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 position of the sun was originally the primary influence in determining
when people went to sleep and when they woke up. Although it no longer
directly influences us, that pattern is so deeply ingrained in our make-up
that our behavior still largely coincides with the positions of the sun.
-- The position of the Moon was originally the primary influence in determining
the cycle of human female physiology. Although it no longer directly influences
us, that pattern is so deeply ingrained in human make-up that the female cycle
still largely coincides with the positions of the Moon.
Answer: mass x height x gravitational field strength (g)
note: gravitational field strength (g) = 10 N/Kg
55 x 15 x 10 = 8250
gpe = 8250j
Explanation:
The total work <em>W</em> done by the spring on the object as it pushes the object from 6 cm from equilibrium to 1.9 cm from equilibrium is
<em>W</em> = 1/2 (19.3 N/m) ((0.060 m)² - (0.019 m)²) ≈ 0.031 J
That is,
• the spring would perform 1/2 (19.3 N/m) (0.060 m)² ≈ 0.035 J by pushing the object from the 6 cm position to the equilibrium point
• the spring would perform 1/2 (19.3 N/m) (0.019 m)² ≈ 0.0035 J by pushing the object from the 1.9 cm position to equilbrium
so the work done in pushing the object from the 6 cm position to the 1.9 cm position is the difference between these.
By the work-energy theorem,
<em>W</em> = ∆<em>K</em> = <em>K</em>
where <em>K</em> is the kinetic energy of the object at the 1.9 cm position. Initial kinetic energy is zero because the object starts at rest. So
<em>W</em> = 1/2 <em>mv</em> ²
where <em>m</em> is the mass of the object and <em>v</em> is the speed you want to find. Solving for <em>v</em>, you get
<em>v</em> = √(2<em>W</em>/<em>m</em>) ≈ 0.46 m/s
Answer:
The box will be moving at 0.45m/s. The solution to this problem requires the knowledge and application of newtons second law of motion and the knowledge of linear motion. The vertical component of the force Fp acts vertically upwards against the directio of motion. This causes a constant upward force of 23sin45° to act on the box. Fhe frictional force of 13N also acts vertically upwards and so two forces act upwards against rhe force of gravity resulting un a net force of 0.7N acting kn the box. This corresponds to an acceleration of 0.225m/s². So in w.0s after i start to push v = 0.45m/s.
Explanation:
