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<span>Answer:
F(x) = ax^2 - bx
or
F(x) = ax² - bx
F(x) = 30x² - 6x
â«F(x)dx = â«(30x² - 6x)dx
as this is evaluated from zero to x
W = 10x³ - 3x² <===ANS
W = 10(0.42³) - 3(0.42²) - [10(0³) - 3(0²)]
W = 0.212 J <===ANS
W = 10(0.72³) - 3(0.72²) - [10(0.42³) - 3(0.42²)]
W = 1.966 J <===ANS</span>
Answer:
warmer air
Explanation:
the particles are more excited which increases the probability that the particles will bump into each other
The kinetic energy of the child at the bottom of the incline is 106.62 J.
The given parameters:
- <em>Mass of the child, m = 16 kg</em>
- <em>Length of the incline, L = 2 m</em>
- <em>Angle of inclination, θ = 20⁰</em>
The vertical height of fall of the child from the top of the incline is calculated as;

The gravitational potential energy of the child at the top of the incline is calculated as;

Thus, based on the principle of conservation of mechanical energy, the kinetic energy of the child at the bottom of the incline is 106.62 J since no energy is lost to friction.
Learn more about conservation of mechanical energy here: brainly.com/question/332163
Answer:
The speed of space station floor is 49.49 m/s.
Explanation:
Given that,
Mass of astronaut = 56 kg
Radius = 250 m
We need to calculate the speed of space station floor
Using centripetal force and newton's second law




Where, v = speed of space station floor
r = radius
g = acceleration due to gravity
Put the value into the formula


Hence, The speed of space station floor is 49.49 m/s.