<span>Answer:The weight of the door creates a CCW torque given by
Tccw = 145 N*3.13 m / 2
You need a CW torque that's equal to that
Tcw = F*2.5 m*sin20</span>
Incomplete question.The complete question is here
Determine the torque applied to the shaft of a car that transmits 225 hp and rotates at a rate of 3000 rpm.
Answer:
Torque=0.51 Btu
Explanation:
Given Data
Power=225 hp
Revolutions =3000 rpm
To find
T( torque )=?
Solution
As
As force moves an object through a distance, work is done on the object. Likewise, when a torque rotates an object through an angle, work is done.
So
Answer:
The minutes hand travels 39.60 cm.
Explanation:
Note: a clock has a shape of a circle, the minutes hand is the radius, and the travel of the minutes hand forms a arc.
Length of an arc = ∅/360(2πr)
L = ∅/360(2πr).................... Equation 1π
Where L = length of an arc, ∅ = angle formed by an arc, r = radius of the arc.
Given: ∅ = 252°, r = 9 cm, π = 3.143.
Substituting these values into equation 1,
L = 252/360(2×3.143×9)
L = 0.7×2×3.143×9
L = 39.60 cm.
Thus the minutes hand travels 39.60 cm.
Answer:
Explanation:
The word 'nun' for thickness, I will interpret in international units, that is, mm.
We will begin by defining the intensity factor for the steel through the relationship between the safety factor and the fracture resistance of the panel.
The equation is,
We know that 
is 33Mpa*m^{0.5} and our Safety factor is 2,
Now we will need to find the average width of both the crack and the panel, these values are found by multiplying the measured values given by 1/2
<em>For the crack;</em>
<em>For the panel</em>
To find now the goemetry factor we need to use this equation
That allow us to determine the allowable nominal stress,
\sigma_{allow} = 208.15Mpa
So to get the force we need only to apply the equation of Force, where
That is the maximum tensile load before a catastrophic failure.
Answer:
you must throw 3 snowballs
Explanation:
We can solve this exercise using the concepts of conservation of the moment, let's define the system as formed by the refrigerator and all the snowballs. Let's write the moment
Initial. Before bumping that refrigerator
p₀ = n m v₀
Where n is the snowball number
Final. When the refrigerator moves
pf = (n m + M) v
The moment is preserved because the forces during the crash are internal
n m v₀ = (n m + M) v
n m (v₀ - v) = M v
n = M/m v/(vo-v)
Let's look for the initial velocity of the balls, suppose the person throws them with the maximum force if it slides in the snow (F = 100N), let's use the second law and Newton
F = m a
a = F / m
The distance the ball travels from zero speed to maximum speed is the extension of the arm (x = 1 m), let's look kinematically for the speed of the balls when leaving the arm
v₁² = v₀² + 2 a x
v₁² = 0+ 2 (100/1) 1
v₁ = 14.14 m / s
This is the initial speed for the crash
v₀ = v = 14.14 m / s
Let's calculate
n = M/m v/ (v₀-v)
n = 10/1 3 / (14.14 -3)
n = 2.7 balls
you must throw 3 snowballs
