Answer:
V = 10 km / 1 hr = 10 km/hr
V = -10 j km / hr if one were to use i, j, k as unit vectors with the usual orientation
Answer:
a) ω = 9.86 rad/s
b) ac = 194. 4 m/s²
c) minimum coefficient of static friction, µs = 19.8
Explanation:
a) angular speed, ω = 2πf, where f is frequency of revolution
1 rps = 6.283 rad/s, π = 3.142
ω = 2 * 3.14 * 0.25 * 6.28
ω = 9.86 rad/s
b) centripetal acceleration, a = rω²
where r is radius in meters; r = 200 cm or 2 m
a = 2 * 9.86²
a = 194. 4 m/s²
c) µs = frictional force/ normal force
frictional force = centripetal force = ma; where a is centripetal acceleration
normal force = mg; where g = 9.8 m/s²
µs = ma/mg = a/g
µs = 194.4 ms⁻²/9.8 ms⁻²
c) minimum coefficient of static friction, µs = 19.8
The velocity of the boat after the package is thrown is 0.36 m/s.
<h3>
Final velocity of the boat</h3>
Apply the principle of conservation of linear momentum;
Pi = Pf
where;
- Pi is initial momentum
- Pf is final momentum
v(74 + 135) = 15 x 5
v(209) = 75
v = 75/209
v = 0.36 m/s
Thus, the velocity of the boat after the package is thrown is 0.36 m/s.
Learn more about velocity here: brainly.com/question/6504879
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The correct answer (d.) epistemology. The major contribution of the French philosopher Rene Descartes was in the realm of epistemology. Epistemology is the theory of knowledge, including methods, validity, and its scope. It could also be concluded to separate justified fact from an opinion.
Answer:
The average induced emf around the border of the circular region is
.
Explanation:
Given that,
Radius of circular region, r = 1.5 mm
Initial magnetic field, B = 0
Final magnetic field, B' = 1.5 T
The magnetic field is pointing upward when viewed from above, perpendicular to the circular plane in a time of 125 ms. We need to find the average induced emf around the border of the circular region. It is given by the rate of change of magnetic flux as :

So, the average induced emf around the border of the circular region is
.