Answer:
°C
Explanation:
= mass of the horseshoe = 0.35 kg
= mass of the water = 1.40 L = 1.40 kg
= mass of the iron pot = 0.45 kg
= specific heat of iron = 450 J kg⁻¹ °C⁻¹
= specific heat of water = 4186 J kg⁻¹ °C⁻¹
= initial temperature of the horseshoe = ?
= initial temperature of the water = 22 °C
= initial temperature of the iron pot = 22 °C
= final temperature = 32 °C
Using conservation of Heat




°C
Answer:
L₀ = L_f , K_f < K₀
Explanation:
For this exercise we start as the angular momentum, with the friction force they are negligible and if we define the system as formed by the disk and the clay sphere, the forces during the collision are internal and therefore the angular momentum is conserved.
This means that the angular momentum before and after the collision changes.
Initial instant. Before the crash
L₀ = I₀ w₀
Final moment. Right after the crash
L_f = (I₀ + mr²) w
we treat the clay sphere as a point particle
how the angular momentum is conserved
L₀ = L_f
I₀ w₀ = (I₀ + mr²) w
w =
w₀
having the angular velocities we can calculate the kinetic energy
starting point. Before the crash
K₀ = ½ I₀ w₀²
final point. After the crash
K_f = ½ (I₀ + mr²) w²
sustitute
K_f = ½ (I₀ + mr²) (
w₀)²
Kf = ½
w₀²
we look for the relationship between the kinetic energy
= 

K_f < K₀
we see that the kinetic energy is not constant in the process, this implies that part of the energy is transformed into potential energy during the collision
Answer:
A
Explanation:
All of the frictions are the same, but weight always goes straight down so it can only be A or B. Since they are going down a slope, then the normal force must be sloped. A is the only one out of A and B with a sloped normal force, so it has to be A
Answer:
Answered
Explanation:
When an igneous rock becomes buried, is subjected to high heat and pressure, and recrystallizes it is formed into Metamorphic rock. Now this rock is eroded, transported, deposited and subsequently lithified to be converted into Sedimentary rock.
The same igneous rock is first converted into Metamorphic and then into sedimentary by the process of weathering.
Work is calculated by multiplying force by the distance that the object had moved. The applied force is 60 N, moving the object by 10 m. Thus, the work does is 600 J. For the friction force which is equal to,
100N x 0.250 = 25.0 N
the work done is,
W = (60 N - 25 N) x 10 m = 350 J
The kinetic energy of the box can be equated to this force. Thus, the answer is also 350 J.