What is the difference between folk dance and ballroom dance

The eye is actually a multiple-lens system, but we can approximate it with a single-lens system for most of our purposes. When t

hram777 [196]

Explanation:

Given that,

The optical power of the equivalent single lens is 45.4 diopters.

(a) The relationship between the focal length and the focal length is given by:

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

$f=\dfrac{1}{45.4}$

f = 0.022 m

or

f = 2.2 cm

(b) We need to find how far in front of the retina is this "equivalent lens" located. It is given by using lens formula as :

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

Here, u = infinity

$\dfrac{1}{v}=\dfrac{1}{2.2}$

v = 2.2 cm

So, at 2.2 cm in front of the retina is this "equivalent lens" located.

Hence, this is the required solution.

5 0

3 years ago

The bowling ball is whizzing down the bowling lane at 4 m/s. If the mass of the bowling ball is 7 kg, what is its kinetic energy

Lisa [10]

Kinetic Energy = 1/2xmassx(velocity)^2
Input values;
K.E=1/2x7kgx(4m/s)^2
K.E.=56J

3 0

3 years ago

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A sample of gas with a volume of 750 ml exerts a pressure of 98 kpa at 30◦c. What pressure will the sample exert when it is comp

Tanzania [10]

Answer:

241 kPa

Explanation:

The ideal gas law states that:

$pV=nRT$

where

p is the gas pressure

V is its volume

n is the number of moles

R is the gas constant

T is the absolute temperature of the gas

We can rewrite the equation as

$\frac{pV}{T}=nR$

For a fixed amount of gas, n is constant, so we can write

$\frac{pV}{T}=const.$

Therefore, for a gas which undergoes a transformation we have

$\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}$

where the labels 1 and 2 refer to the initial and final conditions of the gas.

For the sample of gas in this problem we have

$p_1 = 98 kPa=9.8\cdot 10^4 Pa\\V_1 = 750 mL=0.75 L=7.5\cdot 10^{-4}m^3\\T_1 = 30^{\circ}C+273=303 K\\p_2 =?\\V_2 = 250 mL=0.25 L=2.5\cdot 10^{-4} m^3\\T_2 = -25^{\circ}C+273=248 K$

So we can solve the formula for $p_2$ , the final pressure:

$p_2 = \frac{p_1 V_1 T_2}{T_1 V_2}=\frac{(9.8\cdot 10^4 Pa)(7.5\cdot 10^{-4} m^3)(248 K)}{(303 K)(2.5\cdot 10^{-4} m^3)}=2.41\cdot 10^5 Pa = 241 kPa$

4 0

2 years ago

The work of Subrahmanyan Chandrasekhar is likely to have been most influenced by the work of which of the following scientists?

borishaifa [10]

Answer

Correct Answer is B(Edwin Hubble)

Explantion

The work of Subrahmanyan Chandrasekhar is likely to have been most influenced by Edwin Hubble because both have worked in the feild of astrophysics.

Subrahmanyan Chandrasekhar was an astrophysicist. He belons to India. He is known for his work on the theoretical structure and evolution of stars. Subrahmanyan Chandrasekhar with William Fowler won the Nobel Prize in Physics in 1983 largely for this early work. He has worked in many other areas within theoretical physics and astrophysics.

Edwin Hubble was an astronomer who belongs to America. He is known for Hubble Law. Hubble discovered that many objects which was thought to be clouds of dust and gas were actually galaxies beyond the Milky Way. He has an important contribution in establishing the fields of observational cosmology and extra galactic astronomy

3 0

2 years ago

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TRUE or FALSE:

Aleksandr-060686 [28]

False. Both masses will experience an acceleration which depends on their mass. The small mass will experience a greater acceleration.

5 0

2 years ago