Answer:
μ = 0.350
Explanation:
For the person to able to move the box, the force exerted by the person on the box must equal the force exerted by the box:
In this case, force can be calculated as a product of mass (m) by the acceleration of gravity (g) and the coefficient of static friction (μ):
Therefore, for the person to be able to push the box horizontally, the coefficient of static friction between the box and the floor should not be higher than 0.350.
Answer:
x = v₀ cos θ t , y = y₀ + v₀ sin θ t - ½ g t2
Explanation:
This is a projectile launch exercise, in this case we will write the equations for the x and y axes
Let's use trigonometry to find the components of the initial velocity
sin θ = / v₀
cos θ = v₀ₓ / v₀
v_{y} = v_{oy} sin θ
v₀ₓ = vo cos θ
now let's write the equations of motion
X axis
x = v₀ₓ t
x = v₀ cos θ t
vₓ = v₀ cos θ
Y axis
y = y₀ + t - ½ g t2
y = y₀ + v₀ sin θ t - ½ g t2
v_{y} = v₀ - g t
v_{y} = v₀ sin θ - gt
= v_{oy}^2 sin² θ - 2 g y
As we can see the fundamental change is that between the horizontal launch and the inclined launch, the velocity has components
Answer:
Explanation:
Given that
Length= 2L
Linear charge density=λ
Distance= d
K=1/(4πε)
The electric field at point P
So
Now by integrating above equation
