Answer:
The angular acceleration of CD player is [tex]13.93\ rad/s^2[/tex].
Explanation:
Initial angular speed of a CD player is 0 and final angular speed is 41.8 rad/s. Time to change the angular speed is 3 s.
It is required to find the angular acceleration. The change in angular speed of the CD player divided by time taken is called its angular acceleration. It can be given by :
[tex]a=\dfrac{\omega_f-\omega_i}{t}\\\\a=\dfrac{41.8-0}{3}\\\\a=13.94\ rad/s^2[/tex]
So, the angular acceleration of CD player is [tex]13.93\ rad/s^2[/tex].
What is an independent variable?
A. A variable that is intentionally changed during an experiment
B. A variable that depends on the experimental variable
C. A variable that is not used in an experiment
D. A variable that is unknown during the experiment
Answer:
The answer is A
Explanation:
Independent variables don't have to depend on other factors of the experiment because they're independent
Answer:
A.
Explanation:
Independent variables don't have to depend on other factors of the experiment because they're independent.
I need some help!!!!!!!!!
Answer:
The Object will immediately begin moving toward the left
Explanation:
Because the force of thirteen is greater than ten and applied to the opposite side
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
This is a measure of quantity of matter
Answer:
Mass
Explanation:
Mass is the measure of amount of matter contained within any substance and hence mass determines the weight. Unit of mass is kilogram as per ISI system of units.
Mass is measured through a balance. The more is the mass of an object, the more the balance tilts towards the object side.
Weight is equal to product of mass and the gravitational constant i.e 9.8m/s^2
The main component of all computer memory is
Answer: R.A.M
Explanation:
An astronaut is being tested in a centrifuge. The centrifuge has a radius of 11.0 m and, in starting, rotates according to θ = 0.260t2, where t is in seconds and θ is in radians. When t = 2.40 s, what are the magnitudes of the astronaut's (a) angular velocity, (b) linear velocity, (c) tangential acceleration, and (d) radial acceleration?
Answer:
a) 1.248 rad/s
b) 13.728 m/s
c) 0.52 rad/s^2
d) 17.132m/s^2
Explanation:
You have that the angles described by a astronaut is given by:
[tex]\theta=0.260t^2[/tex]
(a) To find the angular velocity of the astronaut you use the derivative og the angle respect to time:
[tex]\omega=\frac{d\theta}{dt}=\frac{d}{dt}[0.260t^2]=0.52t[/tex]
Then, you evaluate for t=2.40 s:
[tex]\omega=0.52(2.40)=1.248\frac{rad}{s}[/tex]
(b) The linear velocity is calculated by using the following formula:
[tex]v=\omega r[/tex]
r: radius if the trajectory of the astronaut = 11.0m
You replace r and w and obtain:
[tex]v=(1.248\frac{rad}{s})(11.0m)=13.728\frac{m}{s}[/tex]
(c) The tangential acceleration is:
[tex]a_T=\alpha r\\\\\alpha=\frac{\omega^2}{2\theta}=\frac{(1.248rad/s)^2}{2(0.260(2.40s)^2)}=0.52\frac{rad}{s^2}[/tex]
(d) The radial acceleration is:
[tex]a_r=\frac{v^2}{r}=\frac{(13.728m/s)^2}{11.0m}=17.132\frac{m}{s^2}[/tex]
Is mercury (the planet) rocky or gaseous(meaning relating to or having the characteristics of a gas.)
Answer:
Mercury is rocky
Explanation:
Answer:
Rocky
Explanation:
It has no atmosphere so it cannot hold gas.
A steam engine takes in superheated steam at 270 °C and discharges condensed steam from its cylinder at 50 °C. The engine has an efficiency of 30%, and taken in 50 kJ from the hot steam per cycle. If a Carnot engine takes in the same amount of heat per cycle and operates at these temperatures, the work it can turn into is most likely to be:a) 15 kJ. b) 20 kJ. c) 10 kJ. d) 50 kJ.
Answer:
b) 20 kJ
Explanation:
Efficiency of carnot engine = (T₁ - T₂ ) / T₁ Where T₁ is temperature of hot source and T₂ is temperature of sink .
T₁ = 270 + 273 = 543K
T₂ = 50 + 273 = 323 K
Putting the given values of temperatures
efficiency = (543 - 323) / 543
= .405
heat input = 50 KJ
efficiency = output work / input heat energy
.405 = output work / 50
output work = 20.25 KJ.
= 20 KJ .
Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. Suppose that the temperature of a cup of coffee obeys Newton's law of cooling. If the coffee has a temperature of 210 degrees Fahrenheit when freshly poured, and 2.5 minutes later has cooled to 191 degrees in a room at 64 degrees, determine when the coffee reaches a temperature of 156 degrees.
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 .
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
To understand thermal linear expansion in solid materials. Most materials expand when their temperatures increase. Such thermal expansion, which is explained by the increase in the average distance between the constituent molecules, plays an important role in engineering. In fact, as the temperature increases or decreases, the changes in the dimensions of various parts of bridges, machines, etc., may be significant enough to cause trouble if not taken into account. That is why power lines are always sagging and parts of metal bridges fit loosely together, allowing for some movement. It turns out that for relatively small changes in temperature, the linear dimensions change in direct proportion to the temperature.
For instance, if a rod has length L0 at a certain temperature T0 and length L at a higher temperature T, then the change in length of the rod is proportional to the change in temperature and to the initial length of the rod: L - L0 = αL0(T - T0),
or
ΔL = αL0ΔT.
Here, α is a constant called the coefficient of linear expansion; its value depends on the material. A large value of α means that the material expands substantially as the temperature increases; smaller values of α indicate that the material tends to retain its dimensions. For instance, quartz does not expand much; aluminum expands a lot. The value of α for aluminum is about 60 times that of quartz!
Questions:
A) Compared to its length in the spring, by what amount ΔLwinter does the length of the bridge decrease during the Teharian winter when the temperature hovers around -150°C?
B) Compared to its length in the spring, by what amount ΔLsummer does the length of the bridge increase during the Teharian summer when the temperature hovers around 700°C?
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
