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2 years ago

13

- Which of the following does not move as a transverse wave?

1 answer:

Sedaia [141]2 years ago

6 0

Answer:

- Which of the following does not move as a transverse wave?

<em>B. sound waves</em>

because sound waves are longitudinal waves having compressions and rare factions.

Which of these effects describes the change in pitch we hear

moving motorbike goes past?

<em>D</em><em>.</em><em> </em><em>Doppler</em><em> </em><em>eff</em><em>ect</em>

Doppler effect is the phenomenon where there is apparent change in frequency (pitch) and wavelength of a wave due to relative motion of the sound source.

Which of the following does not make use of total internal reflection.

<em>B</em><em>.</em><em> </em><em>Endos</em><em>cop</em><em>e</em>

Endoscope doesn't use total internal reflection since no refraction takes place.

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For protection hermit crabs use gastropod shells to hide their bodies so only their heads are visible. Students research why the

KatRina [158]

Answer:

Due to suffocation and toxic paint.

Explanation:

The population of hermit crabs has been decreasing because of suffocation and poison. Many crabs actually die slowly from suffocation because their gills require high humidity in order to breathe properly. Hermit crabs are slowly poisoned by tap water in which the toxic substances present in the paint that dissolves their shells that leads to the death of hermit crabs.

3 0

2 years ago

Two copper rods are separated by a small gap at B. Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the f

gulaghasi [49]

Answer:

A)3.8196 * $10^{9}$ Newtons B) 2.153 * $10^{9}$ Newtons

Explanation:

Force = Pressure * Area.

Pressure is given as 120GPa

Area = Cross Sectional Area of Rod = Area Of A circle = π $R^{2}$ .

Area Expansivity β = $\frac{Change in Area}{Original Area * Temperature Rise}$

Area Expansivity β = 2α

α = 16.9x10-6/ ° C.

Radius of Rod AB = Half 0f Diameter, 200mm = $\frac{0.2m}{2}$ = 0.1 m

Radius of Rod BC = Half Of Diameter, 150mm = $\frac{0.15}{2}$ = 0.075 m.

Note: To convert from millimeter to meter you divide by 1000.

Area of Rod AB = π * $0.1^{2}$ = 0.0314 $m^{2}$

Area of Rod BC = π * $0.075^{2}$ = 0.0177 $m^{2}$ .

Change in Area = 2α * Original Area * Temperature Rise. Therefore for

Rod AB = 2 * 16.9* $10^{-6}$ * 0.0314 * 30 = 3.183 * $10^{-5}$ $m^{2}$ .

For Rod BC we have = 2 * 16.9* $10^{-6}$ * 0.0177 * 30 = 1.794∈-5 $m^{2}$ .

The forces on each rods will be given by.

Force AB = Pressure * Area of Rod AB

= 120GPa * 3.183 * $10^{-5}$ gives 3.8196 * $10^{9}$ Newtons.

Force BC = Pressure * Area of Rod BC

= 120GPa * 1.794 * $10^{-5}$ gives 2.153 * $10^{9}$ Newtons

7 0

2 years ago

Consider a double-slit with a distance between the slits of 0.04 mm and slit width of 0.01 mm. Suppose the screen is a distance

scZoUnD [109]

Answer:

The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

Explanation:

Given that,

Distance between the slits = 0.04 mm

Width = 0.01 mm

Distance between the slits and screen = 1 m

Wavelength = 600 nm

We need to calculate the distance between the places where the intensity is zero due to the double slit effect

For constructive fringe

First minima from center

$x_{1}=\dfrac{\lambda D}{2d}$

Second minima from center

$x_{2}=\dfrac{3\lambda D}{2d}$

The distance between the places where the intensity is zero due to the double slit effect

$\Delta x_{d}=x_{2}-x_{1}$

$\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}$

$\Delta x_{d}=\dfrac{\lambda D}{d}$

Put the value into the formula

$\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}$

$\Delta x_{d}=0.015 =15\times10^{-3}\ m$

$\Delta x_{d}=15\ mm$

Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

8 0

3 years ago

How did tycho brahe's model of the universe differ from the greek geocentric model?

SCORPION-xisa [38]

Idkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

4 0

3 years ago

As viewed from above in this picture, what direction will the current be in the coil of wire that will cause the loop to rotate

Gala2k [10]

Answer:

When viewed from above, the current in the coil should point towards the top-right corner of the picture.

Explanation:

The current in this coil have only two possible directions: clockwise or counter-clockwise. However, since the diagram shows the coil from above, not from a cross-section, just saying clockwise or counter-clockwise might be ambiguous. The statement that the current is directed towards the top-right corner of the picture is equivalent to saying that when viewed from the lower-right corner of this diagram, the current in the coil is moving clockwise.

Note that at the center of this picture, the current is parallel to the magnetic field- there will be no force on the coil at that position. On the other hand, (also when viewed from above,) at the top-right corner and the lower-left corner of the coil, the current in the coil will be perpendicular to the magnetic field. That's where the force on the coil will be the strongest.

With that in mind, apply the right-hand rule to find the direction of the force on the coil in each of the two possibilities.

Assume that when viewed from above, the current is flowing towards the top-right corner of the picture. Consider the wire near the top-right corner of this coil (as viewed above on this picture.) The current will be going into the picture into the magnetic field. By the right-hand rule, the current on the wire near that point should be pointing towards the bottom of this picture. (Point fingers on the right hand in the direction of the current $I$ . Rotate the right hand such that when curling the fingers, they point in the direction of the magnetic field $B$ . The direction of the right thumb should now point in the direction of the force on the wire $F$ .)

Based on the same assumption, the current in the wires near the bottom left corner of this coil will be pointing out of the picture. By the right hand rule, the magnetic force on the coil in that region should be pointing towards the top of this picture. Combing these two forces, the coil would indeed be rotating around the center of this picture in the direction shown in the diagram.

It can also be shown that if the current points towards the bottom left corner of the picture when viewed from above, the coil will be rotating about the center of this picture in the opposite direction.

7 0

3 years ago