It would have to be c because it is a chemical change. your welcome!!!!!
Answer:
ALL AWNERS HERE
Explanation:
https://quizlet.com/449884025/test-3-physics-2-flash-cards/
Answer:
1.34352 kg
Explanation:
= Mass of water falling = 1 kg
h = Height of fall = 0.1 km
= Change in temperature = 0.1
c = Specific heat of water = 4186 J/kg K
g = Acceleration due to gravity = 9.81 m/s²
= Mass of water in the vessel
Here the potential energy will balance the internal energy

Mass of the water in the vessel is 1.34352 kg
Answer:
f ’= 97.0 Hz
Explanation:
This is an exercise of the doppler effect use the frequency change due to the relative movement of the fort and the observer
in this case the source is the police cases that go to vs = 160 km / h
and the observer is vo = 120 km / h
the relationship of the doppler effect is
f ’= f₀ (v + v₀ / v-
)
let's reduce the magnitude to the SI system
v_{s} = 160 km / h (1000 m / 1km) (1h / 3600s) = 44.44 m / s
v₀ = 120 km / h (1000m / 1km) (1h / 3600s) = 33.33 m / s
we substitute in the equation of the Doppler effect
f ‘= 100 (330+ 33.33 / 330-44.44)
f ’= 97.0 Hz
Answer:
a) see attached, a = g sin θ
b)
c) v = √(2gL (1-cos θ))
Explanation:
In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by
Wₓ = m a
W sin θ = m a
a = g sin θ
b) The diagram is the same, the only thing that changes is the angle that is less
θ' = 9/2 θ
c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.
The easiest way to find linear speed is to use conservation of energy
Highest point
Em₀ = mg h = mg L (1-cos tea)
Lowest point
Emf = K = ½ m v²
Em₀ = Emf
g L (1-cos θ) = v² / 2
v = √(2gL (1-cos θ))