First we calculate the acceleration of the cart:
a = (v₂ - v₁) / (t)
a = (7 m/s - 3 m/s) / (8 s)
a = 4 m/s / 8 s
a = 0.5 m/s²
Using the Newtons Second Law:
F = m×a
F = (10 kg)(0.5 m/s²)
F = 5 N
<h2>The correct option is C</h2>
The answer is "Deflate because volume is directly proportional to temperature"
In a force diagram set-up, we name the angle of inclination theta, g as the acceleration due to gravity. In this case, the forces acting on the box going down is the weight itself impeded by the friction between the box and the inclined plane.
The weight of the box is expressed as mg sin theta
The frictional force is expressed as the normal force times the coefficient of friction that is expressed as mu g cos theta.
By Newton's second law of motion, F = ma = mg sin theta - mu g cos theta
Thus, a = g (sin theta - u cos theta
Answer:
0.705 m/s²
Explanation:
a) The sprinter accelerates uniformly from rest and reaches a top speed of 35 km/h at the 67-m mark.
Using newton's law of motion:
v² = u² + 2as
v = final velocity = 35 km/h = 9.72 m/s, u = initial velocity = 0 km/h, s = distance = 67 m
9.72² = 0² + 2a(67)
134a = 94.484
a = 0.705 m/s²
b) The sprinter maintains this speed of 35 km/h for the next 88 meters. Therefore:
v = 35 km/h = 9.72 m/s, u = 35 km/h = 9.72 m/s, s = 88 m
v² = u² + 2as
9.72² = 9.72² + 2a(88)
176a = 9.72² - 9.72²
a = 0
c) During the last distance, the speed slows down from 35 km/h to 32 km/h.
u = 35 km/h = 9.72 m/s, v = 32 km/h = 8.89 m/s, s = 200 - (67 + 88) = 45 m
v² = u² + 2as
8.89² = 9.72² + 2a(45)
90a = 8.89² - 9.72²
90a = -15.4463
a = -0.1716 m/s²
The maximum acceleration is 0.705 m/s² which is from 0 to 67 m mark.
Answer:
The maximum static frictional force is 40N.
Explanation:
When an object of mass M is on a surface with a coefficient of static friction μ, there is a minimum force that you need to apply to the object in order to "break" the coefficient of static friction and be able to move the object (Called the threshold of motion, once the object is moving we have a coefficient of kinetic friction, which is smaller than the one for static friction).
This coefficient defines the maximum static friction force that we can have.
So if we apply a small force and we start to increase it, the static frictional force will be equal to our force until it reaches its maximum, and then we can move the object and now we will have frictional force.
In this case, we know that we apply a force of 40N and the object just starts to move.
Then we can assume that we are just at the point of transition between static frictional force and kinetic frictional force (the threshold of motion), thus, 40 N is the maximum of the static frictional force.
