Answer:
The magnitude of the uniform magnetic field exerting this torque on the loop is 1.67 T
Explanation:
Given;
radius of the wire, r = 0.45 m
current on the loop, I = 2.4 A
angle of inclination, θ = 36⁰
torque on the coil, τ = 1.5 N.m
The torque on the coil is given by;
τ = NIBAsinθ
where;
B is the magnetic field
Area of the loop is given by;
A = πr² = π(0.45)² = 0.636 m
τ = NIBAsinθ
1.5 = (1 x 2.4 x 0.636 x sin36)B
1.5 = 0.8972B
B = 1.5 / 0.8972
B = 1.67 T
Therefore, the magnitude of the uniform magnetic field exerting this torque on the loop is 1.67 T
Answer:
The correct option is;
Force of Friction
Explanation:
As coach Hogue rode his motorcycle round in circle on the wet pavement, the motorcycle and the coach system tends to move in a straight path but due to intervention by the coach they maintain the circular path
The motion equation is
v = ωr and we have the centripetal acceleration given by
α = ω²r and therefore centripetal force is then
m×α = m × ω²r = m × v²/r
The force required to keep the coach and the motorcycle system in their circular path can be obtained by the impressed force of friction acting towards the center of the circular motion.
Answer:
There are six kinds of forces that act on objects when they come into contact with one another: Normal force, applied force, frictional force, tension force, spring force and resisting force. These forces make objects change their motion or movement , the act of going from one place to another.
Answer:
(A) It will take 22 sec to come in rest
(b) Work done for coming in rest will be 0.2131 J
Explanation:
We have given the player turntable initially rotating at speed of ![33\frac{1}{3}rpm=33.333rpm=\frac{2\times 3.14\times 33.333}{60}=3.49rad/sec](https://tex.z-dn.net/?f=33%5Cfrac%7B1%7D%7B3%7Drpm%3D33.333rpm%3D%5Cfrac%7B2%5Ctimes%203.14%5Ctimes%2033.333%7D%7B60%7D%3D3.49rad%2Fsec)
Now speed is reduced by 75 %
So final speed ![\frac{3.49\times 75}{100}=2.6175rad/sec](https://tex.z-dn.net/?f=%5Cfrac%7B3.49%5Ctimes%2075%7D%7B100%7D%3D2.6175rad%2Fsec)
Time t = 5.5 sec
From first equation of motion we know that '
![\alpha =\frac{\omega -\omega _0}{t}=\frac{2.6175-3.49}{4}=-0.158rad/sec^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%5Cfrac%7B%5Comega%20-%5Comega%20_0%7D%7Bt%7D%3D%5Cfrac%7B2.6175-3.49%7D%7B4%7D%3D-0.158rad%2Fsec%5E2)
(a) Now final velocity ![\omega =0rad/sec](https://tex.z-dn.net/?f=%5Comega%20%3D0rad%2Fsec)
So time t to come in rest ![t=\frac{0-3.49}{-0.158}=22sec](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B0-3.49%7D%7B-0.158%7D%3D22sec)
(b) The work done in coming rest is given by
![\frac{1}{2}I\left ( \omega ^2-\omega _0^2 \right )=\frac{1}{2}\times 0.035\times (0^2-3.49^2)=0.2131J](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7DI%5Cleft%20%28%20%5Comega%20%5E2-%5Comega%20_0%5E2%20%5Cright%20%29%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%200.035%5Ctimes%20%280%5E2-3.49%5E2%29%3D0.2131J)
Answer:
option A is correct because air friction is greater than gravity
Explanation:
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