We use the kinematic equations,
(A)
(B)
Here, u is initial velocity, v is final velocity, a is acceleration and t is time.
Given, ,
and
.
Substituting these values in equation (B), we get
.
Therefore from equation (A),
Thus, the magnitude of the boat's final velocity is 10.84 m/s and the time taken by boat to travel the distance 280 m is 51.63 s
Answer:
0.0367
Explanation:
The loss in kinetic energy results into work done by friction.
Since kinetic energy is given by
KE=0.5mv^{2}
Work done by friction is given as
W= umgd
Where m is the mass of suitacase, v is velocity of the suitcase, g is acceleration due to gravity, d is perpendicular distance where force is applied and u is coefficient of kinetic friction.
Making u the subject of the formula then we deduce that
Substituting v with 1.2 m/s, d with 2m and taking g as 9.81 m/s2 then
Therefore, the coefficient of kinetic friction is approximately 0.0367
Answer:
30298514.82 m/s
Explanation:
M = Mass of star = 2×10³ kg
r = Radius of star = 5×10³ m
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
The object would be moving at a velocity of 30298514.82 m/s