The average speed of the blocks are 0.36 m/s.
Explanation:
Average speed is defined as the ratio of distance covered per unit time. So if it is said that blocks are pulled to 0.9 m in the right side. This means the blocks cover a distance of 0.9 m from the origin and that distance is covered in 2.5 s. Thus, the average speed can be calculated from the change in speed with respect to time. As at time t = 0 , the speed is also zero, and at time t = 2.5 s , the speed will be
Since, in this case, the speed is equal to the average speed of blocks. So the average speed of the blocks will be 0.36 m/s.
Answer : The change in internal energy is, 900 Joules.
Solution : Given,
Heat given to the system = +1400 J
Work done by the system = -500 J
Change in internal energy is equal to the sum of heat energy and work done.
Formula used :
where,
= change in internal energy
q = heat energy
w = work done
As per question, heat is added to the system that means, q is positive and work done by the system that means, w is negative.
Now put all the given values in the above formula, we get
Therefore, the change in internal energy is 900 J.
The change in internal energy depends on the heat energy and work done. As we will change in the heat energy and work done, then changes will occur in the internal energy. Hence, the energy is conserved.
Answer:
A. Bar graph
Explanation:
Bar graphs are used to compare things between different groups or to track changes over time.
Answer:
100Jkg/°C
Explanation:
Given parameters:
Mass of metal = 2kg
Amount of heat energy = 1600J
Initial temperature = 5°C
Final temperature = 13°C
Unknown:
Specific heat capacity of the metal = ?
Solution:
Specific heat capacity of a body is the amount of heat needed to raise the temperature of unit mass of a body by 1°C.
H = m x C x (T₂ - T₁ )
H is the amount of heat
m is the mass
C is the unknown specific heat capacity
T is the temperature
Insert the parameters and solve;
1600 = 2 x C x (13 - 5)
1600 = 16C
C = 100Jkg/°C
The first ionization and second ionization of an atom are similar in following ways:
1. Both ionizations involve atom and energy
2. Both lose an electron.
The difference between first and second ionization of an atom is that both started and ended with different ions/atom.
