1) Calculate the ideal efficiency of a heat engine that takes in energy at 800 K and expels heat to a

What are some non examples of temperature ?

Alja [10]

non examples of temperature are dixionanon , fairinheat, cabrowskin, and lastly ancomthere

6 0

3 years ago

A 103 kg horizontal platform is a uniform disk of radius 1.71 m and can rotate about the vertical axis through its center. A 68.

Andreyy89

Answer:

$I_{total}=220.64 kg*m^{2}$

Explanation:

The moment of inertia of the system is equal to the each population and the platform inertia so

Inertia disk

$I_{disk}=\frac{1}{2}*m_{disk}*(r_{p})^{2}$

Inertia person

$I_{p}=\frac{1}{2}*m_{p}*(r_{p})^{2}$

Inertia dog

$I_{d}=\frac{1}{2}*m_{d}*(r_{d})^{2}$

The Inertia of the system is the sum of each mass taking into account that all exert the force of inertia:

$I_{total}=I_{disk}+I_{p}+I_{d}$

$I_{total}=\frac{1}{2}*103kg*(1.71)^{2}+\frac{1}{2}*68.9kg*(1.09)^{2}+\frac{1}{2}*27.7kg*(1.45)^{2}$

$I_{total}=220.64 kg*m^{2}$

5 0

3 years ago

If the velocity of an object is doubled, its kinetic energy is ______

swat32

Increased by a factor of 4

4 0

3 years ago

g A 1.5-kg mass attached to spring with a force constant of 20.0 N/m oscillates on a horizontal, frictionless track. At t = 0, t

jok3333 [9.3K]

Answer:

(a) f = 0.58Hz

(b) vmax = 0.364m/s

(c) amax = 1.32m/s^2

(d) E = 0.1J

(e) x(t)=0.1m*cos(2π(0.58s^{-1})t)

Explanation:

(a) The frequency of the oscillation, in a spring-mass system, is calculated by using the following formula:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$ (1)

k: spring constant = 20.0N/m

m: mass = 1.5kg

you replace the values of m and k for getting f:

$f=\frac{1}{2\pi}\sqrt{\frac{20.0N/m}{1.5kg}}=0.58s^{-1}=0.58Hz$

The frequency of the oscillation is 0.58Hz

(b) The maximum speed is given by the following relation:

$v_{max}=\omega A=2\pi f A$ (2)

A: amplitude of the oscillations = 10.0cm = 0.10m

$v_{max}=2\pi (0.58s^{-1})(0.10m)=0.364\frac{m}{s}$

The maximum speed of the mass is 0.364 m/s.

The maximum speed occurs when the mass passes trough the equilibrium point of the oscillation.

(c) The maximum acceleration is given by the following formula:

$a_{max}=\omega^2A=(2\pi f)^2 A$

$a_{max}=(2\pi (0.58s^{-1}))(0.10m)=1.32\frac{m}{s^2}$

The maximum acceleration is 1.32 m/s^2

The maximum acceleration occurs where the elastic force is a maximum, that is, where the mass is at the maximum distance from the equilibrium point, that is, the acceleration.

(d) The total energy of the system is calculated with the maximum potential elastic energy:

$E=\frac{1}{2}kA^2=\frac{1}{2}(20.0N/m)(0.10m)^2=0.1J$

The total energy is 0.1J

(e) The displacement as a function of time is:

$x(t)=Acos(\omega t)=Acos(2\pi ft)\\\\x(t)=0.1m\ cos(2\pi(0.58s^{-1})t)$

6 0

3 years ago

Ana walks from 4 m to 200 cm. Which of the following statements is true about

Gemiola [76]

Distance=2m

because 200cm = 2m
so 4m-2m=2m

3 0

3 years ago