An element can be identified by its unique atomic number. When we look in the periodic table, we find that the element with an atomic number of 9292 is uranium. There is only option containing uranium which also confirms the mass number we found. So, the daughter nucleus of the decay is 234^U.
In an alpha decay, a positively charged particle similar to a helium-4 nucleus gets released from the parent nucleus spontaneously. As the composition suggests, an alpha particle consists of two protons and 2 neutrons. The particle does not travel much, but in short range, it carries the most energy.
It's smart to use the thermal energy provided by the radioactive decay to generate electricity. This allows for a stable supply of power without consuming much space which means the saved space can be used for more scientific equipment. The alpha particle, structurally equivalent to the nucleus of a helium atom.
Learn more about nucleus here:
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Answer:
W = 9.93 10² N
Explanation:
To solve this exercise we must use the concept of density
ρ = m / V
the tabulated density of copper is rho = 8966 kg / m³
let's find the volume of the cylindrical tube
V = A L
V = π (R_ext ² - R_int ²) L
let's calculate
V = π (4² - 2²) 10⁻⁴ 3
V = 1.13 10⁻² m³
m = ρ V
m = 8966 1.13 10⁻²
m = 1.01 10² kg
the weight of the tube
W = mg
W = 1.01 10² 9.8
W = 9.93 10² N
Answer:
The Richter scale measures the largest wiggle (amplitude) on the recording, but other magnitude scales measure different parts of the earthquake. The USGS currently reports earthquake magnitudes using the Moment Magnitude scale, though many other magnitudes are calculated for research and comparison purposes.
True if you look up the question Is velocity speed in a certain direction you would’ve gotten the answer but I’m pretty sure it’s true
The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height.
<span>In that particular situation, you can prove it like this: </span>
<span>initial velocity is Vo </span>
<span>launch angle is α </span>
<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>
<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>
<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>
<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>
<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>
<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>
<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
