Answer:
The final velocity of the second car is 57 m/s south.
Explanation:
This is an elastic collision between two train cars. In this case, the total kinetic energy between the two bodies will remain the same.
The formula to apply is :

where ;

Given in the question that;

Apply the formula as;

{14650*18}+{3825*11} = {14650 *6} + {3825 * v₂f}
263700+42075=87900 + 3825v₂f
305775 =87900 + 3825v₂f
305775-87900 = 3825v₂f
217875=3825v₂f
217875/3825 =v₂f
56.96 = v₂f
<u>57 m/s = v₂f { nearest whole number}</u>
Refer to the diagram shown below.
W₁ = (4 kg)*(9.8 m/s²) = 39.2 N
W₂ = (1 kg)*(9.8 m/s²) = 9.8 N
The normal reaction on the 4-kg mass is
N = (39.2 N)*cos(25°) = 35.5273 N
The force acting down the inclined plane due to the weight is
F = (39.2 N)*sin(25°) = 16.5666 N
The net force that accelerates the 4-kg mass at a m/²s down the plane is
F - W₂ = (4 kg)*(a m/s²)
4a = 16.5666 - 9.8
a = 1.6917 m/s²
Answer: 1.69 m/s² (nearest hundredth)
Force and Gravity, is what i think it is.
Answer:
Explanation:
There will be reaction force by each vertical post on horizontal plank . Let it be R₁ and R₂ . R₁ is reaction force by the post nearer to woman
Taking torque of all forces about the end far away from the woman
Torque by reaction force = R₁ x 5.5
= 5.5 R₁ upwards
Torque by weight of woman in opposite direction , downwards
= - 804 x ( 5.5 - 1.55 )
= - 3175.8
Torque by weight of the plank in opposite direction , downwards .
= - 27 x 5.5 / 2
= - 74.25
Torque by R₂ will be zero as it passes through the point about which torque is being taken .
Total torque
= 5.5 R₁ - - 3175.8 - - 74.25 = 0 ( For equilibrium )
5.5 R₁ = 3250
R₁ = 590.9 N .
Answer:
(a): the mug hits the floor 0.752m away from the end of the bar.
(b): the speed of the mug at impact are:
V= 4.87 m/s
direction= 70.82º below the horizontal.
Explanation:
Vx= 1.6 m/s
Vy=?
h= 1.1 m
g= 9.8 m/s²
t is the fall time

t=0.47 sec
Vy= g*t
Vy= 4.6 m/s

V= 4.87 m/s
α= tan⁻¹(Vy/Vx)
α= -70.82º