Gravitational potential energy<span> is </span>energy<span> an object possesses because of its position in a </span>gravitational<span> field. The most common use of </span>gravitational potential energy<span> is for an object near the surface of the Earth where the </span>gravitational<span> acceleration can be assumed to be constant at about 9.8 m/s</span>2<span>.</span>
Answer:
The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s
Explanation:
We are given that
Angular acceleration, 
Diameter of the wheel, d=21 cm
Radius of wheel, 
cm
Radius of wheel, 
1m=100 cm
Magnitude of total linear acceleration, a=
We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.
Tangential acceleration,
Radial acceleration,
We know that
Using the formula
Squaring on both sides
we get
Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s
Answer:
v = 2 v₁ v₂ / (v₁ + v₂)
Explanation:
The body travels the first half of the distance with velocity v₁. The time it takes is:
t₁ = (d/2) / v₁
t₁ = d / (2v₁)
Similarly, the body travels the second half with velocity v₂, so the time is:
t₂ = (d/2) / v₂
t₂ = d / (2v₂)
The average velocity is the total displacement over total time:
v = d / t
v = d / (t₁ + t₂)
v = d / (d / (2v₁) + d / (2v₂))
v = d / (d/2 (1/v₁ + 1/v₂))
v = 2 / (1/v₁ + 1/v₂)
v = 2 / ((v₁ + v₂) / (v₁ v₂))
v = 2 v₁ v₂ / (v₁ + v₂)
