Mass = 0.201kg
Energy = 15J
temperature change = 10C
Energy(E) = mass(m) × specific heat capacity(c) × temperature change(θ)
we can rearrange this to make specific heat capacity the subject
c =
c =
c =7.46268657
Answer:
The speed of the plank relative to the ice is:
Explanation:
Here we can use momentum conservation. Do not forget it is relative to the ice.
(1)
Where:
- m(g) is the mass of the girl
- m(p) is the mass of the plank
- v(g) is the speed of the girl
- v(p) is the speed of the plank
Now, as we have relative velocities, we have:
(2)
v(g/b) is the speed of the girl relative to the plank
Solving the system of equations (1) and (2)
I hope it helps you!
Answer:
K_a = 8,111 J
Explanation:
This is a collision exercise, let's define the system as formed by the two particles A and B, in this way the forces during the collision are internal and the moment is conserved
initial instant. Just before dropping the particles
p₀ = 0
final moment
p_f = m_a v_a + m_b v_b
p₀ = p_f
0 = m_a v_a + m_b v_b
tells us that
m_a = 8 m_b
0 = 8 m_b v_a + m_b v_b
v_b = - 8 v_a (1)
indicate that the transfer is complete, therefore the kinematic energy is conserved
starting point
Em₀ = K₀ = 73 J
final point. After separating the body
Em_f = K_f = ½ m_a v_a² + ½ m_b v_b²
K₀ = K_f
73 = ½ m_a (v_a² + v_b² / 8)
we substitute equation 1
73 = ½ m_a (v_a² + 8² v_a² / 8)
73 = ½ m_a (9 v_a²)
73/9 = ½ m_a (v_a²) = K_a
K_a = 8,111 J
Input work = 9.63×10³ J.
Output work = 3.0×10³ J
By definition,
Efficiency = (Output work)/(Input work)
= (3.0×10³)/(9.63×10³)
= 0.31 = 31%
Answer: 31%
