Answer:
0.62 m/s² at 68° S of E
Explanation:
Net force north = 12 - 70 = -58 N
Net force east = 33 - 10 = 23 N
Net force = √(-58² + 23²) = 62.3939... N
acceleration = F/m = 62.3939/100 = 0.623939... ≈ 0.62 m/s²
θ = arctan(-58/23) = -68.3691... ≈ 68° S of E
Answer:
a) correct answer is C
, b) 14º from the west to the north, c) v_{1g} = 300.79 km / h
Explanation:
This is a relative speed exercise using the addition of speeds.
1) when it is not specified regarding what is being measured, the medicine is carried out with respect to the Z Earth, therefore the correct answer is C
2 and 3) In this case we must compose the speed using the Pythagorean Theorem.
² =
² +
²
where v_{1a} is the speed of the airplane with respect to the air, v_{1g} airplane speed with respect to the Earth, v_{ag} air speed with respect to the Earth
in this case let's clear the speed of the airplane with respect to the Earth
v_{1g} = √(v_{1a}² - v_{ag}²)
v_{1g} = √ (310² - 75²)
v_{1g} = 300.79 km / h
we find the direction of the airplane using trigonometry
sin θ = v_{ag} / v_{1a}
θ = sin⁻¹ (v_{ag} /v_{1a})
θ = sin⁻¹ (75/310)
θ= 14º
the pilot must direct the aircraft at an angle of 14º from the west to the north
<span>the speed of something in a given direction. so i think none of these</span>
Answer: Attach it to clothing, personal flotation device or the person's person.
Explanation:
This particular Question or problem has to do with the rules and regulations or laws governing the use of Personal Watercraft(PWC) in the state of Florida in the United States of America. Hence, one of the rules that is applicable to the use of boats and waterways or the use of personal watercraft in Florida is that in florida, if ones pwc is equipped with an engine cut-off lanyard the person operating it must ATTACH IT TO THE CLOTHING, PERSONAL FLOATATION DEVICE IR THE OPERATORS' PERSON.
Another rule band the use of flammable personal flotation device with Personal Watercraft(PWC) in the state of Florida. All this rules and guildlines are made to limit or minimize hazards.
Potential Energy= 24m * 14kg * 9.8N/kg = 3292.8J