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melamori03 [73]

2 years ago

14

When jeremiah stands in a swimming pool and looks at hid feet, his legs appear to be bent. Which is the term for this phenomenon

?
A. Diffraction
B. Dispersion
C. Reflection
D. Refraction

1 answer:

ValentinkaMS [17]2 years ago

8 0

D: Refraction
Explanation: Refraction occurs when a medium bends the light rays of an object. Like water for instance as an example of a medium.

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Identify the vibrating media in three different<br> types of musical instruments.

mart [117]

Explanation:

Hole. Hole. Different notes can be played on the flute by blocking holes. ...

Drum skin. Drum skin. Hitting the bongo drum makes its tight elastic skin vibrate.

String. String. ...

Sound. hole. ...

Bow. Bow.

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2 years ago

Why does the speed of sound depend on air temperature?

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3 years ago

Read 2 more answers

Suppose you have three identical metal spheres, AA, BB, and CC. Initially sphere AA carries a charge qq and the others are uncha

Sedaia [141]

Complete Question

Suppose you have three identical metal spheres, A, B, and C. Initially sphere A carries a charge q and the others are uncharged. Sphere A is brought in contact with sphere B, and then the two are separated. Spheres CC and BB are then brought in contact and separated. Finally spheres AA and CC are brought in contact and then separated. What is the final charge on the sphere B, in terms of q?

a. 3/8q

b. 1/4q

c. 3/4q

d. q

e. 5/8q

f. 1/3q

g.1/2q

h. 0

Answer:

The correct option is b

Explanation:

From the question we are told that

The charge carried by A is q C

The charge carried by B is 0 C

The charge carried by C is 0 C

When A and B are brought close and then separated the charge carried by A and B is mathematically evaluated as

$\frac{ 0 + q}{2} = \frac{q}{2}$

When C and B are brought close and then separated the charge carried by C and B is mathematically evaluated as

$\frac{0 + \frac{q}{2} }{2} = \frac{q}{4}$

When C and A are brought close and then separated the charge carried by C and A is mathematically evaluated as

$\frac{\frac{q}{4} + \frac{q}{2} }{2} = \frac{3q}{8}$

Looking at these calculation we can see that the charge carried by B is

$\frac{q}{4} C$

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2 years ago

What mineral is most abundant in granite

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3 years ago

A wire of length 6cm makes an angle of 20° with a 3 mT

Crazy boy [7]

Answer:

Approximately $7.3 \times 10^{-3}\; \rm A$ (approximately $7.3\; \rm mA$ ) assuming that the magnetic field and the wire are both horizontal.

Explanation:

Let $\theta$ denote the angle between the wire and the magnetic field.

Let $B$ denote the magnitude of the magnetic field.

Let $l$ denote the length of the wire.

Let $I$ denote the current in this wire.

The magnetic force on the wire would be:

$F = B \cdot l \cdot I \cdot \sin(\theta)$ .

Because of the $\sin(\theta)$ term, the magnetic force on the wire is maximized when the wire is perpendicular to the magnetic field (such that the angle between them is $90^\circ$ .)

In this question:

$\theta = 20^\circ$ (or, equivalently, $(\pi / 9)$ radians, if the calculator is in radian mode.)
$B = 3\; \rm mT = 3 \times 10^{-3}\; \rm T$ .
$l = 6\; \rm cm = 6 \times 10^{-2}\;\rm m$ .
$F = 1.5\times 10^{-4}\; \rm N$ .

Rearrange the equation $F = l \cdot I \cdot \sin(\theta)$ to find an expression for $I$ , the current in this wire.

$\begin{aligned} I &= \frac{F}{l \cdot \sin(\theta)} \\ &= \frac{3\times 10^{-3}\; \rm T}{6 \times 10^{-2}\; \rm m \times \sin \left(20^{\circ}\right)} \\ &\approx 7.3 \times 10^{-3}\; \rm A = 7.3 \; \rm mA\end{aligned}$ .

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2 years ago