La velocidad vertical del tanque después de caer 10 m es 14 m/seg .
La velocidad vertical del tanque se calcula mediante la aplicación de la fórmula de velocidad , la componente vertical Vfy, del movimiento horizontal como se muestra a continuación :
Vfy=?
h = 10 m
Fórmula de Velocidad vertical Vfy:
Vfy² = 2*g*h
Vfy= √(2*9.8m/seg2* 10m )
Vfy= 14 m/seg
The correct one is A. Technician Only.
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Answer:
970 kN
Explanation:
The length of the block = 70 mm
The cross section of the block = 50 mm by 10 mm
The tension force applies to the 50 mm by 10 mm face, F₁ = 60 kN
The compression force applied to the 70 mm by 10 mm face, F₂ = 110 kN
By volumetric stress, we have that for there to be no change in volume, the total pressure applied by the given applied forces should be equal to the pressure removed by the added applied force
The pressure due to the force F₁ = 60 kN/(50 mm × 10 mm) = 120 MPa
The pressure due to the force F₂ = 110 kN/(70 mm × 10 mm) = 157.142857 MPa
The total pressure applied to the block, P = 120 MPa + 157.142857 MPa = 277.142857 MPa
The required force, F₃ = 277.142857 MPa × (70 mm × 50 mm) = 970 kN
The basic relationship between wavelength
, frequency f and speed c of an electromagnetic wave is
where c is the speed of light. Substituting numbers, we find:
Work is calculated as the product of Force, Distance, and
angular motion. In this case, the work done by gravity is perpendicular to the motion
of the cart, so θ = 90°
and W=Fdcosθ
W=35.0 N x 20.0 m x cos90
W=0 J
This means that work done perpendicular to the direction of
the motion is always zero.
