Calculate the heat required when 2.50 mol of a reacts with excess b and a2b according to the reaction: 2a + b + a2b → 2ab + a2 g
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The ball was released from a height of 20 meters
Explanation:
The scenario is as follows:
1. A ball drops from the edge of a cliff.
2. Upon reaching the ground, the energy held in its gravitational potential energy transforms entirely into kinetic energy.
This implies K.E = P.E.
3. The ball impacts the ground at a speed of 20 m/s.
4. The gravitational field strength noted is 10 N/kg.
<pOur goal is to ascertain the height from which the ball was dropped.<pSince the ball was dropped from a cliff, its initial velocity is 0.<p→ K.E =where v is the final velocity, is the initial velocity, and m is the mass.
→ K.E =
→ K.E = 200 m joules when the ball strikes the ground.
<p→ P.E = mg hwhere g is the gravitational field strength, m is mass, and h signifies height.
<p→ g = 10 N/kg.<p→ P.E = m(10)(h)→ P.E = 10m h joules.
<p→ P.E = K.E.→ 10m h = 200 m.
Dividing through by 10m yields:
→ h = 20 meters.
The ball was released from a height of 20 meters.
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Refer to the diagram shown below.
m₁ = 1100 kg represents the mass of the car.
m₂ = 700 kg indicates the combined mass of the trailer and boat.
F = 1900 N is the driving force acting on the vehicle.
N₁ denotes m₁g, the normal force on the car.
N₂ corresponds to m₂g, the normal force on the trailer and boat.
Frictional forces are represented by μN₁ and μN₂, where μ is the coefficient of kinetic friction.
T signifies the force in the connection between the car and the trailer.
Part (a)
Let R₁ signify the total resistance acting against the motion of the car, boat, and trailer.
With the acceleration at 0.550 m/s², it follows that
(m₁ + m₂ kg)*(0.55 m/s²) = F
(1100 + 700 kg)*(0.55 m/s²) = (1900 - R₁) N.
This leads to the equation 990 = 1900 - R.
Therefore, R₁ = 910 N.
Answer: The total resistive force amounted to 910 N.
Part (b)
The trailer and boat experience 80% of the resisting forces.
Let R₂ denote this resistive force.
Thus,
R₂ = 0.8*R₁ = 728 N.
Assuming T is the tension in the hitch connecting the car and trailer, it follows:
T - R₂ = m₂(0.55 m/s²)
(T - 728 N) = (700 kg)*(0.55 m/s²).
This leads to T - 728 = 385.
Thus, T equals 1113 N.
Answer: The tension in the hitch is 1113 N.
m₁ = 1100 kg represents the mass of the car.
m₂ = 700 kg indicates the combined mass of the trailer and boat.
F = 1900 N is the driving force acting on the vehicle.
N₁ denotes m₁g, the normal force on the car.
N₂ corresponds to m₂g, the normal force on the trailer and boat.
Frictional forces are represented by μN₁ and μN₂, where μ is the coefficient of kinetic friction.
T signifies the force in the connection between the car and the trailer.
Part (a)
Let R₁ signify the total resistance acting against the motion of the car, boat, and trailer.
With the acceleration at 0.550 m/s², it follows that
(m₁ + m₂ kg)*(0.55 m/s²) = F
(1100 + 700 kg)*(0.55 m/s²) = (1900 - R₁) N.
This leads to the equation 990 = 1900 - R.
Therefore, R₁ = 910 N.
Answer: The total resistive force amounted to 910 N.
Part (b)
The trailer and boat experience 80% of the resisting forces.
Let R₂ denote this resistive force.
Thus,
R₂ = 0.8*R₁ = 728 N.
Assuming T is the tension in the hitch connecting the car and trailer, it follows:
T - R₂ = m₂(0.55 m/s²)
(T - 728 N) = (700 kg)*(0.55 m/s²).
This leads to T - 728 = 385.
Thus, T equals 1113 N.
Answer: The tension in the hitch is 1113 N.