Answer:
The angle of launch of the rubber band affects the initial velocity. The more the rubber band is stretched the more force it applies to return to equilibrium and the more kinetic energy that results in.
Answer:
Velocity of Afrom B=21m/s
Acceleration of A from B=1.68m/s°2
Explanation:
Given
Radius r=150m
Velocity of a Va= 54km/hr
Va=54*1000/3600=15m/s
Velocity of b Vb=82km/hr
VB=81*1000/3600=22.5mls
The velocity of Car A as observed from B is VBA
VB= VA+VBA
Resolving the vector into X and Y components
For X component= 15cos60=7.5m/s
Y component=22 5sin60=19.48m/s
VBA= √(X^2+Y^2)
VBA= ✓(7.5^2+19.48^2)=21m/s
For acceleration of A observed from B
A=VA^2/r= 15^2/150=1.5m/s
Resolving into Xcomponent=1.5cos60=0.75m/s
Y component=3cos60=1.5
Acceleration BA=√(0.75^2+1.5^2)
1.68m/s
Answer:
when current is passed through a hot plate, it converts <u>electric</u><u> </u>energy into <u>heat</u><u> </u>energy
Answer:
x=?
dt=?
vi=23m/s
vf=0m/s (it stops)
d=0.25m/s^2
time =
vf=vi+d: 0=23m/s+(0.25m/s^2)t
t=92s
displacement=
vf^2=vi^2+2a(dx)
23^2=0^2+2(0.25m/s^2)x =-1058m
Explanation:
you can find time from vf = vi + a(Dt): 0 = 23 m/s + (0.25 m/s/s)t so t = 92 s and you can find the displacement from vf2 = vi2 + 2a(Dx) and find the answer in one step: 232 = 02 + 2(0.25 m/s/s)x so x = -1058 m