Explanation:
6% grade
=> Rises 6 feet for every 100 feet traveled
This would mean that tanx = 6/100.
=> x = tan^-1(6/100) = 3.4°. (1d.p.)
Hence the road is inclined at an angle of 3.4° above the horizontal.
Information I learned from history class Education in the 1950's expanded from previous decades. They no longer focused purely on reading, writing and arithmetic. History and science became a main part of the cirriculum. Also, enrollment skyrocketed as the baby-boomers began enrolling in elementary school. One interesting thing that categorized this generation was the presence of fallout tests. Schools would require the students to go through a fake atomic bomb attack in which they would hide under their desks (which was completely pointless in protecting them from radiation, it was more of an emotional security for the parents and teachers, but scared the hell out of the students). Socially, children were taught to conform and to be normal. Standing out or questioning authority was bad. Sex was taught, though minimally. They explained the penis and vagina. Sexually transmitted diseases were focused on greatly so as to "scare" the students out of premarital sex.
Answer:
The object accelerates downward at 4 m/s² since the tension on the rope is less than weight of the object.
Explanation:
Given;
mass of the object, m = 2 kg
weigh of the object, W = 20 N
tension on the rope, T = 12 N
The acceleration of the object is calculated by applying Newton's second law of motion as follows;
T = F + W
T = ma + W
ma = T - W
(the negative sign indicates deceleration of the object)
The object accelerates downward at 4 m/s² since the tension on the rope is less than weight of the object.
Answer:
θ = sin⁻¹
Explanation:
From one of the equations of motion, v² = u² + 2as.......... equation 1
Since the object thrown was moving against gravity, then the acceleration, a would change to -g and the initial velocity u would change to V₀ sin θ because the object is travelling at angle of θ to the horizontal. By inputting all these parameter into equation 1, we would arrive at:
v² = (u sin θ)² - 2gd
(u sin θ)² = 2gd
d = (u sin θ)²/2g
sin² θ = 2gd
sin θ =
θ = sin⁻¹
