Answer:
Option D. ²²²₉₀Th
Explanation:
Let the unknown be ⁿₘZ. Thus, the equation becomes:
²²⁶₉₂U —> ⁴₂He + ⁿₘZ
Next, we shall determine n, m and Z. This can be obtained as follow:
For n:
226 = 4 + n
Collect like terms
226 – 4 = n
222 = n
n = 222
For m:
92 = 2 + m
Collect like terms
92 – 2 = m
90 = m
m = 90
For Z:
ⁿₘZ => ²²²₉₀Z => ²²²₉₀Th
Therefore, the complete equation becomes:
²²⁶₉₂U —> ⁴₂He + ⁿₘZ
²²⁶₉₂U —> ⁴₂He + ²²²₉₀Th
Thus, the unknown is ²²²₉₀Th
Explanation:
The given data is as follows.
Length of beam, (L) = 5.50 m
Weight of the beam, (
) = 332 N
Weight of the Suki, (
) = 505 N
After crossing the left support of the beam by the suki then at some overhang distance the beam starts o tip. And, this is the maximum distance we need to calculate. Therefore, at the left support we will set up the moment and equate it to zero.
= 0
x = 
= 
= 0.986 m
Hence, the suki can come (2 - 0.986) m = 1.014 from the end before the beam begins to tip.
Thus, we can conclude that suki can come 1.014 m close to the end before the beam begins to tip.
V^2 =U^2 +2AS
V^2 = 25 ^2 + 2x9.5x10
V^2 = 625 + 1900 = 2525
V = 50.25
Answer:
The two factors that affect the type of Shadows formed are stated below ::
- The distance between the source of land and the object.
- The angel where the ray of light falls on the object.
Answer:
16.32 °C
Explanation:
We are given;
Mass of aluminum bowl; m_b = 0.25 kg
Mass of soup; m_s = 0.8 kg
Thus, formula to find the amount of heat energy for a temperature change of 27.6°C to 0°C is;
Q = (m_b•c_b•Δt) + (m_s•c_s•Δt)
Where;
c_b = 0.215 kcal/(kg•°C)
c_s = 1 kcal/(kg•°C)
ΔT = 27.6 - 0 = 27.6°C
Thus;
Q = (0.25 × 0.215 × 27.6) + (0.8 × 1 × 27.6)
Q = 23.5635 Kcal
Now, the energy that exits to be used to freeze the soup is;
Q' = 424 kJ - Q
Let's convert 424 KJ to Kcal
424 KJ = 424/4.184 Kcal = 101.3384 Kcal
Thus;
Q' = 101.3384 - 23.5635
Q' = 77.7749 Kcal
Amount of heat that's removed is given by;
Q_f = Q' - mL
Where;
m = m_s = 0.8 kg
L = 79.8 kcal/kg
Thus;
Q_f = 77.7749 - (0.8 × 79.8)
Q_f = 13.9349 Kcal
Then final temperature will be;
T_f = Q_f/((m_b•c_b) + (m_s•c_s))
T_f = 13.9349/((0.25 × 0.215) + (0.8 × 1))
T_f = 16.32 °C
