A student charges a balloon and then brings it near a metal sphere hanging from the
2 answers:
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Answer:
a)3.5s
b)28.57m/S
c)34.33m/S
d)44.66m/S
Explanation:
Hello!
we will solve this exercise numeral by numeral
a) to find the time the ball takes in the air we must consider that vertically the ball experiences a movement with constant acceleration whose value is gravity (9.81m / S ^ 2), that the initial vertical velocity is zero, we use the following equation for a body that moves with constant acceleration
where
Vo = Initial speed =0
T = time
g=gravity=9.81m/s^2
y = height=60m
solving for time
T=3.5s
b)The horizontal speed remains constant since there is no horizontal acceleration. with the value of the distance traveled (100m) and the time that lasts in the air (3.5s) we estimate the horizontal speed
c)
to find the final vertical velocity we use the equations for motion with constant velocity as follows
Vf=Vo+g.t
Vf=0+(9.81 )(3.5)=34.335m/S
d)Finally, to find the resulting velocity, we add the horizontal and vertical velocities vectorially, this is achieved by finding the square root of the sum of its squares
Answer:
a) -1.25 rev/s² and 23.3 rev
b) 2.67s
Explanation:
a) ω = (500 rev/min)(1min/ 60s) => 8.333 rev/s
ω= (200 rev/min)(1min/ 60s) => 3.333rev/s
time 't'= 4 s
angular acceleration 'α'=?
constant angular acceleration equation is given by,
ω= ω
+ α
t
α= (ω
- ω
)/t => (3.333-8.333)/4
α= -1.25 rev/s²
θ-θ = ω
t + 1/2α
t²
=(8.333)(4) + 1/2 (-1.25)(4)²
=23.3 rev
b) ω=0 (comes to rest)
ω = 3.333 rev/s
α= -1.25 rev/s²
ω= ω
+ α
t
t= (ω - ω
)/α
=> (0- 3.333)/-1.25
t= 2.67s