HELP ILL GIVE MOST BRAINLY AND 50 POINTS
HURRY PLEASE component c it is a compound so it will break
Answer:
a) m=336.18N
b) Vn=16.67m/kmol
Vm=0.1459m^3/kg
Explanation:
To calculate the mass of the octane(m):
Number of mole of octane (n) =0.3kmol(given)
Molarmass of octane (M) =114.23kg/kmol
m=n*M
m=(0.3kmol)*(114.23kg/kmol)
m=34.269kg
To calculate for the weight of octane(W):
W=g*m
W=(9.81m/s^2)*(34.269kg)
W=336.18N
b) For specific volumes of Vn and Vm:
Given volume of octane (V) =5m^3
Vm=V/m
Vm=5m^3/34.269kg
Vm=0.1459m^3/kg
And Vn will be :
Vn=V/m=5m^3/0.3kmol
Vn=16.67m/Kmol
Therefore, the answers are:
a) m=336.18N
b) Vn=16.67m/kmol
Vm=0.1459m^3/kg
The lightning efficiency based on the scenario depicted will be C. 56 lumens/Watt, more efficient.
<h3>How to calculate the lightning efficiency</h3>
The efficiency of the incandescent bulb will be:
= 450/40 = 11.25 lumens per watt.
The efficiency of the LED bulb will be:
= 450/8 = 56 lumens per watt.
In this case, the LED bulb is more efficient than the incandescent bulb.
Therefore, the lighting efficiency will be 56 lumens/Watt, more efficient
Learn more lightning efficiency on:
brainly.com/question/25927632
Answer:
a) The additional time required for the truck to stop is <u>8.5 seconds</u>
b) The additional distance traveled by the truck is <u>230.05 ft</u>
Explanation:
Since the acceleration is constant, the average speed is:
(final speed - initial speed) / 2 = 0.75 v0
Since travelling at this speed for 8.5 seconds causes the vehicle to travel 690 ft, we can solve for v0:
0.75v0 * 8.5 = 690
v0 = 108.24 ft/s
The speed after 8.5 seconds is: 108.24 / 2 = 54.12 ft/s
We can now use the following equation to solve for acceleration:


a = -6.367 m/s^2
Additional time taken to decelerate: 54.12/6.367 = 8.5 seconds
Total distance traveled:

0 - 108.24^2 = 2 * (-6.367) * s
solving for s we get total distance traveled = 920.05 ft
Additional Distance Traveled: 920.05 - 690 = 230.05 ft
Answer:
(4.5125 * 10^-3 kg.m^2)ω_A^2
Explanation:
solution:
Moments of inertia:
I = mk^2
Gear A: I_A = (1)(0.030 m)^2 = 0.9*10^-3 kg.m^2
Gear B: I_B = (4)(0.075 m)^2 = 22.5*10^-3 kg.m^2
Gear C: I_C = (9)(0.100 m)^2 = 90*10^-3 kg.m^2
Let r_A be the radius of gear A, r_1 the outer radius of gear B, r_2 the inner radius of gear B, and r_C the radius of gear C.
r_A=50 mm
r_1 =100 mm
r_2 =50 mm
r_C=150 mm
At the contact point between gears A and B,
r_1*ω_b = r_A*ω_A
ω_b = r_A/r_1*ω_A
= 0.5ω_A
At the contact point between gear B and C.
At the contact point between gears A and B,
r_C*ω_C = r_2*ω_B
ω_C = r_2/r_C*ω_B
= 0.1667ω_A
kinetic energy T = 1/2*I_A*ω_A^2+1/2*I_B*ω_B^2+1/2*I_C*ω_C^2
=(4.5125 * 10^-3 kg.m^2)ω_A^2