Answer:
The 60 kg skater is traveling at 2.63 m/s in the opposite direction from the 45 kg skater.
Explanation:
The velocity of the 60 kg skater can be found by conservation of linear momentum:
(1)
Where:
: is the mass of the first skater = 45 kg
: is the mass of the second skater = 60 kg
: is the initial speed of the first skater = 0 (he is standing still)
: is the initial speed of the second skater = 0 (he is standing still)
: is the final speed of the first skater = 3.5 m/s
: is the final speed of the second skater =?
By replacing the above values into equation (1) and solving for we have:
The minus sign is because the 60 kg skater is moving in the opposite direction from the other skater.
Therefore, the 60 kg skater is traveling at 2.63 m/s in the opposite direction from the 45 kg skater.
I hope it helps you!
Answer:
a) 90 kJ
b) 230.26 kJ
Explanation:
The pressure at the first point 
= 10 bar —> 10 x 102 = 1020 kPa
The volume at the first point 
= 0.1 m^3
The pressure at the second point 
= 1 bar —> 1 x 102 = 102 kPa
The volume at the second point 
= 1 m^3
Process A.
constant volume V = C from point (1) to P = 10 bar.
Constant pressure P = C to the point (2).
Process B.
The relation of the process is PV = C
Required
For process A & B
(a) Sketch the process on P-V coordinates
(b) Evaluate the work W in kJ.
Assumption
Quasi-equilibrium process
Kinetic and potential effect can be ignored.
Solution
For process A.
V=C
There is no change in volume then
The work is defined by
║
V║limit 1--0.1
90 kJ
Process B
PV=C
By substituting with point (1) C = 10^2 x 1= 10^2
The work is defined by
║ ln(V)║limit 1--0.1
=230.26 kJ
Answer:
from the formula of work done =Force *Distance 500*100=50000
Answer:
(1) 1×10⁻⁴
Explanation:
From the question,
α = (ΔL/L)/(ΔT)............. Equation 1
Where α = linear expansivity of the metal plate, ΔL/L = Fractional change in Length, ΔT = Rise in temperature.
Given: ΔL/L = 1×10⁻⁴, ΔT = 10°C
Substitute these values into equation 1
α = 1×10⁻⁴/10
α = 1×10⁻⁵ °C⁻¹ .
β = (ΔA/A)/ΔT................... Equation 2
Where β = Coefficient of Area expansivity, ΔA/A = Fractional change in area.
make ΔA/A the subject of the equation
ΔA/A = β×ΔT.......................... Equation 3
But,
β = 2α.......................... Equation 4
Substitute equation 4 into equation 3
ΔA/A = 2α×ΔT................ Equation 5
Given: ΔT = 5°C, α = 1×10⁻⁵ °C⁻¹
Substitute into equation 5
ΔA/A = ( 2)×(1×10⁻⁵)×(5)
ΔA/A = 10×10⁻⁵
ΔA/A = 1×10⁻⁴
Hence the right option is (1) 1×10⁻⁴
Answer: the volume of the unknown mass
Explanation:
The other choices, molar masses of the reactants, the balanced chemical equation, and the molar masses of the product, are essential for the realization of stoichiometric calculations.
The balanced chemical equations tells the mole ratio of reactants (calcium carbonate) and products (calcium oxide and carbon dioxide).
The molar masses of reactants are used to convert between mass (grams) and moles.
The volume of the unknown mass is not needed when you are working only masses and moles since the relation is given by the molar masses. The volume is required when you are working with gas pressures, molarity contentrations or the information has to be worked from density measures.
