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3 years ago

The bending of light as it passes from one medium to the next is

1 answer:

5 0

This is called refraction. Some common examples of refraction are a straw appearing bent or cut once it enters water in a glass.

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A toy airplane, flying in a horizontal, circular

Goshia [24]

Answer:

8.4 m/s

Explanation:

The toy completes 10 circle in 30 seconds. So its frequency of revolution is

$f=\frac{10}{30 s}=0.33 Hz$

The periof of revolution is the reciprocal of the frequency, so

$T=\frac{1}{f}=\frac{1}{0.33 Hz}=3 s$

The radius of the circular path is

r = 4.0 m

So the total distance covered by the toy in one circle is the length of the circumference:

$2\pi r$

And so the average speed is

$v=\frac{2\pi r}{T}=\frac{2\pi (4.0 m)}{3 s}=8.4 m/s$

4 0

3 years ago

Read 2 more answers

2 kg

pochemuha

Answer:

2500 and kg its is very easy do with method

8 0

3 years ago

a specimen of oil having an initial volume of 5000cm³ is subjected to a pressure of 10⁴N/m² and the volume decreases by 0.20cm³.

inessss [21]

Answer:

B = 2.5 10⁸ Pa

Explanation:

The volume modulus is defined by

B = $- \frac{P}{ \frac{\Delta V}{V} }$

The negative fate is for the module to be positive since the volume change is negative

It is not necessary to reduce the volumes to the SI system, since they are both in the same units

B = $- \frac{10^4}{ \frac{-0.20}{5000} }$ = $\frac{10^4}{4 \ 10^{-5} }$

B = 2.5 10⁸ Pa

4 0

3 years ago

A 0.10 g honeybee acquires a charge of +23 pC while flying.

kari74 [83]

Answer:

a) $\frac{F}{w} =2.347\times 10^{-6}\ N$

b) $E=4.2609\times 10^7\ N.C^{-1}$ parallel to the earth surface.

In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

Explanation:

Given:

mass of the bee, $m=10^{-4}\ kg$

charge acquired by the bee, $q_2=23\times 10^{-12}\ C$

Electrical field near the earth surface, $E=100\ N.C^{-1}$

Now the electric force on the bee:

we know:

$F=\frac{1}{4\pi.\epsilon_0} \times \frac{q_1.q_2}{r^2}$

$F=E.q_2$

$F=100\times 23\times 10^{-12}$

$F=23\times 10^{-10}\ N$

The weight of the bee:

$w=m.g$

$w=10^{-4}\times 9.8$

$w=9.8\times10^{-4}\ N$

Therefore the ratio :

$\frac{F}{w} =\frac{23\times 10^{-10}}{9.8\times10^{-4}}$

$\frac{F}{w} =2.347\times 10^{-6}\ N$

The condition for the bee to hang is its weight must get balanced by the electric force acing equally in the opposite direction.

So,

$F=9.8\times10^{-4}\ N$

$E.q_2=9.8\times10^{-4}\ N$

$E\times 23\times 10^{-12}=9.8\times10^{-4}\ N$

$E=4.2609\times 10^7\ N.C^{-1}$ parallel to the earth surface.

In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.

3 0

3 years ago

A student examines a 20-meter long rectangular stream channel and takes the following measurements: width of stream = 4 meters,

Scrat [10]

Answer:

The discharge of the stream at this location is 40 cubic meters per second.

Explanation:

The discharge is the volume flow rate of the water in the stream. For this purpose we can use the following formula:

Discharge = Volume Flow Rate = (Cross-Sectional Area)(Velocity of Stream)

Volume Flow Rate = (Width of Stream)(Depth of Stream)(Velocity of Stream)

Volume Flow Rate = (4 meters)(2 meters)(5 meters per second)

<u>Volume Flow Rate = 40 cubic meters per second</u>

Therefore, the discharge of the stream at this location is found to be <u>40 cubic meters per second</u>

This result shows that 40 cubic meters volume of water passes or discharges through this point in a time of one second. Hence, this is called the volume flow rate or the discharge of the stream.

3 0

3 years ago