The average velocity or displacement of a particle for the first time interval is <u>Δs / Δt = 6 cm/s.</u>
Solution:
As we know that displacement is calculated in centimeters and the unit of time is second.
The average velocity for the first interval [1,2] is given
Δs / Δt = s (t2) - s (t) / t2 - t1
Δs / Δt = 2sin2 π + 3cos 2 π - ( 2sin π + 3cos π ) / 2 - 1
Δs / Δt = 2(0) + 3(1) - 2(0) - 3 (-1) / 1
Δs / Δt = 6 cm/s
Thus the average velocity or displacement of a particle for the first time interval is Δs / Δt = 6 cm/s
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The complete question is:
The displacement of a particle moving back and forth along a line is given by the following equation s(t) = 2sin π t + 3cos π t. Estimate the instantaneous velocity of the particle when t = 1
G = 9.81 m/sec^2) g = 9.81

<span>Solving for velocity : </span>

<span> = 2gh </span>
<span>v = </span>

<span>v = (2 x 9.81 x 10)^1/2 </span>
<span>v = 196.2 m/sec (answer)</span>
The attractive force between all matter in the universe is gravity.
Answer:
1. Energy = 2880 Joules.
2. Energy = 60 Joules.
3. Quantity of charge = 120 Coulombs.
Explanation:
Given the following data;
1. Voltage = 12 Volts
Current = 0.5 Amps
Time, t = 8 mins to seconds = 8 * 60 = 480 seconds
To find the energy;
Power = current * voltage
Power = 12 * 0.5
Power = 6 Watts
Next, we find the energy transferred;
Energy = power * time
Energy = 6 * 480
Energy = 2880 Joules
2. Charge, Q = 4 coulombs
Potential difference, p.d = 15V
To find the total energy transferred;
Energy = Q * p.d
Energy = 4 * 15
Energy = 60 Joules
3. Voltage = 6 Volts
Current = 1 Amps
Time = 2 minutes to seconds = 2 * 60 = 120 seconds
To find the quantity of charge;
Quantity of charge = current * time
Quantity of charge = 1 * 120
Quantity of charge = 120 Coulombs