Answer:
29.4 N/m
0.1
Explanation:
a) From the restoring Force we know that :
F_r = —k*x
the gravitational force :
F_g=mg
Where:
F_r is the restoring force .
F_g is the gravitational force
g is the acceleration of gravity
k is the constant force
xi , x2 are the displacement made by the two masses.
Givens:
<em>m1 = 1.29 kg</em>
<em>m2 = 0.3 kg </em>
<em>x1 = -0.75 m </em>
<em>x2 = -0.2 m </em>
<em>g = 9.8 m/s^2 </em>
Plugging known information to get :
F_r =F_g
-k*x1 + k*x2=m1*g-m2*g
k=29.4 N/m
b) To get the unloaded length 1:
l=x1-(F_1/k)
Givens:
m1 = 1.95kg , x1 = —0.75m
Plugging known infromation to get :
l= x1 — (F_1/k)
= 0.1
Hola!
Respuesta:
28,64 m/s.
Explicación:
Datos:
Vo: 6 m/s
Altura o distancia recorrida: 40 m
Gravedad: 9,81 m/s²
El ejercicio puede ser resuelto facilmente utilizando la siguiente formula:
Vf²-Vo²=2×g×h.
Es importante hacer notar que al estar lanzando el ladrilo hacia abajo, el sentido del movimiento sigue el sentido de la gravedad, es decir es necesario que tomes el valor de la gravedad como positivo (+) y no negativo (-) como normalmente se usa.
Sustituyendo se tiene:
Vf²= (6 m/s)²+ (2)×(9,81 m/s²)x(40m)
Vf²=820,8 m²/s²
√Vf²=√820,8 m²/s²
Vf= 28,64 m/s
Que tengas un buen día!
Answer:
Now when it will reach at point B then its normal force is just equal to ZERO
Explanation:
Since we need to cross both the loops so least speed at the bottom must be
also by energy conservation this is gained by initial potential energy
so we will have
now we have
here we have
R = 7.5 m
so we have
Now when it will reach at point B then its normal force is just equal to ZERO
now when it reach point C then the speed will be
![mgh - mg(2R_c) = \frac{1}{2]mv_c^2](https://tex.z-dn.net/?f=mgh%20-%20mg%282R_c%29%20%3D%20%5Cfrac%7B1%7D%7B2%5Dmv_c%5E2)
now normal force at point C is given as
