Answer:
a. 45 N. / b. 0.08 m/s^2. / c. 102 N
F = ma
F = 15(3)
F = 45 newtons
F/m = a
20/250 = a
0.08 m/s^2 = a
R = ma
R =1.5(68)
102 N
Answer:
r = 0.5 m
Explanation:
First we find the angular speed of the ball by using its period:
ω = θ/t
For the time period:
ω = angular speed = ?
θ = angular displacement = 2π rad
t = time period = 0.5 s
Therefore,
ω = 2π rad/0.5 s
ω = 12.56 rad/s
Now, for the radius:
v = rω
r = v/ω
where,
v = linear speed = 6.29 m/s
r = radius = ?
r = (6.29 m/s)/(12.56 rad/s)
<u>r = 0.5 m</u>
Answer:
A)A=0.075 m
B)v= 0.21 m/s
Explanation:
Given that
m = 0.75 kg
K= 13.5 N
The natural frequency of the block given as

The maximum speed v given as

A=Amplitude


A=0.075 m
A= 0.75 cm
The speed at distance x



v= 0.21 m/s
a) KE=0.5*mv^2==0.5*145*25^2=45312.5 J
b) PE=mgh=145*9.8*3.5=4973.5 J
c) ME=KE+PE=m(0.5v^2+gh)=62524 J