When the velocity is increasing the acceleration increases too
Answer:
The mass of the heaviest box you will be able to move with this applied force = 61.4 kg
Explanation:
From the diagram attached, the forces acting on the box include the weight of the box, applied force on the box, normal reaction of the surface on the box and the Frictional force in the opposite direction to the applied force.
For the box to be able to move, the applied force must have a horizontal component that at least matches the Frictional force between the box and the surface. This is the force balance in the horizontal direction.
Resolving the applied force into horizontal and vertical components,
Fₓ = 750 cos 25° = 679.73 N
Fᵧ = 750 sin 25° = 316.96 N
Doing a force balance in the vertical axis,
N = (mg + 316.96)
Frictional force = μN = μ (mg + 316.96)
μ = 0.74, g = 9.8 m/s²
Frictional force = Fᵧ
μ (mg + 316.96) = 679.73
0.74(9.8m + 316.96) = 679.73
7.252m + 234.5504 = 679.73
7.252m = 679.73 - 234.5504 = 445.1796
m = (445.1796/7.252)
m = 61.4 kg
Hope this Helps!!!
Answer:
The total mechanical energy of the skydiver is, E = 96402.6 J
Explanation:
Given data,
The mass of the skydiver, m = 100 kg
The speed of the skydiver at 80 m height, v = 60 m/s
The initial velocity of the skydiver, u = 0
Using the III equations of motion,
v² = u² + 2gs
s = v²/2g
Substituting the given values,
s = ½ 60²/ 9.8
= 18.37 m
Hence the initial total distance of the skydiver from the ground initially,
h = s + d
= 18.37 + 80
= 98.37 m
Since the total mechanical energy of a system is conserved, the total mechanical energy of the skydiver at height 'h' is equal to the total mechanical energy at height 'd'.
E = P.E + K.E
= mgh + ½ mu²
= 100 x 9.8 x 98.37 ( ∵ u = 0)
= 96402.6 J
Hence, the total mechanical energy of the skydiver is, E = 96402.6 J
"K+" is the one metal ion among the choices given in the question that produces <span>the light with the highest energy. The correct option among all the options that are given in the question is the third option or the penultimate option. I hope that the answer has actually come to your help.</span>