Answer:
2.16×10⁻⁶ N
Explanation:
Applying,
F = kqq'/r² (coulomb's Law)....................... Equation 1
Where F = electrostatic force, k = coulomb's constant, q = charge on the styrofoam, q' = charge on the grain of salt, r = distance between the charges.
From the question,
Given: q = 0.002 mC = 2.0×10⁻⁶ C, q' = 0.03 nC = 3.0×10⁻¹¹ C, r = 0.5 m
Constant: k = 8.99×10⁹ Nm²/C²
Substitute these values into equation 1
F = (2.0×10⁻⁶)(3.0×10⁻¹¹)(8.99×10⁹)/0.5²
F = 2.16×10⁻⁶ N
Answer:
The maximum permissible propagation delay per flip flop stage is<u> 100 </u>n sec
Explanation:
1024 ripple counter has 10 J-K flip flops(210 = 1024).
So the total delay will be 10×x where x is the delay of each J-K flip flops.
The period of the clock pulse is 1× 10⁻⁶ s.
Now
10x <= 10⁻⁶ s
x <= 100 ns
x= 100 ns for prpoer operation.
pulse train with a frequency of 1 MHz is counted using a modulo-1024 ripple-counter built with J-K flip flops. For proper operation of the counter, the maximum permissible propagation delay per flip flop stage is <u>100 </u>n sec.
Answer: 1608.39 J
Explanation: Given that the
mass M = 42kg
U = 11.5m/s
V = 3.33m/s
how much work did friction do
Work done = Force × distance
Work done = Ma × distance
But acceleration a = V/t
Work done = M × V/t × d
Work done = M × V × d/t
Where d/t = velocity
Therefore,
Work done = M × U × V
Work done = 42 × 11.5 × 3.33
Work done = 1608.39 J
