The bonds that hold atoms together to form molecules are called covalent bonds. They are pretty tough and not easily made or broken apart. It takes energy to make the bonds and energy is released when the bonds are broken.
Answer:
The change in entropy ΔS = 0.0011 kJ/(kg·K)
Explanation:
The given information are;
The mass of water at 20.0°C = 1.0 kg
The mass of water at 80.0°C = 2.0 kg
The heat content per kg of each of the mass of water is given as follows;
The heat content of the mass of water at 20.0°C = h₁ = 83.92 KJ/kg
The heat content of the mass of water at 80.0°C = h₂ = 334.949 KJ/kg
Therefore, the total heat of the the two bodies = 83.92 + 2*334.949 = 753.818 kJ/kg
The heat energy of the mixture =
1 × 4200 × (T - 20) = 2 × 4200 × (80 - T)
∴ T = 60°C
The heat content, of the water at 60° = 251.154 kJ/kg
Therefore, the heat content of water in the 3 kg of the mixture = 3 × 251.154 = 753.462
The change in entropy ΔS = ΔH/T = (753.818 - 753.462)/(60 + 273.15) = 0.0011 kJ/(kg·K).
Answer:
x-axis
Explanation:
when it is at rest, horizontal, on a graph it is the x-axis.
Hope this helped!
Answer:
v = 5.88 10⁷ m / s
Explanation:
For this exercise we use the relation
E = m c²
also indicate that all energy is converted into kinetic energy
E = K = ½ (M-2m) v²
where m is the mass of antimatter and M is the mass of the ship's mass. Factor two is due to the fact that equal amounts of matter and antimatter must be combined
we substitute
m c² = ½ (M-2m) v²
v² =
let's calculate
v =
v =
v = 5.88 10⁷ m / s
Answer:
The speed of player is given by

Explanation:
The time of flight for a projectile motion is given by
(i)
where t is the time of flight, v is the initial speed, and α is the angle.
Now the person must also reach the impact point of ball in the same time as above.
Now the total distance D the player needs to cover is basically R horizontal range of projectile minus the distance d, range R is given by,

Now the distance the player must cover is given by
D= R-d
D=
- d
(ii)
Now the average speed of player is given by
(iii)
Replacing the values of D and T from eq. (i) and (ii) in eq. (iii).

