Answer:
13.33 m/s^2
Explanation:
Velocity^2 then divide that by the radius
Before i Tell you the answer think about it .
Answer:
a.)1.)12m/s
2.)12m/s
3.)7m/s
4.)7m/s
5.)14m/s
6.)0m/s
b.)1.) 3m/s^2
2.)1.71m/s^2
3.).58m/s^2
4.).39m/s^2
5.).70m/s^2
6.)0
c.)1.)48m
2.)84m
3.)84m
4.)126m
5.)280m
6.)suma todos los metros mas 98m y obtendras la distancia en cual el carro se para
Explanation:
vinicial=0m/s
tiempo inicial=0s
Intervalo 1
v=12m/s
t=4s
aceleracion= vf-vi/t
=12-0/4
=12/4=3m/s^2
Intervalo 2
v=12m/s
t=7s
a=12/7
Intervalo 3
v=7m/s
t=12s
a=7/12
Intervalo 4
v=7m/s
t=18s
a=7/18
Intervalo 5
v=14m/s
t=20s
a=14/20
Intervalo 6
v=0m/s
t=27s
a=0/27
Answer:
a) The maximum height the ball will achieve above the launch point is 0.2 m.
b) The minimum velocity with which the ball must be launched is 4.43 m/s or 0.174 in/ms.
Explanation:
a)
For the height reached, we use 3rd equation of motion:
2gh = Vf² - Vo²
Here,
Vo = 3.75 m/s
Vf = 0m/s, since ball stops at the highest point
g = -9.8 m/s² (negative sign for upward motion)
h = maximum height reached by ball
therefore, eqn becomes:
2(-9.8m/s²)(h) = (0 m/s)² - (3.75 m/s²)²
<u>h = 0.2 m</u>
b)
To find out the initial speed to reach the hoop at height of 3.5 m, we again use 3rd eqn. of motion with h= 3.5 m - 2.5m = 1 m (taking launch point as reference), and Vo as unknown:
2(-9.8m/s²)(1 m) = (0 m/s)² - (Vo)²
(Vo)² = 19.6 m²/s²
Vo = √19.6 m²/s²
<u>Vo = 4.43 m/s</u>
Vo = (4.43 m/s)(1 s/1000 ms)(39.37 in/1 m)
<u>Vo = 0.174 in/ms</u>
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Answer:
Mass of raft=m1= 1000kg
Mass of paddle=m2=40kg
Speed of paddle=v2=3 m/s
Speed of raft=recoil speed=v1=?
Formula
V1= -(m2/m1)v2= -(40/1000)3= - 0.12m/s
Explanation:
Here the raft has much greater mass as compare the paddle and its velocity will always be negative due to recoil. 1 and 2 with masses and velocities are subscripts.
