The answers C the molecules in gas move rapidly and all around they are spread out and bounce off each-other
vf^2 = 2ad
vf^2 = 2(9.81)(44m)
vf^2 = 863.28
vf = √863.28
vf = 29.4 - using equations of motion
ME = PE + KE
ME = mgh + 1/2mv^2
ME = (1)(9.81)(44) + 1/2(1)(3^2)
ME = 431.64 + 4.5
ME = 436.14 - using conservation of energy
hope this helps :)
Here we have perfectly inelastic collision. Perfectly inelastic collision is type of collision during which two objects collide, stay connected and momentum is conserved. Formula used for conservation of momentum is:
In case of perfectly inelastic collision v'1 and v'2 are same.
We are given information:
m₁=0.5kg
m₂=0.8kg
v₁=3m/s
v₂=2m/s
v'₁=v'₂=x
0.5*3 + 0.8*2 = 0.5*x + 0.8*x
1.5 + 1.6 = 1.3x
3.1 = 1.3x
x = 2.4 m/s
Answer:
Yes it is related
Explanation:
Consider 2 points in a rectangular co-ordinate system
The distance between them can be found by
d =
This is clear from the figure attached below
Answer:
787528.7 J
Explanation:
Work done: This can be defined as the product of force and distance along the direction of force. The S.I unit of work is Joules (J).
From the question,
W = Tcos∅(d)............. Equation 1
Where W = work done, T = tension in the rope, ∅ = the angle of the rope to the horizontal, d = distance.
But,
d = v(t)..................... equation 2
Where v = velocity, t = time
Substitute equation 2 into equation 1
W = Tcos∅(vt)............. Equation 3
Given: T = 210 N, ∅ = 23°, v = 7 km/h = 1.94 m/s, t = 35 min = 2100 s
Substitute into equation 3
W = 210(cos23°)(1.94×2100)
W = 787528.7 J