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:
3054.4 km/h
Explanation:
Using the conservation of momentum
momentum before separation = 5M × 2980 Km/h where M represent the mass of the module while 4 M represent the mass of the motor
initial momentum = 14900 M km/h
let v be the new speed of the motor so that the
new momentum = 4Mv and the new momentum of the module = M ( v + 94 km/h )
total momentum = 4Mv + Mv + 93 M = 5 Mv + 93M
initial momentum = final momentum
14900 M km/h = 5 Mv + 93M
14900 km/h = 5v + 93
14900 - 93 = 5v
v = 2961.4 km/h
the speed of the module = 2961.4 + 93 = 3054.4 km/h
Answer:
- Fx = -9.15 N
- Fy = 1.72 N
- F∠γ ≈ 9.31∠-10.6°
Explanation:
You apparently want the sum of forces ...
F = 8.80∠-56° +7.00∠52.8°
Your angle reference is a bit unconventional, so we'll compute the components of the forces as ...
f∠α = (-f·cos(α), -f·sin(α))
This way, the 2nd quadrant angle that has a negative angle measure will have a positive y component.
= -8.80(cos(-56°), sin(-56°)) -7.00(cos(52.8°), sin(52.8°))
≈ (-4.92090, 7.29553) +(-4.23219, -5.57571)
≈ (-9.15309, 1.71982)
The resultant component forces are ...
Then the magnitude and direction of the resultant are
F∠γ = (√(9.15309² +1.71982²))∠arctan(-1.71982/9.15309)
F∠γ ≈ 9.31∠-10.6°
Answer:
sometimes harmful and sometimes beneficial
the electrons should be in the outer valence levels/shells.
