That's one of the three changes that are called "acceleration".
The other two are:
-- increase in the magnitude
-- change in direction.
Some might call a decrease in the magnitude "deceleration".
Before coming into conclusion first we have to understand the direction of heat flow.
Heat is the transferred thermal energy from one body to another body due to the temperature difference just like water flows from higher level to lower level.
Whenever two bodies having different temperature come closer to each other heat will flow from hotter body to cooler one if no external work is done. The heat flow may be through any of the ways i.e conduction,radiation or convection. Hence temperature difference is the parameter which gives the direction of heat flow.
The temperature is also considered as a measure of average kinetic energy of the substance.The thermal energy does not give the direction heat flow. Heat may flow from the body having low thermal energy but at higher temperature to the body having higher thermal energy but at low temperature. The reverse does not happen naturally .
In example 1 there is fire and air. Obviously fire is at high temperature and air at low temperature.So heat will flow from object 1 to object 2.
In example 2 there is a metal at 80 degree Celsius and another metal at 12 degree Celsius .So heat will flow from object 1 to object 2
In example 3 we have cooler ocean and warm air. So the heat will flow from object 2 to object 1.
In example 4 we have a tool with high thermal energy and a material with little thermal energy.We already know that thermal energy can not determine the direction of heat flow. Here the temperature of each substance is not given.The kinetic energy is part of thermal energy.So there is the chance of higher kinetic energy of the tool for having higher thermal energy .At that time the heat will flow object 1 to object 2.Otherwise the reverse will occur. So it is a special case.
As per the question only option 4 is correct which tells that heat will flow from object 1 to object 2 in examples 1,2,4, and heat will flow from object 2 to 1 in example 3. Other options violate the fundamental law of thermodynamics.
The Period of the resulting shm will be T=39.7
<u>Explanation:</u>
<u>Given data</u>
m=3kg
d=.06m
k=1200 N/m
Θ=3 °
T=?
we have the formulas,
I = (1/6)Md2
F = ma
F = -kx = -(mω2x)
k = mω2 τ = -d(FgsinΘ)
T=2 x 3.14/ √(m/k)
Solution for the given problem would be,
F=-Kx (where x= dsin Θ)
F=-k dsin Θ
F=-(1200)(.06)sin(3 °)
F=-10.16N
<u>By newton's second law.</u>
F = ma
a= F/m
a=(-10.16N)/3
a=3.38
<u>using the k=mω value</u>
k=mω
ω=k/m
ω=1200/3
ω=400
<u>Using F = -kx value</u>
x = F/-k
x=(-10.16)/1200
x=0.00847m
<u>Restoring the torque value </u>
τ = -dmgsinΘ where( τ = Iα so.).. Iα = -dmgsinΘ α = -(.06)(4)α =
α =(.06)(4)(9.81)sin(4°)
α=-1.781
<u>Rotational to linear form</u>
a = αr
r = .1131 m
a=-1.781 x .1131 m
a=-0.2015233664
<u>Time Period</u>
T=2 x 3.14/ √(m/k)
T=6.28/√(3/1200)
T=6.28/0.158
T=39.7
C) water is the answer i believe
Answer:
can't explain but is correct
