Answer:
114.86%
Explanation:
In both cases, there is a vertical force equal to the sprinter's weight:
Fy = mg
When running in a circle, there is an additional centripetal force:
Fx = mv²/r
The net force is found with Pythagorean theorem:
F² = Fx² + Fy²
F² = (mv²/r)² + (mg)²
F² = m² ((v²/r)² + g²)
F = m √((v²/r)² + g²)
Compared to just the vertical force:
F / Fy
m √((v²/r)² + g²) / mg
√((v²/r)² + g²) / g
Given v = 12 m/s, r = 26 m, and g = 9.8 m/s²:
√((12²/26)² + 9.8²) / 9.8
1.1486
The force is about 114.86% greater (round as needed).
<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:
Both of the stunt professionals will sustain injuries of the same seriousness
Explanation:
We are being told that both stunt professionals are standing from the same height, therefore they will attain the same equivalent speed at the bottom if we are to look at it from the principle of conservation of energy.
Now; According to principle of momentum; the momentum at which the first stunt professional A hits the ground be equal as the momentum with which stunt professional B will hit the wall.
Thus; both of the stunt professionals will sustain injuries of the same seriousness
The city monitors the steady rise of CO from various sources annually. In the year "C: 2019"<span> (rounded off to the nearest integer) will the CO level exceed the permissible limit.
If this isn't the answer, let me know and i'll figure out what it is. But I believe this is it. :) </span>
Answer:
a. The electric field lines are linear and perpendicular to the plates inside a parallel-plate capacitor, and always from positive plate to the negative plate. If a positive charge is released near the positive plate, then<em> it will follow a linear path towards the negative plate under the influence of electrostatic force, F = Eq</em>, where q is the charge of the particle. The electric field inside a parallel plate capacitor is constant and equal to
This can be calculated by Gauss' Law.
A positive charge always follow the electric field lines when released. Another approach is that the positive plate repels the positive charge and negative plate attracts the positive charge. Therefore, the positive charge follows a path towards the negative charge.
b. The particle moves from the higher potential to the lower potential. <em>The direction of motion is the same as the direction of the force that moves the particle, so the work done on the particle by that force is positive.</em>
