Answer:
Taking forces along the plane
F cos θ - M g sin θ -100 = M a net of forces along the plane
F = (M a + M g * .5 + 100) / .866 solving for F
F = (80 * 1.5 + 80 * 9.8 * .5 + 100) / .866 = 707 N
F = 707 N acting along the plane
Fn = F sin θ + M g cos θ forces acting perpendicular to plane
Fn = 707 * 1/2 + 80 * 9.8 * .866 = 1030 Newtons forces normal to plane
(this would give a coefficient of friction of 100 / 1030 = .097 = Fn)
Answer:
Proper weighting
Explanation:
Proper weighing involves the condition of a scuba diver that is fully geared having a near empty tank and the BCD emptied with a held breadth is expected to float at eye level
The fundamental of adequate or good buoyancy of a scuba diver is to ensure proper weighting when diving, With proper weighting, there is more control for the diver when a safety stop is required. There is less need to carry excess weight that increases drag and gas consumption.
Answer:
Part a)
Part b)
Explanation:
Part a)
As we know that electric field intensity due to some given charge distribution is given as
now electric flux through a spherical surface of radius r is given as
now by Guass law we know that
now volume charge density is given as
Part b)
Total charge inside the radius R is given as
Answer:
The new kinetic energy would be 16 times greater than before.
Explanation:
Kinetic energy is found using this formula:
- KE = 1/2mv²
- where KE = kinetic energy (J), m = mass (kg), and v = velocity (m/s)
We can see that kinetic energy is directly proportional to the square of the velocity, meaning that if the speed was increased by 4 times, then the kinetic energy would get increased by a factor of 16.
The velocity just before the ball hits the ground can be found by the equation:
Let's substitute h = 10 m and h = 40 m into this formula.
We can see that the velocity increases by a factor of 4 (10 m → 40 m).
Therefore, this means that the kinetic energy would also be increased by a factor of (4)² = 16. Thus, the answer is D) The new kinetic energy would be 16 times greater than before.
