The time taken by the stone to hit the ground would be 5.12 seconds.
<h3>What are the three equations of motion?</h3>
There are three equations of motion given by Newton
The first equation is given as follows
v = u + at
the second equation is given as follows
S = ut + 1/2×a×t²
the third equation is given as follows
v² - u² = 2×a×s
Keep in mind that these calculations only apply to uniform acceleration.
As given in the problem, a stone is dropped from the helicopter which is ascending at the speed of 19.6 m/s
height(S) = 156.8 meters
initial velocity(u) = -19.6 m/s
acceleration(a) = 9.81 m/s²
By using the second equation of motion given by newton
S = ut + 1/2at²
S = 156.8m ,u= -19.6 m/s , a= 9.81 m/s² and t =? seconds
156.8= -19.6t + 9.81t²
t = 5.12 seconds
Thus, the time taken by the stone to hit the ground would be 5.12 seconds.
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The entire range of anything is called its spectrum.
For em wavelengths, it's the electromagnetic spectrum.
Answer:
The change in entropy ΔS = 0.0011 kJ/(kg·K)
Explanation:
The given information are;
The mass of water at 20.0°C = 1.0 kg
The mass of water at 80.0°C = 2.0 kg
The heat content per kg of each of the mass of water is given as follows;
The heat content of the mass of water at 20.0°C = h₁ = 83.92 KJ/kg
The heat content of the mass of water at 80.0°C = h₂ = 334.949 KJ/kg
Therefore, the total heat of the the two bodies = 83.92 + 2*334.949 = 753.818 kJ/kg
The heat energy of the mixture =
1 × 4200 × (T - 20) = 2 × 4200 × (80 - T)
∴ T = 60°C
The heat content, of the water at 60° = 251.154 kJ/kg
Therefore, the heat content of water in the 3 kg of the mixture = 3 × 251.154 = 753.462
The change in entropy ΔS = ΔH/T = (753.818 - 753.462)/(60 + 273.15) = 0.0011 kJ/(kg·K).
Answer:
The average force the golf club exerts on the ball is 600 N
Explanation:
Newton's second law of motion states that force, F, is directly proportional to the rate of change of momentum produced
F = m× (v₂ - v₁)/(Δt)
The given parameters of the motion of the ball are;
The mass of the ball, m = 45 g = 0.045 kg
The initial velocity of the ball, v₁ = 0 m/s
The speed with which the ball was hit by the golfer, v₂ = 40 m/s
The duration of contact between the golf club and the ball, Δt = 3 ms = 0.003 seconds (s)
By Newton's law of motion, the average force, 'F', which the golf club exerts on the ball is therefore, given as follows;
F = 0.045 kg × (40 m/s - 0 m/s)/(0.003 s) = 600 N
The average force the golf club exerts on the ball = F = 600 N.
