<span>We'll use the momentum-impulse theorem. The x-component of the total momentum in that direction is given by p_(f) = p_(1) + p_(2) + p_(3) = 0.
So p_(1x) = m1v1 = 0.2 * 2 = 0.4 Also p_(2x) = m2v2 = 0 and p_(3x) = m3v3 = 0.1 *v3 where v3 is unknown speed and m3 is the mass of the third particle with the unknown speed
Similarly, the 235g particle, y-component of the total momentum in that direction is given by p_(fy) = p_(1y) + p_(2y) + p_(3y) = 0.
So p_(1y) = 0, p_(2y) = m2v2 = 0.235 * 1.5 = 0.3525 and p_(3y) = m3v3 = 0.1 * v3 where m3 is third particle mass.
So p_(fx) = p_(1x) + p_(2x) + p_(3x) = 0.4 + 0.1v3; v3 = 0.4/-0.1 = - 4
Also p_(fy) = 0.3525 + 0.1v3; v3 = - 0.3525/0.1 = -3.525
So v_3x = -4 and v_3y = 3.525.
The speed is their resultant = âš (-4)^2 + (-3.525)^2 = 5.335</span>
Answer:
A driver.
Explanation:
Using a driver while at least 350 yds away is better than using a iron, because it will be a waste of the par 4 as it is not as powerful as the driver.
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°
