The relatonship between Kalven and celsius______
1 answer:
Answer:
Explanation:
Kelvin's climbing represents the <em>absolute temperature</em>. Temperature is a measure of the molecular kinetic energy of translation. If the molecules move quickly, with the same energy as in the walls of the container, which makes us feel like "heat". If the molecules do not move, the temperature is zero. 0 K.
The Celsius scale has an <em>artificial zero</em>, defined in the solidification temperature of the water. It is very useful to talk about the weather, and about some simpler technical matters. But it is artificial.
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Answer:
V_t=6 = 32 m/s
Explanation:
The origin is at point 0 with the positive motion to the right
The instantaneous acceleration is change of velocity measured at infinitesimal interval of time, so the expression for instantaneous acceleration is:
a=dv/dt
From here we can express dv as:
dv = a dt
Replace a by 2t — 1
dv = (2t — 1) dt
Integrate both sides of equation
v=t
a=t_0
putting these value in integral
<em>v-v_0=(t^2-t)-(t_0^2-t_0)</em>
We know that v_0 = 2 at t_0 = 0, so we'll replace t_0 and v_0 by their values
v — 2 = (t^2 — t) — (0^2 — 0)
From here we can write the expression for v as:
v_t=6=6^2-6+2 (1)
So the velocity at t = 6 s is:
v_t=6 = 32 m/s
V_t=6 = 32 m/s
In order to determine the total distance travelled, we must check how maw times the particle has changed its direction, i.e. how many times its speed was equal to zero
To do that, we'll just replace v by 0 in expression (1)
0 = t^2 — t + 2
The roots of the quadratic equation are:
t_1/2=1± √(1^2-4*2*1)/2
Since 1^2-4*2*1 < 0, the quadratic equation have no real roots, so we can say that the velocity is always positive, i.e. to the right
Now that we have all the details, we can correctly draw the path of the particle
We can see from the sketch that the total distance traveled is:
s^T=Δs_0-1
s^T=| s_1 - s_0 |
Replace s_0 by its value
s^T=| s_1 - 1 | (2)
In order to determine the position of particle at t = 6 s, we'll need to determine the expression for s as function of time
Since we have already wrote expression for v as function of time (step 2), we'll use expression to get the expression for s
v= ds/dt
Multiply both sides of equation by dt
v dt = ds
Replace v by expression (1)
(t^2 — t + 2) dt = ds
Integrate both sides of equation
t=s
b=(s=0)
x=(t^2 — t + 2)
dx=ds
putting these value in integral
(t^3/3-t^2/2+t)-(t_0^3/3-t_0^2/2+t_0)= s-s_0
Since s = 1 m at t = 0, and we want to determine the position s at t = 6, we'll replace so by 1, t_0 by 0 and t by 6
(6^3/3-6^2/2+6)-(0^3/3-0^2/2+0)=s_t=6-1
Image of wheel is missing, so i attached it.
Answer:
ω = 14.95 rad/s
Explanation:
We are given;
Mass of wheel; m = 20kg
T = 20 N
k_o = 0.3 m
Since the wheel starts from rest, T1 = 0.
The mass moment of inertia of the wheel about point O is;
I_o = m(k_o)²
I_o = 20 * (0.3)²
I_o = 1.8 kg.m²
So, T2 = ½•I_o•ω²
T2 = ½ × 1.8 × ω²
T2 = 0.9ω²
Looking at the image of the wheel, it's clear that only T does the work.
Thus, distance is;
s_t = θr
Since 4 revolutions,
s_t = 4(2π) × 0.4
s_t = 3.2π
So, Energy expended = Force x Distance
Wt = T x s_t = 20 × 3.2π = 64π J
Using principle of work-energy, we have;
T1 + W = T2
Plugging in the relevant values, we have;
0 + 64π = 0.9ω²
0.9ω² = 64π
ω² = 64π/0.9
ω = √64π/0.9
ω = 14.95 rad/s
