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3 years ago

A nation undergoing the transition from undeveloped to industrialized is commonly dependent on this to supply its energy needs.

Physics

2 answers:

algol [13]3 years ago

7 0

A) obviously :)))))))))

Ksivusya [100]3 years ago

4 0

The answer would be.. (A) fossil fuels like coal and oil

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In which case does viscosity play a dominant role? Case A: a typical bacterium (size ~ 1 mm1 mm and velocity ~ 20 mm/s20 mm/s) i

My name is Ann [436]

Answer:

Case A

Explanation:

given,

size of bacteria = 1 mm x 1 mm

velocity = 20 mm/s

size of the swimmer = 1.5 m x 1.5 m

velocity of swimmer = 3 m/s

Viscous force

$F = \eta A \dfrac{dv}{dx}$

for the bacteria

$F = \eta \times 10^{-6}\times 20\times 10^{-3}$

$F =2\times 10^{-8} \eta\ N$

for the swimmer

$F = \eta \times 1.5^2\times 3$

$F =6.75 \eta\ N$

from the above force calculation

In case B inertial force that represent mass is more than the inertial force in case of bacteria.

Viscous force is dominant in case of bacteria.

So, In Case A viscous force will be dominant.

5 0

3 years ago

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

In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.

leonid [27]

Answer:

(a): $F_e = 8.202\times 10^{-8}\ \rm N.$

(b): $F_g = 3.6125\times 10^{-47}\ \rm N.$

(c): $\dfrac{F_e}{F_g}=2.27\times 10^{39}.$

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053 $\times 10^{-9}$ m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges $q_1$ and $q_2$ respectively is given by