Answer:
It is impossible to detect underground water from the surface. Dowsing practitioners refuse to explain their secrets.
Explanation:
Answer:
27 cm squared
Explanation:
have a good rest of ur day
Answer:
The tension is 
The horizontal force provided by hinge 
Explanation:
From the question we are told that
The mass of the beam is
The length of the beam is 
The hanging mass is 
The length of the hannging mass is 
The angle the cable makes with the wall is 
The free body diagram of this setup is shown on the first uploaded image
The force
are the forces experienced by the beam due to the hinges
Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero
So

Now about the x-axis the moment is

=> 
Substituting values


Now about the y-axis the moment is

Now the torque on the system is zero because their is no rotation
So the torque above point 0 is





The horizontal force provided by the hinge is

Now substituting for T


Answer: as the public need for equal access for people with disabilities became understood, laws were enacted to mandate assistive technologies inclusion in almost all public spaces
Explanation: I just took the test but my teachers a füçkįñg čūńt so don't know if the this will help
Answer:
Explanation:
Initially no of atoms of A = N₀(A)
Initially no of atoms of B = N₀(B)
5 X N₀(A) = N₀(B)
N = N₀ 
N is no of atoms after time t , λ is decay constant and t is time .
For A
N(A) = N(A)₀ 
For B
N(B) = N(B)₀ 
N(A) = N(B) , for t = 2 h
N(A)₀
= N(B)₀ 
N(A)₀
= 5 x N₀(A) 
= 5 
= 5 
half life = .693 / λ
For A
.77 = .693 / λ₁
λ₁ = .9 h⁻¹
= 5 
Putting t = 2 h , λ₁ = .9 h⁻¹
= 5 
= 30.25
2 x λ₂ = 3.41
λ₂ = 1.7047
Half life of B = .693 / 1.7047
= .4065 hours .
= .41 hours .