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3 years ago

What are the Si prefix values

1 answer:

4 0

centic10-2millim10-3microu [footnote 2]10-6nanon10-9

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Adding energy or increasing the speed of the particles at very low pressure in a solid would cause

weqwewe [10]

D. sublimation is correct

7 0

3 years ago

Where is the near point of an eye for which a contact lens with a power of +2.55 diopters is prescribed?Where is the far point o

tankabanditka [31]

Answer:

(a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

Explanation:

Given that,

Power = 2.55 D

Object distance = 25 cm for near point

Object distance = ∞ for far point

Suppose where is the far point of an eye for which a contact lens with a power of -3.00 D is prescribed for distant vision?

(a) We need to calculate the focal length

Using formula of power

$f =\dfrac{1}{P}$

Put the value into the formula

$f=\dfrac{100}{2.55}$

$f=39.21\ cm$

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{39.21}=\dfrac{1}{v}-\dfrac{1}{-25}$

$\dfrac{1}{39.21}-\dfrac{1}{25}=\dfrac{1}{v}$

$\dfrac{1}{v}=-\dfrac{1421}{98025}$

$v=-68.98\ cm$

The eye's near point is 68.98 cm from the eye.

(b). We need to calculate the focal length

Using formula of power

$f =\dfrac{1}{P}$

Put the value into the formula

$f=-\dfrac{100}{3.00}$

$f=-33.33\ cm$

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{-33.33}=\dfrac{1}{v}-\dfrac{1}{\infty}$

$-\dfrac{1}{33.33}+\dfrac{1}{\infty}=\dfrac{1}{v}$

$\dfrac{1}{v}=-\dfrac{1}{33.33}$

$v=-33.33\ cm$

The eye's far point is 33.33 cm from the eye.

Hence, (a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

5 0

3 years ago

4. Two forces act on a 2 kg object as shown. What is the magnitude of the acceleration of the object?

aleksandrvk [35]

The resultant force on the object is

∑ F = 〈0, 8〉 N + 〈6, 0〉 N = 〈6, 8〉 N

which has a magnitude of

F = √((6 N)² + (8 N)²) = √(100 N²) = 10 N

By Newton's second law, the acceleration has magnitude a such that

F = m a

10 N = (2 kg) a

a = (10 N) / (2 kg)

a = 5 m/s²

so the answer is B.

4 0

2 years ago

The photons of different light waves:

LuckyWell [14K]

Answer: contain different amounts of energy

Explanation:

The energy $E$ of a photon is given by:

$E=h\nu$

Where:

$h=6.626(10)^{-34}\frac{m^{2}kg}{s}$ is the Planck constant

$\nu$ is the frequency of the light which is inversely related to the wavelength.

Now, if we have photons of different light waves, this means we have photons with different frequencies.

As the energy of the photon depends on its frequency:

Photons of different light waves contain different amounts of energy.

8 0

3 years ago

while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity o

MariettaO [177]

Complete question:

while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity of 8.32 meters per second take the speed of sound as 340 meters per second. calculate the frequency reflected off the wall to the bat?

Answer:

The frequency reflected by the stationary wall to the bat is 41 kHz

Explanation:

Given;

frequency emitted by the bat, = 39 kHz

velocity of the bat, $v_b$ = 8.32 m/s