Answer:
T = 0.0088 m²/s
Explanation:
given,
initial piezometric elevation = 12.5 m
thickness of aquifer = 14 m
discharge = 28.24 L/s = 0.02824 m³/s
we know

k = 0.629 mm/sec
Transmissibilty
T = k × H
T = 0.629 × 14 × 10⁻³
T = 0.0088 m²/s
Answer:
Explanation:
Block A sits on block B and force is applied on block A . Block A will experience two forces 1) force P and 2 ) friction force in opposite direction of motion . Block B will experience one force that is force of friction in the direction of motion .
Let force on block A be P . friction force on it will be equal to kinetic friction, that is μ mg , where μ is coefficient of friction and m is mass of block A
friction force = .4 x 2.5 x 9.8
= 9.8 N
net force on block A = P - 9.8
acceleration = ( P - 9.8 ) / 2.5
force on block B = 9.8
acceleration = force / mass
= 9.8 / 6
for common acceleration
( P - 9.8 ) / 2.5 = 9.8 / 6
( P - 9.8 ) / 2.5 = 1.63333
P = 13.88 N .
The triangle <span>in the first law of thermodynamics, represents energy that moves from a hot object to a cooler object.</span>
An object will stay in motion unless acted upon by an UNBALANCED force.
Answer:
1) t = 3.45 s, 2) x = 138 m, 3) v_{y} = -33.81 m /s, 4) v = 52.37 m / s
,
5) θ = -40.2º
Explanation:
This is a projectile exercise, as they indicate that the projectile rolls down the cliff, it goes with a horizontal speed when leaving the cliff, therefore the speed is v₀ₓ = 40 m / s.
1) Let's calculate the time that Taardaen reaches the bottom, we place the reference system at the bottom of the cliff
y = y₀ +
t - ½ g t²
When leaving the cliff the speed is horizontal v_{oy}= 0 and at the bottom of the cliff y = 0
0 = y₀ - ½ g t2
t = √ 2y₀ / g
t = √ (2 60 / 9.8)
t = 3.45 s
2) The horizontal distance traveled
x = v₀ₓ t
x = 40 3.45
x = 138 m
3) The vertical velocity at the point of impact
v_{y} = I go - g t
v_{y} = 0 - 9.8 3.45
v_{y} = -33.81 m /s
the negative sign indicates that the speed is down
4) the resulting velocity at this point
v = √ (vₓ² + v_{y}²)
v = √ (40² + 33.8²)
v = 52.37 m / s
5) angle of impact
tan θ = v_{y} / vx
θ = tan⁻¹ v_{y} / vx
θ = tan⁻¹ (-33.81 / 40)
θ = -40.2º
6) sin (-40.2) = -0.6455
7) tan (-40.2) = -0.845
8) when the projectile falls down the cliff, the horizontal speed remains constant and the vertical speed increases, therefore the resulting speed has a direction given by the angle that is measured clockwise from the x axis