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V^2 = 700J/0.5*35kg
V = square root of 40
v = 6.324 m/s
Answer:
It must be 4 times high.
Explanation:
- Assuming that the car can be treated as a point mass, and that the ramp is frictionless, the total mechanical energy must be conserved.
- This means, that at any time, the following must be true:
- ΔK (change in kinetic energy) = ΔU (change in gravitational potential energy)
⇒
- Let's call v₁, to the final speed of the car, and h₁ to the height of the ramp.
So, at the bottom of the ramp, all the gravitational potential energy
must be equal to the kinetic energy of the car (Defining the bottom of
the ramp as our zero reference for the gravitational potential energy):
(1)
- Now, let's do v₂ = 2* v₁
- Replacing in (1) we get:
(2)
- Dividing (2) by (1), and rearranging terms, we get:
- h₂ = 4* h₁
1 ft =12 in
4 in = 0.333 ft
volume = (п/4)*(0.333)² = 0.087 ft²
vol. flow = spead *volume
=3 ft/s * 0.087 ft²
vol flow = 0.261 ft³/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