What do you mean you need to be more specific
Answer:
time to fall is 3.914 seconds
Explanation:
given data
half distance time = 1.50 s
to find out
find the total time of its fall
solution
we consider here s is total distance
so equation of motion for distance
s = ut + 0.5 × at² .........1
here s is distance and u is initial speed that is 0 and a is acceleration due to gravity = 9.8 and t is time
so for last 1.5 sec distance is 0.5 of its distance so equation will be
0.5 s = 0 + 0.5 × (9.8) × ( t - 1.5)² ........................1
and
velocity will be
v = u + at
so velocity v = 0+ 9.8(t-1.5) ..................2
so first we find time
0.5 × (9.8) × ( t - 1.5)² = 9.8(t-1.5) + 0.5 ( 9.8)
solve and we get t
t = 3.37 s
so time to fall is 3.914 seconds
Let the distance between the towns be d and the speed of the air be s.
distance = speed * time
convert the minutes time into hours.
When flying into the wind, ground speed will be air speed MINUS wind speed, hence the against the wind trip is described by:
d
s−15
=
7
3
return trip is then :
d
s+15
=
7
5
Cross-multiplying both we get the two-variable system:
3d=7∗(s−15)5d=7∗(s+15)
3d=7s−1055d=7s+105
subtract first equation from second equation we get
2d=210d=105km
Substitute the value of d in the above equations for s.
5∗105=7s+1057s=420s=60km/hr
gawaingnpang nkabuhayan, hamon at oportinidad
Answer:
(a) Angular acceleration is 1.112 rad/s².
(b) Average angular velocity is 2.78 rad/s .
Explanation:
The equation of motion in Rotational kinematics is:
θ = θ₀ + 0.5αt²
Here θ is angular displacement at time t, θ₀ is angular displacement at time t=0, t is time and α is constant angular acceleration.
(a) According to the problem, θ is 13.9 rad, θ₀ is zero as it is at rest and t is 5 s. Put these values in the above equation:
13.9 = 0 + 0.5α(5)²
α = 1.112 rad/s²
(b) The equation of average angular velocity is:
ω = Δθ/Δt
ω = 
ω = 2.78 rad/s
