Answer:
a) 578.0 cm²
b) 25.18 km
Explanation:
We're given the density and mass, so first calculate the volume.
D = M / V
V = M / D
V = (6.740 g) / (19.32 g/cm³)
V = 0.3489 cm³
a) The volume of any uniform flat shape (prism) is the area of the base times the thickness.
V = Ah
A = V / h
A = (0.3489 cm³) / (6.036×10⁻⁴ cm)
A = 578.0 cm²
b) The volume of a cylinder is pi times the square of the radius times the length.
V = πr²h
h = V / (πr²)
h = (0.3489 cm³) / (π (2.100×10⁻⁴ cm)²)
h = 2.518×10⁶ cm
h = 25.18 km
The path the bowling ball would most closely follow after leaving the airplane is horizontal direction.
<h3>
Path of the bowling ball</h3>
Based on the law of inertia, which is the reluctance of an object to stop moving once in motion or start moving when it is at rest.
The bowling ball will maintain the path of the airline in the first few seconds of fall, after which it will change its path to vertical direction.
Thus, the path the bowling ball would most closely follow after leaving the airplane is horizontal direction.
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The work done by the applied force on the block against the frictional force is 15.75 J.
<h3>
Work done by the applied force</h3>
The work done by the applied force is calculated as follows;
W = Fd
F - Ff = ma
where;
- F is applied force
- Ff is frictional force
Fcos(37) - μmgsin(37) = ma
Fcos(37) - (0.3)(4)(9.8)sin(37) = 4(0.2)
0.799F - 7.077 = 0.8
F = 9.86 N
W = Fdcosθ
W = 9.86 x 2 x cos(37)
W = 15.75 J
Thus, the work done by the applied force on the block against the frictional force is 15.75 J.
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Answer:
A ball being dropped to the ground