Answer:
192.08J
19.6m/s
Explanation:
Since there will be no potential energy when the ball is on the ground, the change in potential energy is equal to the potential energy at the start when the ball is 19.6m above the ground.
PE=mgh
=(1)(9.8)(19.6)
=192.08J
v²=u²+2as, where v is the final velocity, u is initial velocity, a is acceleration and s is distance. Initial velocity is 0 since it starts at rest.
v²=u²+2as
v²=0²+2(9.8)(19.6)
v=√384.16
=19.6m/s
Answer:
R = 5.28 103 km
Explanation:
The definition of density is
ρ = m / V
V = m /ρ
Where m is the mass and V the volume of the body
The volume of a sphere is
V = 4/3 π r³
Let's replace
4/3 π r³ = m / ρ
R =∛ ¾ m / ρ π
The mass of the planet is
M = 5.5 Me
R = ∛ ¾ 5.5 Me /ρ π
Let's reduce the density to SI units
ρ = 1.76 g / cm³ (1 kg / 10³ g) (10² cm / 1 m)³
ρ = 1.76 10³ kg / m³
Let's calculate
R = ∛ ¾ 5.5 5.97 10²⁴ / (1.76 10³ pi)
R = ∛ 0.14723 10²¹
R = 0.528 10⁷ m
R = 0.528 104 km
R = 5.28 103 km
Answer:
1.8 s
Explanation:
Potential energy = kinetic energy + rotational energy
mgh = ½ mv² + ½ Iω²
For a thin spherical shell, I = ⅔ mr².
mgh = ½ mv² + ½ (⅔ mr²) ω²
mgh = ½ mv² + ⅓ mr²ω²
For rolling without slipping, v = ωr.
mgh = ½ mv² + ⅓ mv²
mgh = ⅚ mv²
gh = ⅚ v²
v = √(1.2gh)
v = √(1.2 × 9.81 m/s² × 4.8 m sin 39.4°)
v = 5.47 m/s
The acceleration down the incline is constant, so given:
Δx = 4.8 m
v₀ = 0 m/s
v = 5.47 m/s
Find: t
Δx = ½ (v + v₀) t
t = 2Δx / (v + v₀)
t = 2 (4.8 m) / (5.47 m/s + 0 m/s)
t = 1.76 s
Rounding to two significant figures, it takes 1.8 seconds.
