Static frictional force = ƒs = (Cs) • (Fɴ)
2.26 = (Cs) • m • g
2.26 = (Cs) • (1.85) • (9.8)
Cs = 0.125
kinetic frictional force = ƒκ = (Cκ) • (Fɴ)
1.49 = (Cκ) • m • g
1.49 = (Cκ) • (1.85) • (9.8)
Cκ = 0.0822
Explanation:
Given parameters:
Distance hopped = 84m
Displacement = 84m due east
Time = 7s
Unknown:
Speed of kangaroo = ?
Velocity of kangaroo = ?
Solution:
To solve this problem,
Speed =
=
= 12m/s
Velocity =
=
= 12m/s due east
Answer:
False -
F = G M1 M2 / R^2
So F depends on M1 and M2 and as long either is not zero there will be a gravitational force between them.
Sorry bro I just need points for my calculus exam
Answer:
See below
Explanation:
See attached diagram
280 km east then 190 km north
Use Pythagorean theorem to find the resultant displacement
d^2 = 280^2 + 190^2
d = 338.4 km
Angle will be arctan ( 190/280) = 34.16 °