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2 years ago

Sometimes you can see a faint reflection in the surface of a shiny plate or cup.Why ?

Physics

1 answer:

bazaltina [42]2 years ago

4 0

Answer:

The degree of reflection whether faint or bright you see on the surface of an object is an indication that light particles had hit the surface. Since light is a wave and as part of its characteristics can get reflected. However, the amount of light reflected by a surface is dependent on the smoothness of the surface which can be shiny or dull, it can also be dependent on the nature of the surface which can be glass, water, and so on. So, from the question, you can see a faint reflection on the surface of a shiny plate or cup because of the smoothness of the surface which reflects the lights that hit it from a particular direction at the same angle.

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Assume a rectangular strip of a material with an electron density of n=5.8x1020 cm-3. The strip is 8 mm wide and 0.8 mm thick an

vampirchik [111]

Answer: I = 111.69 pA

Explanation: The hall effect is all about the fact that when a semiconductor is placed perpendicularly to a magnetic field, a voltage is generated which could be measured at right angle to the current path. This voltage is known as the hall voltage.

The hall voltage of a semiconductor sensor is given below as

V = I×B/qnd

Where V = hall voltage = 1.5mV =1.5/1000=0.0015V

I = current =?,

n= concentration of charge (electron density) = 5.8×10^20cm^-3 = 5.8×10^20/(100)³ = 5.8×10^14 m^-3

q = magnitude of an electronic charge=1.609×10^-19c

B = strength of magnetic field = 5T

d = thickness of sensor = 0.8mm = 0.0008m

By slotting in the parameters, we have that

0.0015 = I × 5/5.8×10^14 × 1.609×10^-19×0.0008

0.0015 = I×5/7.446×10^-8

I = (0.0015 × 7.446×10^-8)/5

I = 111.69*10^(-12)

I = 111.69 pA

3 0

3 years ago

Show the weight of the ladder and draw the missing Frictional force.

madreJ [45]

Since there is no friction between the ladder and the wall, there can be no vertical force component. That's the tricky part ;)

So to find the weight, divide the 100N <em>normal</em> force by earths gravitational acceleration, 9.8m/s^2

$W = \frac{N}{g} = \frac{100N}{9.8m/s^{2}} = \frac{100}{9.8} = 10.2kg$

Then;
Draw an arrow at the base of the ladder pointing towards the wall with a value of 30N, to show the frictional force.

5 0

2 years ago

A 52.5-turn circular coil of radius 5.35 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0

krek1111 [17]

Answer:

5.43 x 10^-3 Nm

Explanation:

N = 52.5, radius, r = 5.35 cm = 0.0535 m, B = 0.455 T, I = 25.3 mA = 0.0253 A

Torque = N I A B Sin theta

Here, theta = 90 degree

Torque = 52.5 x 0.0253 x 3.14 x 0.0535 x 0.0535 x 0.455

Torque = 5.43 x 10^-3 Nm

6 0

3 years ago

What are the strengths and limitations of the doppler and transit methods? What kind of planets are easiest to detect with each

Arturiano [62]

$\huge\mathfrak\red{✔Answer:-}$

Strength: able to detect planets in a wide range of orbits, as long as orbits aren't face on

Limitations: yield only planet's mass and orbital properties

3 0

3 years ago

Help me answer this ASAP

olga_2 [115]

Answer:

The correct answer is Option A.

7 0

3 years ago