| Impedance | = √ [R² +(ωL)²]
R² = 6800² = 4.624 x 10⁷
(ωL)² = (2 · π · f · 2.3 · 10⁻³)²
= 2.0884 x 10⁻⁴ f²
| Z | = √[ (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²) ] = 1.6 x 10⁵
(1.6 x 10⁵)² = (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²)
(2.56 x 10¹⁰) - (4.624 x 10⁷) = 2.0884 x 10⁻⁴ f²
Frequency² = (2.56 x 10¹⁰ - 4.624 x 10⁷) / 2.0884 x 10⁻⁴
= 2.555 x 10¹⁰ / 2.0884 x 10⁻⁴
= 1.224 x 10¹⁴
= 122,400 GHz <== my calculation
11.1 MHz <== online impedance calculator
Obviously, I must have picked up some rounding errors
in the course of my calculation.
The name and strength of the force holding the block up is 50 N upward - Normal force.
The given parameters:
- <em>Mass of the block, m = 5 kg</em>
The weight of the block acting downwards due to gravity is calculated as follows;
W = mg
where;
- <em>g is acceleration due to gravity = 10 m/s²</em>
W = 5 x 10
W = 50 N <em>(</em><em>downwards</em><em>)</em>
Since the block is at rest, an a force equal to the weight of the block must be acting upwards. This force is known as normal reaction.
Fₙ = 50 N <em>(</em><em>upwards</em><em>)</em>
Thus, the name and strength of the force holding the block up is 50 N upward - Normal force.
Learn more about Normal force here: brainly.com/question/14486416
The velocity of B after elastic collision is 3.45m/s
This type of collision is an elastic collision and we can use a formula to solve this problem.
<h3>Elastic Collision</h3>
The data given are;
- m1 = 281kg
- u1 = 2.82m/s
- m2 = 209kg
- u2 = -1.72m/s
- v1 = ?
Let's substitute the values into the equation.
From the calculation above, the final velocity of the car B after elastic collision is 3.45m/s.
Learn more about elastic collision here;
brainly.com/question/7694106
Answer: Some conversions from one system of units to another need to be exact, without increasing or decreasing the precision of the first measurement. This is sometimes called soft conversion. It does not involve changing the physical configuration of the item being measured.
Explanation:
Answer:
Pressure, 
Explanation:
It is given that,
Mass of the woman, m = 55 kg
Diameter of the circular cross section, d = 6 mm
Radius, r = 3 mm = 0.003 m
Let P is the pressure exerted on the floor. It is equal to the force acting on woman per unit area. It is given by :
So, the pressure exerted on the floor is 
. Hence, this is the required solution.
