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coldgirl [10]
3 years ago
5

What factor affects sensation

Physics
1 answer:
Gnesinka [82]3 years ago
8 0

Answer:

D physiological condition

Explanation:

Sensation and perceptions are complimentary to each other but have different roles within the brain. Sensations are the process of experiencing the world with the five senses and sending that information to the brain. Perceptions are the way we interpret sensations.

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The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, what should be the focal length an
yuradex [85]

Answer:

The focal length of the appropriate corrective lens is 35.71 cm.

The power of the appropriate corrective lens is 0.028 D.

Explanation:

The expression for the lens formula is as follows;

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}

Here, f is the focal length, u is the object distance and v is the image distance.

It is given in the problem that the given lens is corrective lens. Then, it will form an upright and virtual image at the near point of person's eye. The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, the corrective lens is used.

Put v= -71.4 cm and u= 24.0 cm in the above expression.

\frac{1}{f}=\frac{1}{24}+\frac{1}{-71.4}

\frac{1}{f}=0.028

f= 35.71 cm

Therefore, the focal length of the corrective lens is 35.71 cm.

The expression for the power of the lens is as follows;

p=\frac{1}{f}

Here, p is the power of the lens.

Put f= 35.71 cm.

p=\frac{1}{35.71}

p=0.028 D

Therefore, the power of the corrective lens is 0.028 D.

3 0
2 years ago
A sample of a gas has a volume of 639 cm3 when the pressure is 75.9 kPa. What is the volume of the gas when the pressure is incr
const2013 [10]

Answer:

388 cm^3

Explanation:

For this problem, we can use Boyle's law, which states that for a gas at constant temperature, the product between pressure and volume remains constant:

pV=const.

which can also be rewritten as

p_1 V_1 = p_2 V_2

In our case, we have:

p_1 = 75.9 kPa is the initial pressure

V_1 = 639 cm^3 is the initial volume

p_2 = 125 kPa is the final pressure

Solving for V2, we find the final volume:

v_2 = \frac{p_1 V_1}{p_2}=\frac{(75.9)(639)}{125}=388 cm^3

7 0
3 years ago
The diver has least gravitational potentail engery at position
ankoles [38]
I'm not sure if a figure or some choices go along with this, but the closer to the sea floor the diver is, the lower the potential energy
7 0
3 years ago
Calculate the wavelength of each frequency of electromagnetic radiation: a. 100.2 MHz (typical frequency for FM radio broadcasti
Natalka [10]

Answer:

a). 100.2 MHz (typical frequency for FM radio broadcasting)

The wavelength of a frequency of 100.2 Mhz is 2.99m.

b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)

The wavelength of a frequency of 1070 khz is 280.3 m.

c. 835.6 MHz (common frequency used for cell phone communication)

The wavelength of a frequency of 835.6 Mhz is 0.35m.

Explanation:

The wavelength can be determined by the following equation:

c = \lambda \cdot \nu  (1)

Where c is the speed of light, \lambda is the wavelength and \nu is the frequency.  

Notice that since it is electromagnetic radiation, equation 1 can be used. Remember that light propagates in the form of an electromagnetic wave.

<em>a). 100.2 MHz (typical frequency for FM radio broadcasting)</em>

Then, \lambda can be isolated from equation 1:

\lambda = \frac{c}{\nu} (2)

since the value of c is 3x10^{8}m/s. It is necessary to express the frequency in units of hertz.

\nu = 100.2 MHz . \frac{1x10^{6}Hz}{1MHz} ⇒ 100200000Hz

But 1Hz = s^{-1}

\nu = 100200000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{100200000s^{-1}}

\lambda = 2.99 m

Hence, the wavelength of a frequency of 100.2 Mhz is 2.99m.

<em>b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)</em>

<em> </em>

\nu = 1070kHz . \frac{1000Hz}{1kHz} ⇒ 1070000Hz

But  1Hz = s^{-1}

\nu = 1070000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{1070000s^{-1}}

\lambda = 280.3 m

Hence, the wavelength of a frequency of 1070 khz is 280.3 m.

<em>c. 835.6 MHz (common frequency used for cell phone communication) </em>

\nu = 835.6MHz . \frac{1x10^{6}Hz}{1MHz} ⇒ 835600000Hz

But  1Hz = s^{-1}

\nu = 835600000s^{-1}

Finally, equation 2 can be used:

\lambda = \frac{3x10^{8}m/s}{835600000s^{-1}}

\lambda = 0.35 m

Hence, the wavelength of a frequency of 835.6 Mhz is 0.35m.

6 0
3 years ago
What is the oxidation state of a hydrogen atom bound to an iron atom.?
tresset_1 [31]
the answer is rust so the answer is rust
8 0
3 years ago
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