Answer:
a) 113N
b) 0.37
Explanation:
a) Using the Newton's second law:
\sum Fx =ma
Since the crate doesn't move (static), acceleration will be zero. The equation will become:
\sum Fx = 0
\sumFx = Fm - Ff = 0.
Fm is the applied force
Ff is the frictional force
Since Fm - Ff = 0
Fm = Ff
This means that the applied force is equal to the force of friction if the crate is static.
Since applied force is 113N, hence the magnitude of the static friction force will also be 113N
b) Using the formula
Ff = nR
n is the coefficient of friction
R is the reaction = mg
R = 31.2 × 9.8
R = 305.76N
From the formula
n = Ff/R
n = 113/305.76
n = 0.37
Hence the minimum possible value of the coefficient of static friction between the crate and the floor is 0.37
<span>c. They help discard some myths about objects in space.
When we learn something new, then we know more stuff,
and that helps us avoid ignorance and superstition.
The REAL question is:
</span><span>Why should we continue to send robotic spacecrafts,
and NOT spacecrafts piloted by people ?</span>
Answer:
d²x/dt² = - 4dx/dt - 4x is the required differential equation.
Explanation:
Since the spring force F = kx where k is the spring constant and x its extension = 2.45 equals the weight of the 4 kg mass,
F = mg
kx = mg
k = mg/x
= 4 kg × 9.8 m/s²/2.45 m
= 39.2 kgm/s²/2.45 m
= 16 N/m
Now the drag force f = 16v where v is the velocity of the mass.
We now write an equation of motion for the forces on the mass. So,
F + f = ma (since both the drag force and spring force are in the same direction)where a = the acceleration of the mass
-kx - 16v = 4a
-16x - 16v = 4a
16x + 16v = -4a
4x + 4v = -a where v = dx/dt and a = d²x/dt²
4x + 4dx/dt = -d²x/dt²
d²x/dt² = - 4dx/dt - 4x which is the required differential equation
Answer:
Rectangular path
Solution:
As per the question:
Length, a = 4 km
Height, h = 2 km
In order to minimize the cost let us denote the side of the square bottom be 'a'
Thus the area of the bottom of the square, A = 
Let the height of the bin be 'h'
Therefore the total area, 
The cost is:
C = 2sh
Volume of the box, V = 
(1)
Total cost, 
(2)
From eqn (1):
Using the above value in eqn (1):
Differentiating the above eqn w.r.t 'a':
For the required solution equating the above eqn to zero:
a = 4
Also
The path in order to minimize the cost must be a rectangle.
