Answer:
The acceleration of the snowball is 0.3125
Explanation:
The initial speed of the snowball up the hill, u = 0
The speed the snowball reaches, v = 5 m/s
The length of the hill, s = 40 m
The equation of motion of the snowball given the above parameters is therefore;
v² = u² + 2·a·s
Where;
a = The acceleration of the snowball
Plugging in the values, we have;
5² = 0² + 2 × a × 40
∴ 2 × 40 × a = 5² = 25
80 × a = 25
a = 25/80 = 5/16
a = The acceleration of the snowball = 5/16 m/s².
The acceleration of the snowball = 5/16 m/s² = 0.3125 m/s² .
Answer:
640.5
Explanation: i got it right on acellus
Answer:
t ’=
, v_r = 1 m/s t ’= 547.19 s
Explanation:
This is a relative velocity exercise in a dimesion, since the river and the boat are going in the same direction.
By the time the boat goes up the river
v_b - v_r = d / t
By the time the boat goes down the river
v_b + v_r = d '/ t'
let's subtract the equations
2 v_r = d ’/ t’ - d / t
d ’/ t’ = 2v_r + d / t
In the exercise they tell us
d = 1.22 +1.45 = 2.67 km= 2.67 10³ m
d ’= 1.45 km= 1.45 1.³ m
at time t = 69.1 min (60 s / 1min) = 4146 s
the speed of river is v_r
t ’=
t ’=
In order to complete the calculation, we must assume a river speed
v_r = 1 m / s
let's calculate
t ’=
t ’= 547.19 s
Answer:
pirate walk 259 feet
displacement vector = 111 i - 234 j
direction is along south of east at angle 64.62° , anticlockwise
total travel 345 feet
Explanation:
given data
walk east = 111 feet
walk south = 234 feet
to find out
How far did the pirate walk and displacement vector and distance and direction relative to east
solution
we consider here distance AB is 111 feet and than he turn right i.e south distance BC is 234 feet so
so angle BAC will be
tan θ = 
θ = 64.62
and AC distance will be
AC = 
AC = 259 feet
so pirate walk 259 feet
and
displacement vector is express as
displacement vector = AC ( cosθ i + sinθ j )
displacement vector = 259 ( cos64.62 i + sin64.62 j )
displacement vector = 111 i - 234 j
and
so direction is along south of east at angle 64.62° , anticlockwise