The exponential decay function is shown as:
At = Ao*2^(-t/h)
Where:
At = Final amount
Ao = original amount
t = time
h = years for half-life to occur
We are not given the original and final amounts, but we are told that after the certain amount of time had passed, only 5% had remained. So if we were to isolate At and Ao in one side of the equation, such that: At/Ao = 0.05, we can solve for the missing variable, which is t.
Substituting:
<span>At/Ao = 2^(-t/h)
</span>0.05 = <span>2^(-t/20)
t = 86.44 years
Therefore, the pottery was made approximately 86.44 years ago.</span>
Answer:
Velocity of Object with 2 kg= 3.390 m/s
Velocity of Object with 3 kg= 3.404 m/s
Explanation:
From the picture, it can be seen that object B is initially at rest while object A is travelling at a speed of 5m/s. After the collision, Object A moves at an angle of 65 degrees while object B moves at an angle of 37 degrees.
We also know that momentum of a closed system is conserved.
Initial momentum along the x-axis = 2*5.5 = 11
Initial momentum along y-axis = 0
Final momentum along x-axis= a*Cos(65)*2 +b*Cos(37) *3= 11 (a is the velocity of object A of 2 kg after collision where as b is the velocity of object B of 3 kg after collision. velocity is multiplied by cosines of the angle from x axis to give the horizontal component of the velocities).
Final momentum along y-axis = a*Sin(65)*2 - b*Sin(37)*3 =0 (We can see that vertical components of velocity are opposite in direction to each other)
Solve both the equations simultaneously for a and b.
They lack a cell nucleus.
Answer:
277.78 hours
Explanation:
The formula for calculating the amount of charge is expressed as;
Q = It
I is the current
t is the time
Given
I =0.05A
Q = 50,000C
Required
Time t
Recall that: Q = It
t = Q/I
t = 50,000/0.05
t = 1,000,000secs
Convert to hours
1,000,000secs = 1,000,000/3600
1,000,000secs = 277.78 hours
Hence it will take 277.78 hours for the charge to flow through the diode
Answer:
C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀) we see that for the same t v₁> v₂
Explanation:
You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.
Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.
Stone 1
y₁ = v₀₁ t + ½ g t²
y₁ = 0 + ½ g t²
Rock2
It comes out a little later, let's say a second later, we can use the same stopwatch
t ’= (t-t₀)
y₂ = v₀₂ t ’+ ½ g t’²
y₂ = 0 + ½ g (t-t₀)²
y₂ = + ½ g (t-t₀)²
Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to
S = y₁ -y₂
S = ½ g t²– ½ g (t-t₀)²
S = ½ g [t² - (t²- 2 t to + to²)]
S = ½ g (2 t t₀ - t₀²)
S = ½ g t₀ (2 t -t₀)
This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.
For t <to. The rock y has not left and the distance increases
For t> = to. the ratio (2t/to-1)> 1 therefore the distance increases as time
passes
Now we can analyze the different statements
A) false. The difference in height increases over time
B) False S increases
C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t v₁> v₂
