Use Charles Law: V1/T1 = V2/T2
0.30 m^3/27 C = V2/127 C
27V2 = 127 * 0.3
V2= 38.1/27 = 1.4 m^3
Answer:
b
Explanation:
bbbbbbbbbbbbbbvgh c tyvftj xf
To find the surface area of a single cube we first nees to take the cube root of 8cm3 which is 2.
Now we know that the length of each side is 2 and we can find the area of one side by doing 2x2 which is 4.
To find the total surface area of one cube we do 4 times 6 side giving us a total of 24cm2.
To find the total surface area of the 8 individual cubes, we multiply 24cm2 by 8 to give us a total of 192cm2.
Now to find the total surface area of the one large cube, we know that each side of one of the small cubes is 4cm2 and the large cube is set up so that there are two levels of four cubes right on top of each other. So, the total area of each side of the large cube is 4cm2 times 4 which gives us 16cm2.
Then we multiply 16cm2 by 6 sides to give us a total surface area of 96cm2.
The ratio of the surface area of the single large cube comapred to the total surface area of the single cubes is 96:192
We can further simplify this ratio:
96:192
48:96
24:48
12:24
6:12
3:6
1:2
Answer:
v = 57.2 m/s
Explanation:
The average velocity of the train can be defined as the total distance covered by the train divided by the time taken by the train to cover that distance. Therefore, we will use the following formula to find the average velocity of the train:
v = s/t
where,
s = distance covered = 460 km = (460 km)(1000 m/1 km) = 4.6 x 10⁵ m
t = time taken to cover the distance = 2 h 14 min
Now, we convert it into minutes:
t = (2 h)(60 min/1 h) + 14 min
t = 120 min + 14 min = (134 min)(60 s/1 min)
t = 8040 s
Therefore, the value of velocity will be:
v = (4.6 x 10⁵ m)/8040 s
<u>v = 57.2 m/s</u>
Answer:
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Explanation:
The additional information to the question is embedded in the diagram attached below:
The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m
Balancing the equilibrium about point A;
F(1.1) - mg (1.25) = 
- 1200(9.8)(1.25) = 1200a(0.35)
- 14700 = 420 a ------- equation (1)
--------- equation (2)
Replacing equation 2 into equation 1 ; we have :

1320 a - 14700 = 420 a
1320 a - 420 a =14700
900 a = 14700
a = 14700/900
a = 16.33 m/s²
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²