Answer:
a) 
b) 
Explanation:
a)
In order to solve this problem, we need to start by remembering how the acceleration is related to the velocity of a particle. We have the following relation:
in other words, the acceleration is defined to be the derivative of the velocity function with respect to time. So let's take our speed function:
u=20-2x
if we take its derivative we get:
du=-2dx
this is the same as writting:
we also know that velocity is defined to be:
so we get that:
a=-2u
when substituting we get that:
a=-2(20-2x)
when expanding we get:
a=-40+4x
and now we can use this equation to find our acceleration at x=3, so:
a=-40+4(3)
a=-40+12
b)
the same applies to this problem with the difference that this will be the rate of change of the temperature per m. So we proceed and take the derivative of the temperature function:
T=200-5x
so the rate of change is 
Answer:
C. The intensity control
Explanation:
The focus adjust the sharpness of the trace. The Intensity knob adjusts the brightness of the trace by adjusting the potential of a grid controlling number of electrons reaching the screen. The level control varies the voltage required to generate a trigger. Slope control determines whether the trigger point is on the rising or the falling edge of a signal.
The correct option is therefore the intensity control.
Answer:
The break force that must be applied to hold the plane stationary is 12597.4 N
Explanation:
p₁ = p₂, T₁ = T₂
The heat supplied = 
× Heating value of jet fuel
The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s
The heat supplied = 
· 
= 20 kg/s
The heat supplied = 20* = 21,350 kJ/s
= 1.15 kJ/kg
T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K
p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa
p₃ = p₂ = 855 kPa
T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K
T₄ = 1413.3 - 204.88 = 1208.42 K
T₅ = 1208.42*(2/2.333) = 1035.94 K
= √(1.333*287.3*1035.94) = 629.87 m/s
The total thrust = × 
= 20*629.87 = 12597.4 N
Therefore;
The break force that must be applied to hold the plane stationary = 12597.4 N.
Answer:
The governing ratio for thin walled cylinders is 10 if you use the radius. So if you divide the cylinder´s radius by its thickness and your result is more than 10, then you can use the thin walled cylinder stress formulas, in other words:
- if
then you have a thin walled cylinder
or using the diameter:
- if
then you have a thin walled cylinder
Answer:
1.176
Explanation:
When the bullets impact the mass they become embedded on it, it is a plastic collision, therefore momentum is conserved.
v2 * (M + mb) = v1 * mb
Where
v1: muzzle velocity of the bullet
M: mass of the bob
mb: mass of the bullet
v2: mass of the bob with the bullet after being hit
v2 = v1 * mb / (M + mb)
Upon being impacted the bob will acquire speed v2, this implies a kinetic energy. The bob will then move and raise a height h. Upon acheiving the maximum height it will have a speed of zero. At that point all kinetic energy will be converted into potential energy.
Ek = 1/2 (M + mb) * v2^2
Ep = (M + mb) * g * h
Ek = Ep
1/2 (M + mb) * v2^2 = (M + mb) * g * h
1/2 * (v1 * mb / (M + mb))^2 = g * h
1/2 * v1^2 * mb^2 / (M + mb)^2 = g * h
v1^2 = g *h * (M+ mb)^2 / (1/2 * mb^2)
The height h that it reaches is related to the length L of the pendulum arm and the angle it forms with the vertical.
h = L * (1 - cos(a))
For the 9 mm:
For the 0.44 caliber:
The ratio is 460 / 391 = 1.176
