Answer:
a) t = 19.6 s, b) fr = 1.274 10⁴ N
Explanation:
This is a Newton's second law problem
Y Axis
for the cabin
N₁-W₁ = 0
N₁ = W₁
for the trailer
N₂- W₂ = 0
N₂ = W₂
X axis
for the cabin plus trailer, where friction is only in the cabin
fr = (m₁ + m₂) a
the friction force equation is
fr = μ N
we substitute
μ N₁ = (m₁ + m₂) a
μ m₁ g = (m₁ + m₂) a
a = μ g 
let's calculate
a = 0.65 9.8 
a = 1,274 m / s²
a) to find the stopping distance we can use kinematics
Let's slow down the sI system
v₀ = 90 km / h (1000 m / 1km) (1h / 3600s) = 25 m / s
v = v₀ - a t
when it is stopped its speed is zero
0 = v₀ - at
t = v₀ / a
t = 25 / 1.274
t = 19.6 s
b) the friction force is
fr = 0.65 2000 9.8
fr = 1.274 10⁴ N
This is the braking force and also the forces that couple the cars.
Answer:
i dont think anything will happen so maybe A: the wall is larger than you.
Explanation:
i think what is needed is for people to see that alt. energy sources are very useful and inexpensive.
[Assuming that you've written 3.40 kg in 'a', and not 3.90 kg]
(a) 3,400 g x <u>0.001</u> = 3.40 kg [converting grams to kilograms]
(b) 220 cm x <u>0.01</u> = <u>2.2</u> m [converting centimeters to meters]
(c) 9.42 kg x <u>1000</u> = <u>9420</u> g [converting kilograms to grams]
(d) 6.53 m x <u>100</u> = <u>653</u> cm [converting meters to centimeters]
Answer:
x = 3.6 [m]
Explanation:
This problem can be easily solved using a static analysis of forces acting on the ladder, taking into account the respective distances. For easy understanding, a free body diagram should be made.
We perform a sum of force on the X-axis equal to zero, to find that the force exerted by the wall is equal to the friction force on the floor.
Then we perform a summation of forces on the Y axis, to determine that the normal force exerted by the floor is equal to the weight of the ladder.
We know that the friction force is equal to the product of normal force by the coefficient of friction.
In this way, by relating the friction force to the equations deduced above we can find the force exerted by the wall.
Then we make a summation of moments around the base point of the ladder, the equation realized can be seen in the attached image.
In the last analysis we can find the relationship between the horizontal and vertical distance of the ladder, with respect to the wall and the floor.
Then with the complementary analysis of the Pythagorean theorem we can find another additional equation.
The result of the greater distance is 3.6 [m]
