Answer:
60mph=26.8224meters per second
Explanation:
Answer:
The average velocity is
and
respectively.
Explanation:
Let's start writing the vertical position equation :

Where distance is measured in meters and time in seconds.
The average velocity is equal to the position variation divided by the time variation.
= Δx / Δt = 
For the first time interval :
t1 = 5 s → t2 = 8 s
The time variation is :

For the position variation we use the vertical position equation :

Δx = x2 - x1 = 1049 m - 251 m = 798 m
The average velocity for this interval is

For the second time interval :
t1 = 4 s → t2 = 9 s


Δx = x2 - x1 = 1495 m - 125 m = 1370 m
And the time variation is t2 - t1 = 9 s - 4 s = 5 s
The average velocity for this interval is :

Finally for the third time interval :
t1 = 1 s → t2 = 7 s
The time variation is t2 - t1 = 7 s - 1 s = 6 s
Then


The position variation is x2 - x1 = 701 m - (-1 m) = 702 m
The average velocity is

Most likely solid. Molecules move in a chaotic matter, which a Scottish botanist found out in the 19th century. In gases, this chaotic movement is the most “dramatic” and they move around freely; in liquids it’s a bit less noticeable, and in solids the particles oscillate around locked positions. Movement in plasma is similar to gas.
The force between the two point charge when they are separated by 18 cm is 3 N
<h3>How do I determine the force when they are 18 cm apart?</h3>
Coulomb's law states as follow:
F = Kq₁q₂ / r²
Cross multiply
Fr² = Kq₁q₂
Kq₁q₂ => constant
F₁r₁² = F₂r₂²
Where
- F₁ and F₂ are the initial and new force
- r₁ and r₂ are the initial and new distance apart
With the above formula, we can obtain the force between the two point charge when they are 18 cm apart. Details below:
- Initial distance apart (r₁) = 6 cm
- Initial force of attraction (F₁) = 27 N
- New distance apart (r₂) = 18 cm
- New force of attraction (F₂) =?
F₁r₁² = F₂r₂²
27 × 6² = F₂ × 18²
972 = F₂ × 324
Divide both side by 324
F₂ = 927 / 324
F₂ = 3 N
Thus, the force when they are 18 cm apart is 3 N
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Answer:
Explanation:
Given
mass of ethanol 
mass of aluminium cup 
both are at an initial temperature of 
specific heat of ethanol 
specific heat of aluminium 
specific heat of ice 
specific heat of water 
Latent heat of fusion 
suppose m is the mass of ice added
Heat loss by Al cup and ethanol after
is reached

Heat gained by ice such that ice is melted and reached a temperature of 

Comparing 1 and 2 we get

Thus 23.65 gm of ice is added