La cinemática permite encontrar la respuesta para la aceleracion del cuerpo en el cañón es:
a = 1,8 10⁶ m/s²
La cinemática estudia el movimiento de los cuerpos, buscando relaciones entre la posicion, la velocidad y la aceleración.
v² = v₀² + 2 a x
Donde v y v₀ son la velocidad actual e inicial, respectivamente, a es la aceleracion y x la distancia recorrida.
Indica que la longitud de cañon es x= 18 m la velocidad de salida es
v= 29000 km/h ( ) (
s) = 8,055,56 m/s.
La velocidad inicial del proyectil es cero.
a =
a =
a = 1,8 10⁶ m/s²
En conclusión usando la cinemática podemos encontrarla respuesta para la aceleracion del cuerpo en el cañón es:
a = 1,8 10⁶ m/s²
Aprender mas aquí: brainly.com/question/19793086
This question involves the concepts of orbital velocity and orbital radius.
The orbital velocity of ISS must be "7660.25 m/s".
The orbital velocity of the ISS can be given by the following formula:
where,
v = orbital velocity = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of Earth = 5.97 x 10²⁴ kg
R = orbital radius = radius of earth + altitude = 63.78 x 10⁵ m + 4.08 x 10⁵ m
R = 67.86 x 10⁵ m
Therefore,
<u>v = 7660.25 m/s</u>
Learn more about orbital velocity here:
Answer:
vB = 15.4 m/s
Explanation:
Principle of conservation of energy:
Because there is no friction the mechanical energy is conserve
ΔE = 0
ΔE : mechanical energy change (J)
K : Kinetic energy (J)
U: Potential energy (J)
K = (1/2)mv²
U = m*g*h
Where :
m: mass (kg)
v : speed (m/s)
h : hight (m)
Ef - Ei = 0
(K+U)final - (K+U)initial =0
(K+U)final = (K+U)initial
((1/2)mv²+m*g*h)final = ((1/2)mv²+m*g*h)initial , We divided by m both sides of the equation:
((1/2)vB² + g*hB = (1/2 )vA²+ g*hA
(1/2) (vB)² + (9.8)*(14.7) = 0 + (9.8)(26.8 )
(1/2) (vB)² = (9.8)(26.8 ) - (9.8)*(14.7)
(vB)² = (2)(9.8)(26.8 - 14.7)
(vB)² = 237.16
vB = 15.4 m/s : speed of the cart at B
Answer:
Explanation:
-The chemical formula for Molybdenum (V) Dichromate is
-There are 21 moles of oxygen per one mole of Molybdenum (V) Dichromate
-We apply Avogadro's constant to find the number of atoms of oxygen:
Hence, there are
Answer:
a)W=8.333lbf.ft
b)W=0.0107 Btu.
Explanation:
<u>Complete question</u>
The force F required to compress a spring a distance x is given by F– F0 = kx where k is the spring constant and F0 is the preload. Determine the work required to compress a spring whose spring constant is k= 200 lbf/in a distance of one inch starting from its free length where F0 = 0 lbf. Express your answer in both lbf-ft and Btu.
Solution
Preload = F₀=0 lbf
Spring constant k= 200 lbf/in
Initial length of spring x₁=0
Final length of spring x₂= 1 in
At any point, the force during deflection of a spring is given by;
F= F₀× kx where F₀ initial force, k is spring constant and x is the deflection from original point of the spring.
Change to lbf.ft by dividing the value by 12 because 1ft=12 in
100/12 = 8.333 lbf.ft
work required to compress the spring, W=8.333lbf.ft
The work required to compress the spring in Btu will be;
1 Btu= 778 lbf.ft
?= 8.333 lbf.ft----------------cross multiply
(8.333*1)/ 778 =0.0107 Btu.