Answer:
∑ τ =0, L₀ = ![L_{f}](https://tex.z-dn.net/?f=L_%7Bf%7D)
Explanation:
In a circular turning movement, when the arms are extended and then contracted in two possibilities:
- They are lowered the force of gravity is what pulls them, the tension of the muscle becomes zero to allow this movement.
In this movement the force is vertical(gravity) and the movement of the center of mass of each arm is vertical, so that the work is the weight value of the arm by the distance traveled by the center of mass.
- Another possibility is that the arms have stuck to the body, in this case the person's muscles perform the force, this force is horizontal and the displacement is the horizontal of the center of mass of the arms from the extended position to the contracted
In these movements the torque of the external force is equal for each arm, but in the opposite direction, so they are canceled where a net torque of zero, this causes the angular momentum to be preserved, which changes is the moment of inertia of the system and therefore you must also change the angular velocity to keep your product constant
∑ τ =0
L₀ = ![L_{f}](https://tex.z-dn.net/?f=L_%7Bf%7D)
I₀ w₀ = I w
A charged particle moving in a magnetic field experiences a force equal to:
![\vec{F}=q\vec{v}\times \vec{B}](https://tex.z-dn.net/?f=%5Cvec%7BF%7D%3Dq%5Cvec%7Bv%7D%5Ctimes%20%5Cvec%7BB%7D)
Thus, the magnitude of the force that the proton experiences is given by:
![F=qvBsin\theta](https://tex.z-dn.net/?f=F%3DqvBsin%5Ctheta)
The magnetic field is perpendicular to the proton's velocity, therefore, we have
. Replacing the given values, we obtain:
![F=1.6*10^{-19}C(300\frac{m}{s})(1T)sin(90^\circ)\\F=4.8*10^{-17}N](https://tex.z-dn.net/?f=F%3D1.6%2A10%5E%7B-19%7DC%28300%5Cfrac%7Bm%7D%7Bs%7D%29%281T%29sin%2890%5E%5Ccirc%29%5C%5CF%3D4.8%2A10%5E%7B-17%7DN)
Answer:
Resultant force = 8.6N
Explanation:
Using Pythogorus' theorem
![{a}^{2} + {b}^{2} = {c }^{2} \\ {7}^{2} + {5}^{2} = {r}^{2} \\ 49 + 25 = {r}^{2} \\ 74 = {r}^{2} \\ \sqrt{74} = \sqrt{ {r }^{2} } \\ 8.6 = r \\](https://tex.z-dn.net/?f=%20%20%7Ba%7D%5E%7B2%7D%20%20%2B%20%20%7Bb%7D%5E%7B2%7D%20%20%3D%20%20%7Bc%20%7D%5E%7B2%7D%20%20%5C%5C%20%20%7B7%7D%5E%7B2%7D%20%20%2B%20%20%7B5%7D%5E%7B2%7D%20%20%3D%20%20%7Br%7D%5E%7B2%7D%20%5C%5C%2049%20%2B%2025%20%3D%20%20%7Br%7D%5E%7B2%7D%20%20%5C%5C%2074%20%3D%20%20%7Br%7D%5E%7B2%7D%20%20%5C%5C%20%20%20%5Csqrt%7B74%7D%20%20%3D%20%20%20%5Csqrt%7B%20%7Br%20%7D%5E%7B2%7D%20%7D%20%20%5C%5C%208.6%20%3D%20r%20%5C%5C%20)
Resultant force = 8.6N
Answer:
I just noticd i dont speak this launguage
Explanation:
Answer:
N
N
Explanation:
= 1 A
= 4 A
= distance between the two wire = 5 m
= Force per unit length acting between the two wires
Force per unit length acting between the two wires is given as
![F = \frac{\mu _{o}}{4\pi }\frac{2i_{1}i_{2}}{r}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7B%5Cmu%20_%7Bo%7D%7D%7B4%5Cpi%20%7D%5Cfrac%7B2i_%7B1%7Di_%7B2%7D%7D%7Br%7D)
![F = (10^{-7})\frac{2(1)(4)}{5}](https://tex.z-dn.net/?f=F%20%3D%20%2810%5E%7B-7%7D%29%5Cfrac%7B2%281%29%284%29%7D%7B5%7D)
N
= distance of each wire from the midpoint = 2.5 m
Magnetic field midway between the two wires is given as
![B = \frac{\mu _{o}}{4\pi } \left \left ( \frac{2i_{2}}{r'} \right - \frac{2i_{1}}{r'} \right \right ))](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7B%5Cmu%20_%7Bo%7D%7D%7B4%5Cpi%20%7D%20%5Cleft%20%5Cleft%20%28%20%5Cfrac%7B2i_%7B2%7D%7D%7Br%27%7D%20%5Cright%20-%20%5Cfrac%7B2i_%7B1%7D%7D%7Br%27%7D%20%5Cright%20%5Cright%20%29%29)
![B = (10^{-7}) \left \left ( \frac{2(4)}{2.5} \right - \frac{2(1)}{2.5} \right \right ))](https://tex.z-dn.net/?f=B%20%3D%20%2810%5E%7B-7%7D%29%20%5Cleft%20%5Cleft%20%28%20%5Cfrac%7B2%284%29%7D%7B2.5%7D%20%5Cright%20-%20%5Cfrac%7B2%281%29%7D%7B2.5%7D%20%5Cright%20%5Cright%20%29%29)
![B = 2.4\times 10^{-7}](https://tex.z-dn.net/?f=B%20%3D%202.4%5Ctimes%2010%5E%7B-7%7D)