Answer:
37.125 m
Explanation:
Using the equation of motion
s=ut+0.5at^{2} where s is distance, u is initial velocity, t is time and a is acceleration
<u>Distance during acceleration</u>
Acceleration, a=\frac {V_{final}-V_{initial}}{t} where V_{final} is final velocity and V_{initial} is initial velocity.
Substituting 0.0 m/s for initial velocity and 4.5 m/s for final velocity, acceleration will be
a=\frac {4.5 m/s-0 m/s}{4.5 s}=1 m/s^{2}
Then substituting u for 0 m/s, t for 4.5 s and a for 1 m/s^{2} into the equation of motion
s=0*4.5+ 0.5*1*4.5^{2}=0+10.125
=10.125 m
<u>Distance at a constant speed</u>
At a constant speed, there's no acceleration and since speed=distance/time then distance is speed*time
Distance=4.5 m/s*6 s=27 m
<u>Total distance</u>
Total=27+10.125=37.125 m
Answer:
assume nitrogen is an ideal gas with cv=5R/2
assume argon is an ideal gas with cv=3R/2
n1=4moles
n2=2.5 moles
t1=75°C <em>in kelvin</em> t1=75+273
t1=348K
T2=130°C <em>in kelvin</em> t2=130+273
t2=403K
u=пCVΔT
U(N₂)+U(Argon)=0
<em>putting values:</em>
=>4x(5R/2)x(Tfinal-348)=2.5x(3R/2)x(T final-403)
<em>by simplifying:</em>
Tfinal=363K
Answer:
W = 0.135 N
Explanation:
Given:
- y (x, t) = 8.50*cos(172*x -2730*t)
- Weight of string m*g = 0.0126 N
- Attached weight = W
Find:
The attached weight W given that Tension and W are equal.
Solution:
The general form of standing mechanical waves is given by:
y (x, t) = A*cos(k*x -w*t)
Where k = stiffness and w = angular frequency
Hence,
k = 172 and w = 2730
- Calculate wave speed V:
V = w / k = 2730 / 172 = 13.78 m/s
- Tension in the string T:
T = Y*V^2
where Y: is the mass per unit length of the string.
- The tension T and weight attached W are equal:
T = W = Y*V^2 = (w/L*g)*V^2
W = (0.0126 / 1.8*9.81)*(13.78)^2
W = 0.135 N