From the information given, cannon ball weighs 40 kg and has a potential energy of 14000 J.
We need to find its height.
We will use the formula P.E = mgh
Therefore h = P.E / mg
where P.E is the potential energy,
m is mass in kg,
g is acceleration due to gravity (9.8 m/s²)
h is the height of the object's displacement in meters.
h = P.E. / mg
h = 14000 / 40 × 9.8
h = 14000 / 392
h = 35.7
Therefore the canon ball was 35.7 meters high.
Energy is neither created nor destroyed
E in = E out
750 J in so 750 J out
If gravity did not affect the path of a horizontally thrown ball the ball would eventually slow due to air resistance. If the ball were somehow thrown in a vacuum devoid of gravity and air, it would technically continue in a straight line forever.
<span>Force is needed to change an objects state of motion and it is always a part of a mutual action that involves another force. The force can be push or pull.</span>
A ball that moves when kicked <span>describes a force acting on an object.
</span><span>The force is a vector quantity, because it that needs both magnitude and direction in order to be defined.</span>
Answer:
4.92°
Explanation:
The banking angle θ = tan⁻¹(v²/rg) where v = designated speed of ramp = 30 mph = 30 × 1609 m/3600 s = 13.41 m/s, r = radius of curve = 700 ft = 700 × 0.3048 m = 213.36 m and g = acceleration due to gravity = 9.8 m/s²
Substituting the variables into the equation, we have
θ = tan⁻¹(v²/rg)
= tan⁻¹((13.41 m/s)²/[213.36 m × 9.8 m/s²])
= tan⁻¹((179.8281 m²/s)²/[2090.928 m²/s²])
= tan⁻¹(0.086)
= 4.92°
