To find the work done on the particle, the following is the
solution:
Dw = F dx
W = integral over the path ( F(x) dx)
W = integral from 0 to 1 (e^(-x/5 + 5) dx)
W = -5e^(-x/5 + 5) from 0 to 1
W = 135 J
The work done is 135 J.
Answer:
5.90 ft/s^2
Explanation:
There are mixed units in this question....convert everything to miles or feet
and hr to s
28 mi / hr = 41.066 ft/s
Displacement = vo t + 1/2 at^2
599 = 41.066 (8.9) + 1/2 a (8.9^2)
solve for a = ~ 5.90 ft/s^2
Answer:
2352645198509.9604 m/s²
Explanation:
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
M = Mass of black hole = 
= 10000+100 m
= Distance between the nose and the center of the black hole = 10000 m
The difference in the gravitational field in this system is given by
The acceleration is 2352645198509.9604 m/s²
We need to see what forces act on the box:
In the x direction:
Fh-Ff-Gsinα=ma, where Fh is the horizontal force that is pulling the box up the incline, Ff is the force of friction, Gsinα is the horizontal component of the gravitational force, m is mass of the box and a is the acceleration of the box.
In the y direction:
N-Gcosα = 0, where N is the force of the ramp and Gcosα is the vertical component of the gravitational force.
From N-Gcosα=0 we get:
N=Gcosα, we will need this for the force of friction.
Now to solve for Fh:
Fh=ma + Ff + Gsinα,
Ff=μN=μGcosα, this is the friction force where μ is the coefficient of friction. We put that into the equation for Fh.
G=mg, where m is the mass of the box and g=9.81 m/s²
Fh=ma + μmgcosα+mgsinα
Now we plug in the numbers and get:
Fh=6*3.6 + 0.3*6*9.81*0.8 + 6*9.81*0.6 = 21.6 + 14.1 + 35.3 = 71 N
The horizontal force for pulling the body up the ramp needs to be Fh=71 N.
A model of the solar system with the sun in the middle
