1.47x10^5 Joules
The gravitational potential energy will be the mass of the object, multiplied by the height upon which it can drop, multiplied by the local gravitational acceleration. And since it started at the top of a 60.0 meter hill, halfway will be at 30.0 meters. So
500 kg * 30.0 m * 9.8 m/s^2 = 147000 kg*m^2/s^ = 147000 Joules.
Using scientific notation and 3 significant figures gives 1.47x10^5 Joules.
At STP, 1 mole of an ideal gas occupies a volume of about 22.4 L. So if <em>n</em> is the number of moles of this gas, then
<em>n</em> / (19.2 L) = (1 mole) / (22.4 L) ==> <em>n</em> = (19.2 L•mole) / (22.4 L) ≈ 0.857 mol
If the sample has a mass of 12.0 g, then its molecular weight is
(12.0 g) / <em>n</em> ≈ 14.0 g/mol
Answer: they could get a piece of the coral and run tests on it
Explanation:
The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.
Explanation:
"Rank the magnitude of the torque the signs exert about the point at which the rod is attached to the side of the building."
Torque is the cross product of the radius vector and force vector.
τ = r × F
The magnitude of the torque is the force times the perpendicular distance. In other words, it is the force times distance times the sine of the angle between them.
τ = F r sin θ
Let's assume the length of each rod is L meters. In each case, the weight is a vertical force, so the perpendicular distance is the horizontal distance between the wall and the sign.
A) τ = (50 kg × 10 m/s²) (L sin 30° m)
τ = 250L Nm
B) τ = (100 kg × 10 m/s²) (L m)
τ = 1000L Nm
C) τ = (50 kg × 10 m/s²) (L sin 60° m)
τ ≈ 433L Nm
D) τ = (90 kg × 10 m/s²) (L m)
τ = 900L Nm
Ranked from greatest to least:
B > D > C > A