The horizontal force : f = k*N
k- coefficient of friction
k = f /N
N = m * g = 45 kg * 9.81 m/s² = 441.45 N
k = 25 N : 441.45 N = 0.057
Answer C) 0.057
Answer:
The force pulling the roller along the ground is 128.55 N
Explanation:
A force of 200 N acting at an angle of 50° with the ground level
This force is pulled a garden roller
We need to find the force pulling the roller along the ground
The force that pulling the roller along the ground is the horizontal
component of the force acting
→ The force acting is 200 N at direction 50° with ground (horizontal)
→ The horizontal component = F cosФ
→ F = 200 N , Ф = 50
→ The horizontal component = 200 cos(50) = 128.55 N
128.55 N is the horizontal component of the force that pulling the
roller along the ground
<em>The force pulling the roller along the ground is 128.55 N</em>
Explanation:
If a body does not cover a equal distance at a equal interval of time it is said to be non uniform motion...
V = d ÷ t --> bc d=vt
V = (76+54)÷(2+5) = 130÷7 = 18.57km/hr
Answer:
v = -v₀ / 2
Explanation:
For this exercise let's use kinematics relations.
Let's use the initial conditions to find the acceleration of the electron
v² = v₀² - 2a y
when the initial velocity is vo it reaches just the negative plate so v = 0
a = v₀² / 2y
now they tell us that the initial velocity is half
v’² = v₀’² - 2 a y’
v₀ ’= v₀ / 2
at the point where turn v = 0
0 = v₀² /4 - 2 a y '
v₀² /4 = 2 (v₀² / 2y) y’
y = 4 y'
y ’= y / 4
We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.
v² = v₀² -2a y’
v² = 0 - 2 (v₀² / 2y) y / 4
v² = -v₀² / 4
v = -v₀ / 2
We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.
