Answer:
Explanation:
The problem is based on Newton's law of cooling .
According to Newton's law
dQ / dt = k ( T - T₀ ) ,
dT / dt = k' ( T - T₀ ) ; dT / dt is rate of fall of temperature.
T is average temperature of hot body , T₀ is temperature of surrounding .
In the first case rate of fall of temperature = (210 - 191) / 2.5
= 7.6 degree / s
average temperature T = (210 + 191) /2
= 200.5
Putting in the equation
7.6 = k' ( 200.5 - 64 )
k' = 7.6 / 136.5
= .055677
In the second case :---
In the second case, rate of fall of temperature = (191 - 156) / t
= 35 / t , t is time required.
average temperature T = (156 + 191) /2
= 173.5
Putting in the equation
35 / t = .05567 ( 173.5 - 64 )
t = 5.74 minute .
The instantaneous speed of a particle moving along one straight line is v(t) = ate−6t, where the speed v is measured in meters per second, the time t is measured in seconds, and the magnitude of the constant a is measured in meters per second squared. What is its maximum speed, expressed as a multiple of a? (Do not include units in your answer.)
Answer:
v_max = (1/6)e^-1 a
Explanation:
You have the following equation for the instantaneous speed of a particle:
[tex]v(t)=ate^{-6t}[/tex] (1)
To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:
[tex]\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))][/tex] (2)
where you have use the derivative of a product.
Next, you equal the expression (2) to zero in order to calculate t:
[tex]a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}[/tex]
For t = 1/6 you obtain the maximum speed.
Then, you replace that value of t in the expression (1):
[tex]v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a[/tex]
hence, the maximum speed is v_max = ((1/6)e^-1)a
A Texas cockroach of mass 0.157 kg runs counterclockwise around the rim of a lazy Susan (a circular disk mounted on a vertical axle) that has a radius 14.9 cm, rotational inertia 5.92 x 10-3 kg·m2, and frictionless bearings. The cockroach's speed (relative to the ground) is 2.92 m/s, and the lazy Susan turns clockwise with angular velocity ω0 = 3.89 rad/s. The cockroach finds a bread crumb on the rim and, of course, stops. (a) What is the angular speed of the lazy Susan after the cockroach stops? (b) Is mechanical energy conserved as it stops?
Answer:
-7.23 rad/s
Explanation:
Given that
Mass of the cockroach, m = 0.157 kg
Radius of the disk, r = 14.9 cm = 0.149 m
Rotational Inertia, I = 5.92*10^-3 kgm²
Speed of the cockroach, v = 2.92 m/s
Angular velocity of the rim, w = 3.89 rad/s
The initial angular momentum of rim is
Iw = 5.92*10^-3 * 3.89
Iw = 2.3*10^-2 kgm²/s
The initial angular momentum of cockroach about the axle of the disk is
L = -mvr
L = -0.157 * 2.92 * 0.149
L = -0.068 kgm²/s
This means that we can get the initial angular momentum of the system by summing both together
2.3*10^-2 + -0.068
L' = -0.045 kgm²/s
After the cockroach stops, the total inertia of the spinning disk is
I(f) = I + mr²
I(f) = 5.92*10^-3 + 0.157 * 0.149²
I(f) = 5.92*10^-3 + 3.49*10^-3
I(f) = 9.41*10^-3 kgm²
Final angular momentum of the disk is
L'' = I(f).w(f)
L''= 9.41*10^-3w(f)
Using the conservation of total angular momentum, we have
-0.068 = 9.41*10^-3w(f) + 0
w(f) = -0.068 / 9.41*10^-3
w(f) = -7.23 rad/s
Therefore, the speed of the lazy Susan after the cockroach stops is -7.23 and is directed in the opposite direction of the initial lazy Susan angular speed
b)
The mechanical energy of the cockroach is not converted as it stops
