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
Answer:
M=28.88 gm/mol
Explanation:
Given that
T= 95 K
P= 1.6 atm
V= 4.87 L
m = 28.6 g
R=0.08206L atm .mol .K
We know that gas equation for ideal gas
P V = n R T
P=Pressure , V=Volume ,n=Moles,T= Temperature ,R=gas constant
Now by putting the values
P V = n R T
1.6 x 4.87 = n x 0.08206 x 95
n=0.99 moles
We know that number of moles given as
M=Molar mass
M=28.88 gm/mol
Complete Question:
A coin is dropped off of a building landing on its side. It hits with a pressure of 400 N/m². It hits with a force of 0.1N. Calculate the area of the coin?
Answer:
Area = 0.00025 m²
Explanation:
Given the following data;
Pressure = 400N/m²
Force = 0.1N
To find the area of the coin;
Pressure = Force/area
Area = Force/pressure
Substituting into the equation, we have;
Area = 0.1/400
Area = 0.00025 m²
Answer:
Base units are defined units based on specific objects or events in the physical world. Derived units are defined by combining base units.
Base units are defined by a particular process of measuring a base quantity whereas derived units are defined as algebraic combinations of base units. For example, length is a base quantity in both SI and the English system, but the meter is a base unit in the SI system only.
