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3 years ago

Which diagram best illustrates what happens when electromagnetic waves strike a reflective material?

Physics

2 answers:

Ghella [55]3 years ago

8 0

The last diagram below represents what happenes when electromagnetic waves strike a reflective material. The last diagram is showed in the image below.

AysviL [449]3 years ago

6 0

Answer:

The second diagram does that

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Calculate the correct fuse that should be used for a 230V,1KW electric hair dryer.

zzz [600]

Answer: 4 A

Explanation:

Given

Voltage $V=230\ V$

Power $P=1\ kW$

Power is given by $P=VI\\$

Insert the values

$\Rightarrow 1000=230\times I\\\\\Rightarrow I=\dfrac{1000}{230}\\\\\Rightarrow I=3.84\ A$

The rating of fuse is slightly higher than the normal operating conditions. Therefore, a 4 A fuse should be used here.

6 0

3 years ago

Why do some people suffer for nose bleed at higher altitudes?

Novosadov [1.4K]

Answer:because of small amount of oxygen

Explanation:

If you get higher by the atmosphere the air will start to decrease

8 0

3 years ago

4. A meter has a resistance of 100 Ω and gives a full scale deflection when it carries a current of 25 μA. (a) What resistor, Rx

frez [133]

Answer:

A=50mΩ

B≅50mΩ

Explanation:

A) To answer this question we have to use the Current Divider Rule. that rule says:

$Ix=.\frac{Req}{Rx} *Itotal$ (1)

Itotal represents the new maximun current, 50mA, Ix is the current going through the 100 ohms resistor, and Req. is the equivalent resitor.

We now have a set of two resistor in parallel, so:

$Req.=\frac{1}{\frac{1}{R1}+\frac{1}{R2} }$ (2)

where R1 is the resitor we have to calculate, and R2 is the 100 ohms resistor (25 uA).

substituting and rearranging (2)

$Req.=\frac{ 100*R1}{R1+100}$ (3)

Now substituting (3) in (1).

$25*10^{-6} =\frac{\frac{ 100*R1}{R1+100}}{100} *50*10^{-3}$

solving this, The value of R1 is: 50mΩ

This value of R1 will guaranty that the ammeter full reflection willl be at 50mA.

Given that R2 (100ohm) it too much bigger than 50mΩ, the equivalent resistor will tend to 50mΩ

If you substitude this values on (2) Req. will be 49.97 mΩ.

7 0

3 years ago

A block is released from the top of a frictionless incline plane as pictured above. If the total distance travelled by the block

kotegsom [21]

Complete Question

The diagram for this question is showed on the first uploaded image (reference homework solutions )

Answer:

The velocity at the bottom is $v = 11.76 \ m/ s$

Explanation:

From the question we are told that

The total distance traveled is $d = 1.2 \ m$

The mass of the block is $m_b = 0.3 \ kg$

The height of the block from the ground is h = 0.60 m

According the law of energy

$PE = KE$

Where PE is the potential energy which is mathematically represented as

$PE = m * g * h$

substituting values

$PE = 3 * 9.8 * 0.60$

$PE = 17.64 \ J$

KE is the kinetic energy at the bottom which is mathematically represented as

$KE = \frac{1}{2} * m v^2$

$\frac{1}{2} * m* v ^2 = PE$

substituting values

=> $\frac{1}{2} * 3 * v ^2 = 17.64$

=> $v = \sqrt{ \frac{ 17.64}{ 0.5 * 3 } }$

=> $v = 11.76 \ m/ s$

4 0

4 years ago

A box with a mass of 0.6-kg falls from a very tall building. While free-falling it is subjected to a frictional drag force given

rjkz [21]

Answer:

1.03m/s

Explanation:

According to the question

Fdrag = bv²

now Fdrag = weight of the box

that is Fdrag = mg

where m = mass of the box, 0.6 Kg

and g = acceleration due to gravity

thus,

Fdrag = bv² becomes

mg = bv²

making v the subject of formula

$v = \sqrt{\frac{mg}{b} }$

$v = \sqrt{\frac{0.6 * 9.8}{5.5} }$

v = 1.03m/s

8 0

3 years ago

Read 2 more answers