To calculate the ideal mechanical advantage for an inclined plane, divide th length of the incline by the height of the incline.
Therefore; IMA = L/h
L= 3.0 m, while h =1.0 m
IMA = 3/1
= 3
Therefore the IMA of the ramp is 3
This means the ramp increases the force that is being exerted by 3 times.
Answer:
to overcome the out of friction we must increase the angle of the plane
Explanation:
To answer this exercise, let's propose the solution of the problem, write Newton's second law. We define a coordinate system where the x axis is parallel to the plane and the other axis is perpendicular to the plane.
X axis
fr - Wₓ = m a (1)
Y axis
N-
= 0
N = W_{y}
let's use trigonometry to find the components of the weight
sin θ = Wₓ / W
cos θ = W_{y} / W
Wₓ = W sin θ
W_{y} = W cos θ
the friction force has the formula
fr = μ N
fr = μ Wy
fr = μ mg cos θ
from equation 1
at the point where the force equals the maximum friction force
in this case the block is still still so a = 0
F = fr
F = (μ mg) cos θ
We can see that the quantities in parentheses with constants, so as the angle increases, the applied force must be less.
This is the force that balances the friction force, any force slightly greater than F initiates the movement.
Consequently, to overcome the out of friction we must increase the angle of the plane
the correct answer is to increase the angle of the plane
Answer:
Explanation:
The cross product of two vectors is given by

Where, θ be the angle between the two vectors and \widehat{n} be the unit vector along the direction of cross product of two vectors.
Here, K x i = - j
As K is the unit vector along Z axis, i is the unit vector along X axis and j be the unit vector along axis.
The direction of cross product of two vectors is given by the right hand palm rule.
So, k x i = j
j x i = - k
- j x k = - i
i x i = 0
Answer:
Take-off velocity = v = 81.39[m/s]
Explanation:
We can calculate the takeoff speed easily, using the following kinematic equation.

where:
a = acceleration = 4[m/s^2]
x = distance = 750[m]
vi = initial velocity = 25 [m/s]
vf = final velocity
![v_{f}=\sqrt{(25)^{2}+(2*4*750) } \\v_{f}=81.39[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3D%5Csqrt%7B%2825%29%5E%7B2%7D%2B%282%2A4%2A750%29%20%7D%20%5C%5Cv_%7Bf%7D%3D81.39%5Bm%2Fs%5D)
Answer:
Explanation:
The volume of a sphere is:
V = 4/3 * π * a^3
The volume charge density would then be:
p = Q/V
p = 3*Q/(4 * π * a^3)
If the charge density depends on the radius:
p = f(r) = k * r
I integrate the charge density in spherical coordinates. The charge density integrated in the whole volume is equal to total charge.





Since p = k*r
Q = p*π^2*r^3 / 2
Then:
p(r) = 2*Q / (π^2*r^3)