y = 75.9 m
Explanation:
y = -(1/2)gt^2 + v0yt + y0
If we put the origin of our coordinate system at the point where a body is launched, then y0 = 0.
y = -(1/2)(9.8 m/s^2)(3 s)^2 + (40 m/s)(3 s)
= -44.1 m + 120 m
= 75.9
Answer:
c. You would weigh less on planet A because the distance between
you and the planet's center of gravity would be smaller.
Explanation:
The statement that best describes your weight on each planet is that you would weigh less on planet A because the distance between you and the planet's center of gravity would be smaller.
- This is based on Newton's law of universal gravitation which states that "the force of gravity between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distances between them".
Since weight is dependent on the force of gravity and mass, the planet with more gravitational pull will have masses on them weighing more.
- Since the distance between the person and the center of the planet is smaller, therefore, the weight will be lesser.
Answer:
This can be used to find out the speed of the returned journey. The equation means speed = returned distance ÷ time.
Explanation:
Answer:
v = 10 m/s
Explanation:
Let's assume the wheel does not slip as it accelerates.
Energy theory is more straightforward than kinematics in my opinion.
Work done on the wheel
W = Fd = 45(12) = 540 J
Some is converted to potential energy
PE = mgh = 4(9.8)12sin30 = 235.2 J
As there is no friction mentioned, the remainder is kinetic energy
KE = 540 - 235.2 = 304.8 J
KE = ½mv² + ½Iω²
ω = v/R
KE = ½mv² + ½I(v/R)² = ½(m + I/R²)v²
v = √(2KE / (m + I/R²))
v = √(2(304.8) / (4 + 0.5/0.5²)) = √101.6
v = 10.07968...
