Answer:
0.8 N
Explanation:
From coulomb's law,
Formula:
F = kqq'/r²........................ Equation 1
Where F = Force of repulsion, k = coulomb's constant, q = first positive charge, q' = second positive charge, r = distance between the charge.
Given: q = 20 μC = 20×10⁻⁶ C, q' = 100 μC = 100×10⁻⁶ C, r = 150 cm = 1.5 m.
Constant: k = 9×10⁹ Nm²/C²
Substitute these values into equation 1
F = (20×10⁻⁶ )( 100×10⁻⁶)(9×10⁹)/1.5²
F = 1800×10⁻³/2.25
F = 1.8/2.25
F = 0.8 N
Answer:
<h3>38,673.9N</h3>
Explanation:
According to newton's second law:
Force = mass * acceleration
Given
Mass = 873kg
acceleration = 44.66m/s²
Magnitude of the force is expressed as;
F = ma
F = 873 * 44.6
F = 38,673.9N
<em>Hence the magnitude of the net force exerted on the dragster during this time is 38,673.9N</em>
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
If you need to learn more about displacement click here:
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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