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3 years ago

15

The most common type of lighting unit utilized in street lighting is:

1 answer:

Juli2301 [7.4K]3 years ago

3 0

<span>Option C. The sodium vapor lamp is a gas discharge lamp that uses sodium vapor for the production of light. It is an efficient lighting source that provides lumens per watt and produces bright yellow. It has a monochromatic light that makes its applications are reduced. It is used in street lighting, streets, highways, etc.</span>

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A virtual image formed by convex lens is always

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Answer:

2. erect and magnified

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Which one of the following substances is a liquid fuel used in rocket engines?

melomori [17]

There are none on the list you included with your question.

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a 150 N force is used to pull a wooden box across a wooden surface at a constant velocity. what is the mass of the box?

IRINA_888 [86]

Answer:

The mass of the box:

m = 60 kg

Explanation:

Given:

F = 150 N

g = 10 m/s²

_________

m - ?

Coefficient of friction wood on wood:

μ = 0.25

Friction force:

F₁ = μ*m*g

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1 year ago

Cohort studies allow the investigator to examine multiple outcomes and multiple exposures. Consider the following three exposure

mojhsa [17]

Answer:

See the explanation for the answer.

Explanation:

<u>Cohort study for smoking</u>

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1. People at risk of lung cancer due to smoking

2. Influence of outdoor air pollution in lung cancer

3. Variation in tobacco smoking and heavy tobacco smoking was observed.

4. Occupational factors and housing factor were observed.

<u>Cohort study for low vitamin D</u>

<u>Outcomes that were examined</u>

1. People at risk of vitamin D deficiency.

2. Observing value in summer.

3. Correlating vitamin D value with weather and age

4. 25-hydroxy vitamin D was observed.

<u>Cohort study for cold weather</u>

<u>Outcomes that were examined</u>

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3 0

3 years ago

How much kinetic energy does a system containing 3 moles of an ideal gas at 300 K possess? What is the heat capacity at constant

ElenaW [278]

Answer:

1. K.E = 11.2239 kJ ≈ 11.224 kJ

2. $C_{V} = 37.413 JK^{- 1}$

3. $Q = 10.7749 kJ$

Solution:

Now, the kinetic energy of an ideal gas per mole is given by:

K.E = $\frac{3}{2}mRT$

where

m = no. of moles = 3

R = Rydberg's constant = 8.314 J/mol.K

Temperature, T = 300 K

Therefore,

K.E = $\frac{3}{2}\times 3\times 8.314\times 300$

K.E = 11223.9 J = 11.2239 kJ ≈ 11.224 kJ

Now,

The heat capacity at constant volume is:

$C_{V} = \frac{3}{2}mR$

$C_{V} = \frac{3}{2}\times 3\times 8.314 = 37.413 JK^{- 1}$

Now,

Required heat transfer to raise the temperature by $15^{\circ}$ is:

$Q = C_{V}\Delta T$

$\Delta T = 15^{\circ} = 273 + 15 = 288 K$

$Q = 37.413\times 288 = 10774.9 J = 10.7749 kJ$

8 0

3 years ago