Answer:

I dont the child mass...you should substitute that value to (m)
then you can get your answer
Answer:
18.1 × 10⁻⁶ A = 18.1 μA
Explanation:
The current I in the wire is I = ∫∫J(r)rdrdθ
Since J(r) = Br, in the cylindrical wire. With width of 10.0 μm, dr = 10.0 μm. r = 1.20 mm. We have a differential current dI. We integrate first by integrating dθ from θ = 0 to θ = 2π.
So, dI = J(r)rdrdθ
dI/dr = ∫J(r)rdθ = ∫Br²dθ = Br²∫dθ = 2πBr²
Now I = (dI/dr)dr at r = 1.20 mm = 1.20 × 10⁻³ m and dr = 10.0 μm = 0.010 mm = 0.010 × 10⁻³ m
I = (2πBr²)dr = 2π × 2.00 × 10⁵ A/m³ × (1.20 × 10⁻³ m)² × 0.010 × 10⁻³ m = 0.181 × 10⁻⁴ A = 18.1 × 10⁻⁶ A = 18.1 μA
Since the slowest instruction in the SCA executes in 12.5 ns, the maximum system clock frequency is 80 MHz
To answer the question, we need to know what frequency is.
<h3>What is frequency?</h3>
Frequency is the number of oscillations per second of a wave.
It is given by f = 1/T where T = period of wave
Now, given that the slowest instruction in the SCA executes in t = 12.5 ns, we need to calculate maximum system clock frequency, f.
<h3>What is the maximum system clock frequency?</h3>
So, f = 1/t
= 1/12.5 ns
= 1/(12.5 × 10⁻⁹ s)
= 1/12.5 × 10⁹ Hz
= 0.08 × 10⁹ Hz
= 80 × 10⁻³ × 10⁹ Hz
= 80 × 10⁶ Hz
= 80 MHz
So, the maximum system clock frequency is 80 MHz
Learn more about maximum system clock frequency here:
brainly.com/question/14636488
#SPJ11
Answer:
It isn't safe for the truck to cross the bridge because the pressure exerted by the truck on the bridge is greater than the maximum tolerable pressure for the bridge, 562.5 Pa > 450 Pa.
Explanation:
Pressure is expressed in Force/Area.
So, for the truck, force exerted = 54000 N
Area covered = 30 × 3.2 = 96 m²
Pressure exerted by the truck = 54000/96 = 562.5 Pa
The pressure exerted by the truck on the bridge is greater than the maximum tolerable pressure for the bridge, 562.5 > 450, hence, it isn't safe for the truck to cross the bridge.