with Earth  the moment of inertia of the system will increase because the mass of the system increases. Moment of inertia depends upon mass and its distribution around the axis.

d ) Increasing the number of blades on a propeller increases the moment of inertia , because both mass and mass distribution around axis of rotation increases.

e ) It is not possible  that  a body has the same moment of inertia for all possible axes because a body can not remain symmetrical about all axes possible. Sphere has same moment of inertia about all axes passing through its centre.

f ) To maximize the moment of inertia of a flywheel while minimizing its weight, the  shape and distribution of mass should be such that maximum mass of the body may be situated at far end of the body from axis of rotation . So flywheel must have thick outer boundaries and this should be

attached with axis with the help of thin rods .

g ) When the body is rotating at the same place , its translational kinetic   energy is zero but its rotational energy can be increased

at the same place.

3 0
3 years ago
What is the easiest way to increase the magnetic force acting on the rotor in an induction motor?
Schach [20]

Answer:

Explanation:

Magnets are of two major forms namely the permanent magnet and the temporary magnets. Temporary magnets magnetizes and demagnetize easily while permanent magnets does not magnetizes and demagnetize easily.

This permanents magnets are applicable in loudspeakers, generators, induction motor etc.

To increase the

The following will tend to increase the magnetic force acting on the rotor in an induction motor.

1. Increasing the strength of the bar magnet. Increase in strength of the magnet will lead to increase in the magnetic force acting on the rotor.

2. Increase in the magnetic line of force also known as the magnetic flux around the magnet will also increase the magnetic force acting on the rotor.

6 0
3 years ago
Two circular coils are concentric and lie in the same plane. The inner coil contains 170 turns of wire, has a radius of 0.0095 m
PIT_PIT [208]

Answer:

Current in outer circle will be 15.826 A

Explanation:

We have given number of turns in inner coil N_I=170

Radius of inner circle r_i=0.0095m

Current in the inner circle I_i=8.9A

Number of turns in outer circle N_o=150

Radius of outer circle r_o=0.015m

We have to find the current in outer circle so that net magnetic field will zero

For net magnetic field current must be in opposite direction as in inner circle

We know that magnetic field is given due to circular coil is given  by

B=\frac{N\mu _0I}{2r}

For net magnetic field zero

\frac{N_I\mu _0I_I}{2r_I}=\frac{N_O\mu _0I_0}{2r_O}

So \frac{170\times \mu _0\times 8.9}{2\times 0.0095}=\frac{150\times \mu _0I_O}{2\times 0.015}

I_O=15.92A

4 0
3 years ago
How will the solubility of a gas solute change if the pressure above the solution is reduced?
Hoochie [10]
If the pressure above a solution containing a gas solute is reduced, the limit of the gas's solubility will decrease.
6 0
3 years ago
Read 2 more answers
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