Answer:
Explanation:
One prolonged blast every two minutes is an indication of incoming of a power-driven vessel underway. This blast is done to make aware other boats, about your location, in the conditions of poor visibility. The other boats will have to slow down their speed, in order to avoid collision.
Answer:
Time interval;Δt ≈ 37 seconds
Explanation:
We are given;
Angular deceleration;α = -1.6 rad/s²
Initial angular velocity;ω_i = 59 rad/s
Final angular velocity;ω_f = 0 rad/s
Now, the formula to calculate the acceleration would be gotten from;
α = Change in angular velocity/time interval
Thus; α = Δω/Δt = (ω_f - ω_i)/Δt
So, α = (ω_f - ω_i)/Δt
Making Δt the subject, we have;
Δt = (ω_f - ω_i)/α
Plugging in the relevant values to obtain;
Δt = (0 - 59)/(-1.6)
Δt = -59/-1.6
Δt = 36.875 seconds ≈ 37 seconds
Answer:
(a) 42 N
(b)36.7 N
Explanation:
Nomenclature
F= force test line (N)
W : fish weight (N)
Problem development
(a) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled in at constant speed
We apply Newton's first law of equlibrio because the system moves at constant speed:
∑Fy =0
F-W= 0
42N -W =0
W = 42N
(b) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled with an acceleration whose magnitude is 1.41 m/s²
We apply Newton's second law because the system moves at constant acceleration:
m= W/g , m= W/9.8 , m:fish mass , W: fish weight g:acceleration due to gravity
∑Fy =m*a
m= W/g , m= W/9.8 , m:fish mass , W: fish weight g:acceleration due to gravity
F-W= ( W/9.8 )*a
42-W= ( W/9.8 )*1.41
42= W+0.1439W
42=1.1439W
W= 42/1.1439
W= 36.7 N
A) The ball on the small ball is far smaller than the force on the basketball.
B) The total momentum before and after the collision remains constant.
C) We know momentum is conserved so we do:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0.1 x 5 + 0.6 x 0 = 0.1 x -4 + 0.6 x v₂
v₂ = 1.5 m/s
