"Fluid intelligence involves being able to think and reason abstractly and solve problems. This ability is considered independent of learning, experience, and education. Examples of the use of fluid intelligence include solving puzzles and coming up with problem-solving strategies."
- Verywell Mind
If a coin is dropped at a relatively low altitude, it's acceleration remains constant. However, if the coin is dropped at a very high altitude, air resistance will have a significant effect. The initial acceleration of the coin will be the greatest. As it falls down, air resistance will counteract the weight of the coin. So, the acceleration will decrease. Although the acceleration decreases, the coin still accelerates, that is why it falls faster. When the air resistance fully counters the weight of the coin, the acceleration will become zero and the coin will fall at a constant speed (terminal velocity). So, the answer should be, The acceleration decreases until it reaches 0. The closest answer is.
a. The acceleration decreases.
Answer:
Acceleration = 5.77 m/s²
Distance cover in 13 seconds = 487.56 meter
Explanation:
Given:
Final velocity of mobile device = 75 m/s
initial velocity of mobile device = 0 m/s
Time taken = 13 seconds
Find:
Acceleration
Distance cover in 13 seconds
Computation:
v = u + at
75 = 0 + (a)(13)
13a = 75
a = 5.77
Acceleration = 5.77 m/s²
s = ut + (1/2)(a)(t²)
s = (0)(t) + (1/2)(5.77)(13²)
Distance cover in 13 seconds = 487.56 meter
Answer:
The tomato won't hit the car
Explanation:
According to the statement, the car moves at constant speed behind the truck fully loaded with tomatoes, and in the same direction. When a tomato falls from the top of the truck, it should not hit the car as the tomato falls due to the force of gravity, while horizontally has the same speed and in the same direction as the truck. So we assume that the tomato will fall to the road without touching the car.
Have a nice day!
Answer:
Explanation:
a )
change in the gravitational potential energy of the bear-Earth system during the slide = mgh
= 45 x 9.8 x 11
= 4851 J
b )
kinetic energy of bear just before hitting the ground
= 1/2 m v²
= .5 x 45 x 5.8²
= 756.9 J
c ) If the average frictional force that acts on the sliding bear be F
negative work done by friction
= F x 11 J
then ,
4851 J - F x 11 = 756.9 J
F x 11 = 4851 J - 756.9 J
= 4094.1 J
F = 4094.1 / 11
= 372.2 N