1. The correct answer among the choices provided is the third option. Measuring the temperature increase of water from doing work by stirring it is an experiment generally regarded as being first carried out by James Joule.
2. Joule's experiment directly shows that heat is a form of energy. He wanted to make a different way of measuring energy.
Answer:
we got time and velocity over time.
so the distance is again the area underneath the graph
for a triangle with known base and height it's
4*10 / 2
distance traveled is 20
deceleration occurs when velocity decreases. that happens from t=2 till t=4
in 2 time-units we loose 10 units of velocity, so we decelerate by 5 units per 1 time
a (from t=2 to t=4) = -5v/t
<em><u>One</u></em><em><u> </u></em><em><u>newton</u></em><em><u> </u></em><em><u>force</u></em><em> </em><em>is</em><em> </em><em>defined as t</em><em>h</em><em>e</em><em> </em><em>force</em><em> </em><em>that</em><em> </em><em>is</em><em> necessary to provide a mass of one kilogram with an acceleration of one metre per second per second. One newton is equal to a force of 100,000 dynes in the centimetre-gram-second (CGS) system, or a force of about 0.2248 pound </em><em>i</em><em>n</em><em> </em><em>the</em><em> </em><em>f</em><em>o</em><em>o</em><em>t</em><em>-</em><em>p</em><em>o</em><em>u</em><em>n</em><em>d</em><em>-</em><em> </em><em>s</em><em>e</em><em>c</em><em>o</em><em>n</em><em>d</em><em> </em><em>system</em><em>.</em>
the answer is the third one down
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)