Answer:
d = 3.5*10^4 m
Explanation:
In order to calculate the displacement of the airplane you need only the information about the initial position and final position of the airplane. THe initial position is at the origin (0,0,0) and the final position is given by the following vector:
The displacement of the airplane is obtained by using the general form of the Pythagoras theorem:
(1)
where x any are the coordinates of the final position of the airplane and xo and yo the coordinates of the initial position. You replace the values of all variables in the equation (1):
hence, the displacement of the airplane is 3.45*10^4 m
Answer:
C
Explanation:
To solve this question, we will need to develop an expression that relates the diameter 'd', at temperature T equals the original diameter d₀ (at 0 degrees) plus the change in diameter from the temperature increase ( ΔT = T):
d = d₀ + d₀αT
for the sphere, we were given
D₀ = 4.000 cm
α = 1.1 x 10⁻⁵/degrees celsius
we have D = 4 + (4x(1.1 x 10⁻⁵)T = 4 + (4.4x10⁻⁵)T EQN 1
Similarly for the Aluminium ring we have
we were given
d₀ = 3.994 cm
α = 2.4 x 10⁻⁵/degrees celsius
we have d = 3.994 + (3.994x(2.4 x 10⁻⁵)T = 3.994 + (9.58x10⁻⁵)T EQN 2
Since @ the temperature T at which the sphere fall through the ring, d=D
Eqn 1 = Eqn 2
4 + (4.4x10⁻⁵)T =3.994 + (9.58x10⁻⁵)T, collect like terms
0.006=5.18x10⁻⁵T
T=115.7K
Answer:
184 feets
Explanation:
Given the data:
time (sec) __ velocity (ft/sec)
0 __________30
1 __________ 54
2 __________56
3 __________34
4 __________ 8
5 __________ 2
6 __________22
Using left end approximation:
(0,1) ___ f(0) = 30
(1,2) ___ f(1) = 54
(2,3) ___f(2) = 56
(3,4) ___f(3) = 34
(4,5) ___f(4) = 8
(5,6) __ f(5) = 2
Hence, the Total distance traveled during the 6 second interval is:
Change ; dT = 1
1 * (30 + 54 + 56 + 34 + 8 + 2) = 184
