A group of individuals living in a particular geographic area is termed population.
Answer:
C.As the two objects touch, thermal energy flows as heat from the warmer block to the colder block until particles in both blocks move at the same rate and reach the same temperature.
Explanation:
Heat is the transfer of thermal energy from an object at higher temperature to an object at colder temperature.
The temperature of an object is a measure of how fast the particles in the object move: the higher its temperature, the faster the particles move, the higher the average kinetic energy of the particles in the object. As a result, the particles of the object at higher temperature tend to transfer more energy (called thermal energy) to the particles of the object at colder temperature by colliding with them: this process continues until the particles of the colder object reach the same average kinetic energy as the particles of the warmer object, and this means that the two objects have reached the same temperature.
Explanation:
The given data is as follows.
mass = 0.20 kg
displacement = 2.6 cm
Kinetic energy = 1.4 J
Spring potential energy = 2.2 J
Now, we will calculate the total energy present present as follows.
Total energy = Kinetic energy + spring potential energy
= 1.4 J + 2.2 J
= 3.6 Joules
As maximum kinetic energy of the object will be equal to the total energy.
So, K.E = Total energy
= 3.6 J
Also, we know that
K.E = 
or, v = 
= 
= 
= 6 m/s
thus, we can conclude that maximum speed of the mass during its oscillation is 6 m/s.
Answer:
a) T ’= 0.999 s
, b) t = 3596.4 s
Explanation:
The angular velocity of a simple pendulum is
w = √g / L
The angular velocity, frequency and period are related
w = 2π f = 2π / T
2π / T = √ g / L
T = 2π √ L / g
L = T² g / 4π²
L = 1² 9.8 / 4π²
L = 0.248 m
To know the effect of the temperature change let's use the thermal expansion ratios
ΔL = α L ΔT
ΔL = 24 10⁻⁶ 0.248 (-4 - 20)
ΔL = 142.8 10⁻⁶ m
Lf - L = -142. 8 10⁻⁶
Lf = 142.8 10⁻⁶ + 0.248
Lf = 0.2479 m
Let's calculate new period
T ’= 2π √ L / g
T ’= 2π √ (0.2479 / 9.8)
T ’= 0.999 s
We can see that the value of the period is reduced so that the clock is delayed
b) change of time in 1 hour
When the clock is at 20 ° C in one hour it performs 3600 oscillations, for the new period the time of this number of oscillations is
t = 3600 0.999
t = 3596.4 s
Therefore the clock is delayed almost 4 s
