Answer:
Part a)
![T_L = 155.4 N](https://tex.z-dn.net/?f=T_L%20%3D%20155.4%20N)
Part b)
![T_R = 379 N](https://tex.z-dn.net/?f=T_R%20%3D%20379%20N)
Explanation:
As we know that mountain climber is at rest so net force on it must be zero
So we will have force balance in X direction
![T_L cos65 = T_R cos80](https://tex.z-dn.net/?f=T_L%20cos65%20%3D%20T_R%20cos80)
![T_L = 0.41 T_R](https://tex.z-dn.net/?f=T_L%20%3D%200.41%20T_R)
now we will have force balance in Y direction
![mg = T_L sin65 + T_Rsin80](https://tex.z-dn.net/?f=mg%20%3D%20T_L%20sin65%20%2B%20T_Rsin80)
![514 = 0.906T_L + 0.985T_R](https://tex.z-dn.net/?f=514%20%3D%200.906T_L%20%2B%200.985T_R)
Part a)
so from above equations we have
![514 = 0.906T_L + 0.985(\frac{T_L}{0.41})](https://tex.z-dn.net/?f=514%20%3D%200.906T_L%20%2B%200.985%28%5Cfrac%7BT_L%7D%7B0.41%7D%29)
![514 = 3.3 T_L](https://tex.z-dn.net/?f=514%20%3D%203.3%20T_L)
![T_L = 155.4 N](https://tex.z-dn.net/?f=T_L%20%3D%20155.4%20N)
Part b)
Now for tension in right string we will have
![T_R = \frac{T_L}{0.41}](https://tex.z-dn.net/?f=T_R%20%3D%20%5Cfrac%7BT_L%7D%7B0.41%7D)
![T_R = 379 N](https://tex.z-dn.net/?f=T_R%20%3D%20379%20N)
Perfectly inelastic collision is type of collision during which two objects collide, stay connected and momentum is conserved. Formula used for conservation of momentum is:
![m_{1} * v_{1} + m_{2} * v_{2} = m_{1} * v'_{1} + m_{2} * v'_{2}](https://tex.z-dn.net/?f=%20m_%7B1%7D%20%2A%20v_%7B1%7D%20%2B%20%20m_%7B2%7D%20%2A%20v_%7B2%7D%20%3D%20%20m_%7B1%7D%20%2A%20v%27_%7B1%7D%20%2B%20%20m_%7B2%7D%20%2A%20v%27_%7B2%7D)
In case of perfectly inelastic collision v'1 and v'2 are same.
We have following information:
m₁=3 kg
m₂=? kg
v₁=x m/s
v₂=0 m/s
v'1 = v'2 = 1/3 * v₁
Now we insert given information and solve for m₂:
3*v₁ + 0*? = 3*1/3*v₁ + m₂*1/3*v₁
3v₁ = v₁ + m₂*1/3*v₁
2v₁ = m₂*1/3*v₁
2 = m₂*1/3
m₂= 6kg
Mass of second mud ball is 6kg.
Answer:
it's D. Make the column wider
Explanation:
B if not please comment back .
Answer:
398.3 m, 334.2 m
Explanation:
The magnitude of the displacement vector is
v = 520 m
And its direction is
measured as north of east.
The x-component of this vector is given by:
![v_x = v_0 cos \theta = (520)cos 40^{\circ}=398.3 m](https://tex.z-dn.net/?f=v_x%20%3D%20v_0%20cos%20%5Ctheta%20%3D%20%28520%29cos%2040%5E%7B%5Ccirc%7D%3D398.3%20m)
While the y-component is given by
![v_y = v_0 sin \theta =(520)sin 40^{\circ}=334.2 m](https://tex.z-dn.net/?f=v_y%20%3D%20v_0%20sin%20%5Ctheta%20%3D%28520%29sin%2040%5E%7B%5Ccirc%7D%3D334.2%20m)