Answer: 2
Explanation:
Each shell can contain only a fixed number of electrons: The first shell can hold up to two electrons, the second shell can hold up to eight (2 + 6) electrons, the third shell can hold up to 18 (2 + 6 + 10) and so on. The general formula is that the nth shell can in principle hold up to 2(n2) electrons.
You will use the height of the bridge from the ground.
Solution:
Formula to be used is y=Viy(t)+g(t^2)/2
Where:
Vi=initial velocity which is 0 m/s
y=10 m
Gravitational acceleration or g =9.8m/s^2
T= time you need
Substitute all the given to the formula
10m=(0m/s)(t)+(9.8m/s^2)(t^2)/2
10mx2=9.8m/s^2(t^2)
Now isolate the variable you want to find which is T or time
10mx2/9.8m/s^2=t^2
20m/9.8m/s^2=t^2
Square root of 2.04= square root of t^2
T=1.43 secs
The answer is 1.43 seconds
Answer:
0.75%
Explanation:
Measured value of melting point of potassium thiocyanate = 174.5 °C
Actual value of melting point of potassium thiocyanate = 173.2 °C
<em>Error in the reading = |Experimental value - Theoretical value|</em>
<em>= |174.5 - 173.2|</em>
<em>= |1.3|</em>
<em>Percentage error = (Error / Theoretical value) × 100</em>
<em>= (1.3 / 173.2)×100</em>
<em>= 0.75 %</em>
∴ Percentage error in the reading is 0.75%
The answer I found was parabola?
Answer:
Approximately 
(assuming that the projectile was launched at angle of 
above the horizon.)
Explanation:
Initial vertical component of velocity:
.
The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing 
is the same as the altitude 
at which this projectile was launched: 
.
Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 
(upwards,) the vertical velocity right before landing would be 
(downwards.) The change in vertical velocity is:
.
Since there is no drag on this projectile, the vertical acceleration of this projectile would be 
. In other words, 
.
Hence, the time it takes to achieve a (vertical) velocity change of 
would be:
.
Hence, this projectile would be in the air for approximately .
