Answer:
Explanation:
Circle motion
Weight=4lb
Radius=3ft
Velocity Vb= 6ft/s
Velocity Vr=2ft/s
Velocity Vo=12
V2=√Vo²+Vr²
12=√Vo²+2²
Square both side
144=Vo²+4
Vo²=140
Vo=11.83ft/s
Applying conservation of angular momentum
Ha1=Hb2
MbVbr1=MbVor2
r2=Vbr1/Vo
r2=6×3/11.83
r2=1.52ft
The require time is written as.
∆r=Vrt
t=∆r/Vr
t=r1-r2/Vr
t=3-1.52/2
t=0.74sec
Answer:
include the statements pls so i can choose wich one it is and tell you
Explanation:
Answer:
The y-component of the electric force on this charge is 
Explanation:
<u>Given:</u>
- Electric field in the region,
- Charge placed into the region,
where, 
are the unit vectors along the positive x and y axes respectively.
The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,
Thus, the y-component of the electric force on this charge is
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
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