<span>When reading a buret, the initial reading should be taken from the top of the glassware and the final volume should still taken at the top. If the buret is completely, the initial volume for most buret would be zero. though, there are some where their initial starts at 50 decreasing to zero.</span>
Answer:
option B
Explanation:
given,
height of building = 0.1 km
ball strikes horizontally to ground at = 65 m
speed at which the ball strike = ?
vertical velocity = 0 m/s
time at which the ball strike
t = 4.53 s
vertical velocity at the time 4.53 s = g × t = 9.8 × 4.53 = 44.39 m/s
horizontal velocity = 
=14.35 m/s
speed of the ball = 
= 46.65 m/s
hence, the speed of the ball just before it strike the ground = 47 m/s
The correct answer is option B
Answer:
v_average = (d₂-d₁) / Δt
this average velocity is not necessarily the velocity of the extreme points,
Explanation:
To resolve the debate, it must be shown that the two have part of the reason, the space or distance between the two points divided by time is the average speed between the points.
v_average = (d₂-d₁) / Δt
this average velocity is not necessarily the velocity of the extreme points, in the only case that it is so is when there is no acceleration.
Therefore neither of them is right.
Answer:
.409 N
Explanation:
For this to balance, the moments around the fulcrum must sum to zero.
On the left you have .21 ( is that down? I will assume it is)
Counterclockwise moments :
.21 * 40 + 1.0 * 20
Clockwise moments :
.5 * 20 + F * 45
these moments must equal each other
.21*40 + 1 *20 = .5 * 20 + F * 45
F = .409 N
Answer:
Index of expansion: 4.93
Δu = -340.8 kJ/kg
q = 232.2 kJ/kg
Explanation:
The index of expansion is the relationship of pressures:
pi/pf
The ideal gas equation:
p1*v1/T1 = p2*v2/T2
p2 = p1*v1*T2/(T2*v2)
500 C = 773 K
20 C = 293 K
p2 = 35*0.1*773/(293*1.3) = 7.1 bar
The index of expansion then is 35/7.1 = 4.93
The variation of specific internal energy is:
Δu = Cv * Δt
Δu = 0.71 * (20 - 500) = -340.8 kJ/kg
The first law of thermodynamics
q = l + Δu
The work will be the expansion work
l = p2*v2 - p1*v1
35 bar = 3500000 Pa
7.1 bar = 710000 Pa
q = p2*v2 - p1*v1 + Δu
q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg
