Answer:
T1 = 417.48N
T2 = 361.54N
T3 = 208.74N
Explanation:
Using the sin rule to fine the tension in the strings;
Given
amass = 42.6kg
Weight = 42.6 * 9.8 = 417.48N
The third angle will be 180-(60+30)= 90 degrees
Using the sine rule
W/Sin 90 = T3/sin 30 = T2/sin 60
Get T3;
W/Sin 90 = T3/sin 30
417.48/1 = T3/sin30
T3 = 417.48sin30
T3 = 417.48(0.5)
T3 = 208.74N
Also;
W/sin90 = T2/sin 60
417.48/1 = T2/sin60
T2 = 417.48sin60
T2 = 417.48(0.8660)
T2 = 361.54N
The Tension T1 = Weight of the object = 417.48N
Answer
is: V<span>an't
Hoff factor (i) for this solution is 1,81.
Change in freezing point from pure solvent to
solution: ΔT =i · Kf · b.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
b - molality, moles of solute per
kilogram of solvent.
</span><span>b = 0,89 m.
ΔT = 3°C = 3 K.
i = </span>3°C ÷ (1,86 °C/m · 0,89 m).
i = 1,81.
Answer:
A) The speed of the water must be 8.30 m/s.
B) Total kinetic energy created by this maneuver is 70.12 Joules.
Explanation:
A) Mass of squid with water = 6.50 kg
Mass of water in squid cavuty = 1.55 kg
Mass of squid = 
Velocity achieved by squid = 
Momentum gained by squid = 
Mass of water = 
Velocity by which water was released by squid = 
Momentum gained by water but in opposite direction = 
P = P'
B) Kinetic energy does the squid create by this maneuver:
Kinetic energy of squid = K.E =
Kinetic energy of water = K.E' = 
Total kinetic energy created by this maneuver:
Answer:
M.A = load/ effort
1200N/400N
= 3
velocity ratio= radius of wheel/radius of Axle
40cm/10cm
=4
efficiency= 3/4*100
75%
Answer:
The specific heat for the metal is 0.466 J/g°C.
Explanation:
Given,
Q = 1120 Joules
mass = 12 grams
T₁ = 100°C
T₂ = 300°C
The specific heat for the metal can be calculated by using the formula
Q = (mass) (ΔT) (Cp)
ΔT = T₂ - T₁ = 300°C - 100°C = 200°C
Substituting values,
1120 = (12)(200)(Cp)
Cp = 0.466 J/g°C.
Therefore, specific heat of the metal is 0.466 J/g°C.
