The correct answer for the question that is being presented above is this one: "(d) the magnitude of the normal force on the box from the ramp." A box is on a ramp that is at angle θ to the horizontal. As θ is increased from zero, and before the box slips, do the following increase, decrease, or remain the same and <span> the magnitude of the normal force on the box from the ramp.</span>
You can use the points of answer:
B and F
Due to conservation of energy, half way the potential energy will be 1.5J so the remaining 1.5J is kinetic energy.
Vo = 89 m/s
angle: 40°
=> Vox = Vo * cos 40° = 89 * cos 40°
=> Voy = Vo. sin 40° = 89 * sin 40°
x-movement: uniform => x =Vox * t = 89*cos(40)*t
x = 300 m => t = 300m / [89m/s*cos(40) = 4.4 s
y-movement: uniformly accelerated => y = Voy * t - g*t^2 /2
y = 89m/s * sin(40) * (4.4s) - 9.m/s^2 * (4.4)^2 / 2 = 156.9 m = height the ball hits the wall.
Answer:
The height of the hill is, h = 38.42 m
Explanation:
Given,
The horizontal velocity of the soccer ball, Vx = 15 m/s
The range of the soccer ball, s = 42 m
The projectile projected from a height is given by the formula
S = Vx [Vy + √(Vy² + 2gh)] / g
Therefore,
h = S²g/2Vx² (Since Vy = 0)
Substituting the values
h = 42² x 9.8/ (2 x 15²)
= 38.42 m
Hence, the height of the hill is, h = 38.42 m
