Answer:
Given that
T= 0.43 s
Radius of the ball path's , r=2.1 m
a)
We know that
f= 1/T
Here f= frequency
T= Time period
Now by putting the values
f= 1/T
T= 0.43 s
f= 1/0.43
f=2.32 Hz
b)
We know that
V= ω r
ω = 2 π f
ω=Angular speed
V= Linear speed
ω = 2 π f=ω = 2 x π x 2.32 =14.60 rad/s
V= ω r= 14.60 x 2.1 = 30.66 m/s
c)
Acceleration ,a
a =ω ² r
a= 14.6 ² x 2.1 = 447.63 m/s²
We know that g = 10 m/s²
So
a= a/g= 447.63/10 = 44.7 g m/s²
a= 44.7 g m/s²
Answer:
60 cycles
Explanation:
The first thing we must do to solve the problem is to find how many cycles are presented in 1cm by multiplying the frequency by the base time of the
K=time base=2ms/cm=2x10-3s/cm
f=frecuency=3000s^-1
N=fk
N=(3000)(2x10^-3)=6cycles/cm
Ntot=6x10=60cycles
Answer:
3 m/s
Explanation:
Speed in m/s would just be 60 m / 20 s = 3 m/s
Evaporation is the process of the hydrological cycle maintained by the ocean's large surface area.