Answer:
D. Forces between molecules
Explanation:
Specific heat capacity of water can be defined as the amount of heat a gram of water must lose or absorb in order to change its temperature by a degree Celsius. It is measured in Joules per kilogram per degree Celsius (J/kg°C). Generally, the specific heat capacity of water is 4.182J/kg°C and is the highest among liquids.
Mathematically, the specific heat capacity of a substance is given by the formula;
Where;
Q represents the heat capacity or quantity of heat.
m represents the mass of an object.
c represents the specific heat capacity of water.
dt represents the change in temperature.
Cohesion is a property of water and it typically refers to the attraction between molecules of water which holds them together.
In Science, the property which helps to explain differences in the specific heat capacities of two substances is the forces between molecules.
This ultimately implies that, the more closely bonded the atoms of a substance are, the higher or greater would be the substance's specific heat capacity. Thus, it varies for the various states of matter i.e solid, liquid and gas.
Answer:
195
Explanation:
he can only apply as much force as his body mass so it would be 195
Answer:
<em>The grapefruit dropped 2.54 m and hit the ground at 7.06 m/s</em>
Explanation:
<u>Free Fall Motion
</u>
A free-falling object falls under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. Free-falling objects do not encounter air resistance.
If an object is dropped from rest in a free-falling motion, it falls with a constant acceleration called the acceleration of gravity, which value is 
The final velocity of a free-falling object after a time t is given by:
vf=g.t
The distance traveled by a dropped object is:
Given a grapefruit free falls from a tree and hits the ground t=0.72 s later, we can calculate the height it fell from:
y = 2.54 m
The final speed is computed below:
vf = 7.06 m/s
The grapefruit dropped 2.54 m and hit the ground at 7.06 m/s
Answer:IM SORRY BUT I D K BUT I HOPE THAT YOU HAVE BETTER DAY AND REALAX AND THINK OF HAPPY THOUGHTS :D
Explanation:IM HE MYSTERY MAN WHOOSH??????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????
Answer:
V₁ = √ (gy / 3)
Explanation:
For this exercise we will use the concepts of mechanical energy, for which we define energy n the initial point and the point of average height and / 2
Starting point
Em₀ = U₁ + U₂
Em₀ = m₁ g y₁ + m₂ g y₂
Let's place the reference system at the point where the mass m1 is
y₁ = 0
y₂ = y
Em₀ = m₂ g y = 2 m₁ g y
End point, at height yf = y / 2
= K₁ + U₁ + K₂ + U₂
= ½ m₁ v₁² + ½ m₂ v₂² + m₁ g 
+ m₂ g
Since the masses are joined by a rope, they must have the same speed
= ½ (m₁ + m₂) v₁² + (m₁ + m₂) g
= ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g
How energy is conserved
Em₀ =
2 m₁ g y = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g
2 m₁ g y = ½ (3m₁) v₁² + (3m₁) g y / 2
3/2 v₁² = 2 g y -3/2 g y
3/2 v₁² = ½ g y
V₁ = √ (gy / 3)
