Answer:
Explanation:
1. FALL PROTECTION-GENERAL REQUIREMENTS (29 CFR 1926.501) 6,010 VIOLATIONS
2. HAZARD COMMUNICATION (29 CFR 1910.1200). 3,671
3. SCAFFOLDING (29 CFR 1926.451). 2,813
Answer:
the answer is b because it does not show evidence
Answer: d) the presence of solidified lava flows on the Moon
Explanation:
A geological activity means an occurrence of event such as volcanic eruption, earthquake, sedimentation, erosion etc. The revolution of the Moon around the Earth
, the axial tilt of the Moon or the phases of the Moon are not surface features. hence, these events cannot provide the evidence of geological activity in the past of Moon.
The surface features of moon such as Mares, Craters, mountains, Rays and rills are the proof of some geological activity on the Moon. Mares are the dark patches on the moon's surface formed of solidified lava. Due to negligible atmosphere on the moon, the meteors strike its surface and cause craters to form. Thus, the correct answer is d.
Answer:

Explanation:
The root mean square velocity of the gas at an equilibrium temperature is given by the following formula:

where,
v = root mean square velocity of molecules:
R = Universal Gas Constant
T = Equilibrium Temperature
M = Molecular Mass of the Gas
Therefore,
For T = T₁ :

For T = T₂ :

Since both speeds are given to be equal. Therefore, comparing both equations, we get:

Answer:
20 metres
Explanation:
<em>Speed</em><em> </em><em>=</em><em> </em><em>distance</em><em> </em><em>÷</em><em> </em><em>time</em>
<em> </em>
<em>
</em>
If we substitute the values:

<em>
</em>