<span>Since the wheel start from rest. angular acceleration,
θ=1/2αt²
14=1/2α x 8.7²
α= 0.3699 rad/s²
moment of inertia of loop= mr²= 4.1x0.37=1.517 kgm²
torque=T= lα
T= 0.5611Nm= 0.56Nm to significant figure
Disc
moment of inertia of disc= 1/2mr²
Required torque value= 0.28Nm
So,
I= 1/2X 4.1X 0.37²= 0.280 Kgm²
T= Iα = 0.280 X 0.3699= 0.10 to two significant figure</span>
Answer:
Friction force is 0.1375 N
Solution:
As per the question:
Radius of the metal disc, R = 4.0 cm = 0.04 m
Magnetic field, B = 1.25 T
Current, I = 5.5 A
Force on a current carrying conductor in a magnetic field is given by considering it on a differential element:
Integrating the above eqn:
(1)
Now the torque is given by:
(2)
From eqn (1) and (2):
Thus the Frictional force is given by:
Answer:
The final speed will be 8.02 rad/s.
Explanation:
Given that
radius ,r=5.9 m
Total acceleration = 18
Tangential acceleration = 12
Here particle is moving in circular path with varying linear speed.So there will be two acceleration ,one is radial acceleration and other one is tangential acceleration.
12=α x 5.9
ω=2.2 rad/s
θ=2πn
θ=2 x π x 1/8 rad
θ=π /4 rad
So the final speed will be 8.02 rad/s.
Yes it does ! Uh huh. Right you are. Truer words are seldom written.
You have quoted the law quite accurately but also incompletely.
Do you have a question to ask ?
Given : Acceleration of the ball is 9.8 m / s² . The pulling force is 1.4 N .
To Find : The mass of the ball .
Solution : It is given that a ball accelerates download with a acceleration of 9.8 m / s² . And the acting force is 1.4 N .
We know that , Force acting is equal to the product of mass and acceleration . That is ,
⇒ Force = mass × accelerⁿ
⇒ 1.4 N = mass × 9.8 m / s²
⇒ mass = 1.4 N / 98 m/s²
⇒ mass = 14N / 98 m/s².
⇒ mass = 1/7 kg .
<u>Hence</u><u> </u><u>the </u><u>mass</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>object</u><u> is</u><u> </u><u>1</u><u>/</u><u>7</u><u> </u><u>kg</u><u> </u><u>≈</u><u> </u><u>0</u><u>.</u><u>1</u><u>4</u><u> </u><u>kg</u><u>.</u>