The vertical weight carried by the builder at the rear end is F = 308.1 N
<h3>Calculations and Parameters</h3>
Given that:
The weight is carried up along the plane in rotational equilibrium condition
The torque equilibrium condition can be used to solve
We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person
This would lead to:
F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)
F(1cos20)= 197/2(3.10sin20 + 2 cos 20)
Fcos20= 289.55
F= 308.1N
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Answer:
a) P=0.25x10^-7
b) R=B*N2*E
c) N=1.33x10^9 photons
Explanation:
a) the spontaneous emission rate is equal to:
1/tsp=1/3 ms
the stimulated emission rate is equal to:
pst=(N*C*o(v))/V
where
o(v)=((λ^2*A)/(8*π*u^2))g(v)
g(v)=2/(π*deltav)
o(v)=(λ^2)/(4*π*tp*deltav)
Replacing values:
o(v)=0.7^2/(4*π*3*50)=8.3x10^-19 cm^2
the probability is equal to:
P=(1000*3x10^10*8.3x10^-19)/(100)=0.25x10^-7
b) the rate of decay is equal to:
R=B*N2*E, where B is the Einstein´s coefficient and E is the energy system
c) the number of photons is equal to:
N=(1/tsp)*(V/C*o)
Replacing:
N=100/(3*3x10^10*8.3x10^-19)
N=1.33x10^9 photons
Answer:a=v-u/t
=23-8/3
=5m/s hope you got your answer
Explanation:
Speaking that the trash can absorb the momentum. It only should go 5
