Answer:
<em>866.1 N</em>
Explanation:
The torque on the flywheel = 300 N-m
The force from the hydraulic cylinder will generate a moment on CA about point A.
The part of this moment that will be at point B about A must be proportional to the torque on the cylinder which is 300 N-m
we know that moment = F x d
where F is the force, and
d is the perpendicular distance from the turning point = 1 m
Equating, we have
300 = F x 1
F = 300 N this is the frictional force that stops the flywheel
From F = μN
where F is the frictional force
μ is the coefficient of static friction = 0.4
N is the normal force from the hydraulic cylinder
substituting, we have
300 = 0.4 x N
N = 300/0.4 = 750 N
This normal force calculated is perpendicular to CA. This actual force, is at 30° from the horizontal. To get the force from the hydraulic cylinder R, we use the relationship
N = R sin (90 - 30)
750 = R sin 60°
750 = 0.866R
R = 750/0.866 = <em>866.1 N</em>
Answer:
0.064 mg/kg/day
6.25% from water, 93.75% from fish
Explanation:
Density of water is 1 kg/L, so the concentration of the chemical in the water is 0.1 mg/kg.
The BCF = 10³, so the concentration of the chemical in the fish is:
10³ = x / (0.1 mg/kg)
x = 100 mg/kg
For 2 L of water and 30 g of fish:
2 kg × 0.1 mg/kg = 0.2 mg
0.030 kg × 100 mg/kg = 3 mg
The total daily intake is 3.2 mg. Divided by the woman's mass of 50 kg, the dosage is:
(3.2 mg/day) / (50 kg) = 0.064 mg/kg/day
b) The percent from the water is:
0.2 mg / 3.2 mg = 6.25%
And the percent from the fish is:
3 mg / 3.2 mg = 93.75%
Answer:
it allows your dash board to light up you MPH RPM and all the other numbers on the spadomter
Explanat
Answer:
μ=0.329, 2.671 turns.
Explanation:
(a) ln(T2/T1)=μβ β=angle of contact in radians
take T2 as greater tension value and T1 smaller, otherwise the friction would be opposite.
T2=5000 lb and T1=80 lb
we have two full turns which makes total angle of contact=4π radians
μ=ln(T2/T1)/β=(ln(5000/80))/4π
μ=0.329
(b) using the same relation as above we will now compute the angle of contact.
take greater tension as T2 and smaller as T1.
T2=20000 lb T1=80 lb μ=0.329
β=ln(20000/80)/0.329=16.7825 radians
divide the angle of contact by 2π to obtain number of turns.
16.7825/2π =2.671 turns
Answer:
(a) 2.39 MPa (b) 3.03 kJ (c) 3.035 kJ
Explanation:
Solution
Recall that:
A 10 gr of air is compressed isentropically
The initial air is at = 27 °C, 110 kPa
After compression air is at = a450 °C
For air, R=287 J/kg.K
cv = 716.5 J/kg.K
y = 1.4
Now,
(a) W efind the pressure on [MPa]
Thus,
T₂/T₁ = (p₂/p₁)^r-1/r
=(450 + 273)/27 + 273) =
=(p₂/110) ^0.4/1.4
p₂ becomes 2390.3 kPa
So, p₂ = 2.39 MPa
(b) For the increase in total internal energy, is given below:
ΔU = mCv (T₂ - T₁)
=(10/100) (716.5) (450 -27)
ΔU =3030 J
ΔU =3.03 kJ
(c) The next step is to find the total work needed in kJ
ΔW = mR ( (T₂ - T₁) / k- 1
(10/100) (287) (450 -27)/1.4 -1
ΔW = 3035 J
Hence, the total work required is = 3.035 kJ
