No force is required to lift that balloon. In fact, force is required to hold it down, and if you let go, it's up, up, and away.
Since the balloon's density is less than the density of the air around it, it's lighter than the air it displaces, there is a net upward buoyant force acting on it, and it floats up !
<span>A. Chemical energy to chemical energy</span>
Somersaulting- for longer distances.It bends the narrow end in the direction it wants to go & takes grip with tentacles. It releases the broad end and straightens up. like this it continues. looping- for shorter distances.
Hope this helps
The concept of this problem is the Law of Conservation of Momentum. Momentum is the product of mass and velocity. To obey the law, the momentum before and after collision should be equal:
m₁ v₁ + m₂v₂ = m₁v₁' + m₂v₂', where
m₁ and m₂ are the masses of the proton and the carbon nucleus, respectively,
v₁ and v₂ are the velocities of the proton and the carbon nucleus before collision, respectively,
v₁' and v₂' are the velocities of the proton and the carbon nucleus after collision, respectively,
m(164) + 12m(0) = mv₁' + 12mv₂'
164 = v₁' + 12v₂' --> equation 1
The second equation is the coefficient of restitution, e, which is equal to 1 for perfect collision. The equation is
(v₂' - v₁')/(v₁ - v₂) = 1
(v₂' - v₁')/(164 - 0) = 1
v₂' - v₁'=164 ---> equation 2
Solving equations 1 and 2 simultaneously, v₁' = -138.77 m/s and v₂' = +25.23 m/s. This means that after the collision, the proton bounced to the left at 138.77 m/s, while the stationary carbon nucleus move to the right at 25.23 m/s.
Answer:
Q = 40.1 degrees
Explanation:
Given:
- The weight of the timber W = 670 N
- Water surface level from pivot y = 2.1 m
- The specific density of water Y = 9810 N / m^3
- Dimension of timber = (0.15 x 0.15 x 0.0036) m
Find:
- The angle of inclination Q that the timber makes with the horizontal.
Solution:
- Calculate the Flamboyant Force F_b acting upwards at a distance x along the timber, which is unknown:
F_b = Y * V_timber
F_b = 9810*0.15*0.15*x
F_b = 226.7*x N
- Take static equilibrium conditions for the timber, and take moments about the pivot:
(M)_p = 0
W*0.5*3.6*cos(Q) - x/2 * F_b*cos(Q) = 0
- Plug values in:
670*0.5*3.6 - x^2 * 0.5*226.7 = 0
x^2 = 1206 / 113.35
x = 3.26 m
- Now use the value of x and vertical height y to compute the angle of inclination to be:
sin(Q) = y / x
sin(Q) = 2.1 / 3.26
Q = sin^-1 (0.6441718)
Q = 40.1 degrees
