Answer:

Explanation:
Given that,
Mass of a brick, m = 3.1 kg
The dimensions of the brick 225 m x 112 m x 75 m
We need to find the maximum pressure created by the brick. We know that, the force acting per unit area is called pressure exerted. It is given by the formula as follows :

F = mg
A = area with minimum dimensions i.e. 112 m x 75 m
Pressure is maximum when the area is least.
So,

So, the maximum pressure created by brick is
.
Answer:
The upper limit of metamorphism occurs at the pressure and temperature of wet partial melting of the rock in question. Once melting begins, the process changes to an igneous process rather than a metamorphic process. During metamorphism the protolith undergoes changes in texture of the rock and the mineral make up of the rock.
Explanation:
i hope this helps
Answer:
≈19.144°C.
Explanation:
all the details are in the attachment.
Note, that c₁, m₁, t₁ are the parameters of the sample of brass; c₂, m₂ and t₂ are the parameters of the sample of water.
P.S. change the provided design according Your requirements.
Answer:
v = 8.09 m/s
Explanation:
For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.
Let's calculate the energy
starting point. Higher
Em₀ = U = m gh
final point. To go down the slope
Em_f = K = ½ m v²
The work of the friction force is
W = fr L cos 180
to find the friction force let's use Newton's second law
Axis y
N - W_y = 0
N = W_y
X axis
Wₓ - fr = ma
let's use trigonometry
sin θ = y / L
sin θ = 11/110 = 0.1
θ = sin⁻¹ 0.1
θ = 5.74º
sin 5.74 = Wₓ / W
cos 5.74 = W_y / W
Wₓ = W sin 5.74
W_y = W cos 5.74
the formula for the friction force is
fr = μ N
fr = μ W cos θ
Work is friction force is
W_fr = - μ W L cos θ
Let's use the relationship of work with energy
W + ΔU = ΔK
-μ mg L cos 5.74 + (mgh - 0) = 0 - ½ m v²
v² = - 2 μ g L cos 5.74 +2 (gh)
v² = 2gh - 2 μ gL cos 5.74
let's calculate
v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74
v² = 215.6 -150.16
v = √65.44
v = 8.09 m/s