Spinning a marshmallow over a fire is effective maybe if you hang it over the fire and heat it up equally on each side
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
Answer:
Explanation:
We shall apply conservation of momentum law in vector form to solve the problem .
Initial momentum = 0
momentum of 12 g piece
= .012 x 37 i since it moves along x axis .
= .444 i
momentum of 22 g
= .022 x 34 j
= .748 j
Let momentum of third piece = p
total momentum
= p + .444 i + .748 j
so
applying conservation law of momentum
p + .444 i + .748 j = 0
p = - .444 i - .748 j
magnitude of p
= √ ( .444² + .748² )
= .87 kg m /s
mass of third piece = 58 - ( 12 + 22 )
= 24 g = .024 kg
if v be its velocity
.024 v = .87
v = 36.25 m / s .
The largest possible displacement on a circular track is the straight-line distance between the starting point and the point directly opposite it, half-way around the circle. That's the diameter of the track ... 204 meters.