Asked by Emanuel silver(Ag)
A spinning top has a diameter of 10 cm. A point on the outer rim of the top moves through an angle of 8¶ radians each second. A) what is the angular velocity of the point? B) what is the distance moved by the point in 5 seconds? C) what is the velocity of the point? D) what is the acceleration of the point?
Answers
Damon
R = 0.05 meter
A. You already said omega = 8 pi radians/second
C. omega R = v = 8 pi (0.05)
B. theta = angle moved = (8 pi)(5)
so distance = R Theta = (0.05)(8 pi)(5)
D. Ac = R omega^2 = 0.05 (64 pi^2)
A. You already said omega = 8 pi radians/second
C. omega R = v = 8 pi (0.05)
B. theta = angle moved = (8 pi)(5)
so distance = R Theta = (0.05)(8 pi)(5)
D. Ac = R omega^2 = 0.05 (64 pi^2)
ela
step by step
Rezika
I donot gate explained answer
Israel Bekele
Why did you say omega is 8πrad/SCE?
Hanan neb
a .8 rad / sec
b. ፀ = tw = 5s x 8pi rad/s = 40pi rad
c. V= r w= 0.05 x 8 = 0.4
d. a= v squre/r =0.16/ 0.05 = 3.2
b. ፀ = tw = 5s x 8pi rad/s = 40pi rad
c. V= r w= 0.05 x 8 = 0.4
d. a= v squre/r =0.16/ 0.05 = 3.2
Mohamed
Knowledge
Abiyot degu
A, 8pi rad/s
B, 40rad.
B, 40rad.
Mesfin
to get answer
Nuresselam
a,8¶r per s=4r/s b/s 1revo =360°=2¶r so
W=4rad/s
b,teta= wt=4rad/s*5s=20rad
C,v=wr=4rad/s*0.05m= 0.2m/s
d,a=w^2*r=16*0.05=0.8m/s^2
W=4rad/s
b,teta= wt=4rad/s*5s=20rad
C,v=wr=4rad/s*0.05m= 0.2m/s
d,a=w^2*r=16*0.05=0.8m/s^2