Answer:3.87*10^-4
Explanation:
To determine the mass reduction, delta mass Xe, of the xenon nucleus due to its decay, we first use the provided wavelength of the gamma radiation to calculate its frequency via c = freq*wavelength.
From C=f*lambda we set up: 3*10^8=f*3.44*10^-12.
Solving gives frequency F=0.87*10^20 Hz.
Next, we calculate the emitted energy using the equation E=hf, which translates to E=f*Planck's constant.
Thus, E=0.87*10^20*6.62*10^-34, resulting in E=575.94*10^(-16).
This energy is then converted from joules to MeV.
Utilizing the formula E=mc^2, with c^2 = 931.5 MeV/u, enables us to find the reduction in mass, yielding
3.87*10^-4 u.
Answer:
The car that is the furthest from the finish line is: Car III (Choice C).
Explanation:
Here, we seek the car with the lowest overall average speed throughout the race. Thus, the one in last place inherently possesses the slowest average speed.
Since Car III is significantly behind Cars I and II, Choice A and B cannot be correct. Choice D is also not valid, as the positions of the cars are not the same. Lastly, Choice E is incorrect due to sufficient evidence demonstrating that Choice C has the lowest average speed.
B) 14.0 N
To address this inquiry, we need to evaluate the kinetic energy of the box before and after crossing the rough section. The kinetic energy is given by the formula:
E = 0.5 M V^2
where
E = Energy
M = Mass
V = velocity
Now, utilizing the known data, we compute the energy prior and post.
Before:
E = 0.5 M V^2
E = 0.5 * 13.5kg * (2.25 m/s)^2
E = 6.75 kg * 5.0625 m^2/s^2
E = 34.17188 kg*m^2/s^2 = 34.17188 joules
After:
E = 0.5 M V^2
E = 0.5 * 13.5kg * (1.2 m/s)^2
E = 6.75 kg * 1.44 m^2/s^2
E = 9.72 kg*m^2/s^2 = 9.72 Joules
Hence, the box consumed energy equal to 34.17188 J - 9.72 J = 24.451875 J over a length of 1.75 meters. Next, we will calculate the loss per meter by dividing the energy loss by the distance traversed.
24.451875 J / 1.75 m = 13.9725 J/m = 13.9725 N
When we round to one decimal point, we arrive at 14.0 N, which corresponds with option “B.”
Transverse waves propagate in a direction that is at right angles to the movement of the particles (or the medium involved). Hence, the particles would be shifting from east to west, which is perpendicular to the north-south direction of the wave.
