Could be very slow since they’re basically going against the current which is hard so will be going slow
Answer: Hence, ( 30,20 ) will not maximize the profit as it lies inside the solution region.
Explanation:
Answer:
47.76°
Explanation:
Magnitude of dipole moment = 0.0243J/T
Magnetic Field = 57.5mT
kinetic energy = 0.458mJ
∇U = -∇K
Uf - Ui = -0.458mJ
Ui - Uf = 0.458mJ
(-μBcosθi) - (-μBcosθf) = 0.458mJ
rearranging the equation,
(μBcosθf) - (μBcosθi) = 0.458mJ
μB * (cosθf - cosθi) = 0.458mJ
θf is at 0° because the dipole moment is aligned with the magnetic field.
μB * (cos 0 - cos θi) = 0.458mJ
but cos 0 = 1
(0.0243 * 0.0575) (1 - cos θi) = 0.458*10⁻³
1 - cos θi = 0.458*10⁻³ / 1.397*10⁻³
1 - cos θi = 0.3278
collect like terms
cosθi = 0.6722
θ = cos⁻ 0.6722
θ = 47.76°
Answer:
Kinetic energy, E = 133.38 Joules
Explanation:
It is given that,
Mass of the model airplane, m = 3 kg
Velocity component, v₁ = 5 m/s (due east)
Velocity component, v₂ = 8 m/s (due north)
Let v is the resultant of velocity. It is given by :


Let E is the kinetic energy of the plane. It is given by :


E = 133.38 Joules
So, the kinetic energy of the plane is 133.38 Joules. Hence, this is the required solution.
Question
Initially, the baton is spinning about a line through its center at angular velocity 3.00 rad/s. What is its angular momentum? Express your answer in kilogram meters squared per second.
Answer:

Explanation:
The angular momentum L of the baton moving about an axis perpendicular to it, passing through the center of the baton is,

Here, l is the length of the baton.
Substitute 0.120 kg for m, 3 rads/s for ![\omega[\tex] and 0.8 m for l [tex]\begin{array}{c}\\L = \frac{1}{{12}}m{l^2}\omega \\\\ = \frac{1}{{12}}\left( {0.120{\rm{ kg}}} \right){\left( {{\rm{80}}{\rm{.0 cm}}} \right)^2}{\left( {\frac{{1 \times {{10}^{ - 2}}{\rm{m}}}}{{1{\rm{ cm}}}}} \right)^2}\left( {{\rm{3}}{\rm{.00 rad/s}}} \right)\\\\ = 0.0192{\rm{ kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\\\end{array}](https://tex.z-dn.net/?f=%5Comega%5B%5Ctex%5D%20and%200.8%20m%20for%20l%20%5Btex%5D%5Cbegin%7Barray%7D%7Bc%7D%5C%5CL%20%3D%20%5Cfrac%7B1%7D%7B%7B12%7D%7Dm%7Bl%5E2%7D%5Comega%20%5C%5C%5C%5C%20%3D%20%5Cfrac%7B1%7D%7B%7B12%7D%7D%5Cleft%28%20%7B0.120%7B%5Crm%7B%20kg%7D%7D%7D%20%5Cright%29%7B%5Cleft%28%20%7B%7B%5Crm%7B80%7D%7D%7B%5Crm%7B.0%20cm%7D%7D%7D%20%5Cright%29%5E2%7D%7B%5Cleft%28%20%7B%5Cfrac%7B%7B1%20%5Ctimes%20%7B%7B10%7D%5E%7B%20-%202%7D%7D%7B%5Crm%7Bm%7D%7D%7D%7D%7B%7B1%7B%5Crm%7B%20cm%7D%7D%7D%7D%7D%20%5Cright%29%5E2%7D%5Cleft%28%20%7B%7B%5Crm%7B3%7D%7D%7B%5Crm%7B.00%20rad%2Fs%7D%7D%7D%20%5Cright%29%5C%5C%5C%5C%20%3D%200.0192%7B%5Crm%7B%20kg%7D%7D%20%5Ccdot%20%7B%7B%5Crm%7Bm%7D%7D%5E%7B%5Crm%7B2%7D%7D%7D%7B%5Crm%7B%2Fs%7D%7D%5C%5C%5Cend%7Barray%7D)