Answer:
The winding density of the solenoid, n = 104 turns/m
Explanation:
Given that,
Length of the solenoid, l = 0.7 m
Radius of the circular cross section, r = 5 cm = 0.05 m
Energy stored in the solenoid, 
Current, I = 0.4 A
To find,
The winding density of the solenoid.
Solution,
The expression for the energy stored in the solenoid is given by :
Where
L is the self inductance of the solenoid
n is the winding density of the solenoid
n = 104 turns/m
So, the winding density of the solenoid is 104 turns/m
We are given a mercury atom in the ground state which absorbs 20 eV of energy. It is then ionized by losing an electron. We need to calculate the kinetic energy that the electron has after ionization.
The initial energy is 20 eV = 20 J/C
The electron charge is = 1.60217662 × 10-19<span> coulombs
To determine the kinetic energy, we can use this equation:
KE = 20 Joules / Coulombs * </span>1.60217662 × 10-19<span> coulombs
KE = 1.25x10^20 Joules
Therefore, the amount of kinetic energy that the electron has after ionization is </span>1.25x10^20 Joules or 1.25x10^17 kJ. <span />
Answer:
407 steps
Explanation:
From the question,
P = mgh/t........... Equation 1
Where P = power, m = mass, g = acceleration due to gravity, h = height, t = time.
Make h the subject of the equation
h = Pt/mg............. Equation 2
Given: P = 746 W, t = 1 minute = 60 seconds, m = 70 kg.
Constant: g = 9.8 m/s²
Substitute into equation 2
h = 746(60)/(70×9.8)
h = 44760/686
h = 65.25 m
h = 6525 cm
number of steps = 6525/16
number of steps = 407 steps
Answer:
a) W_total = 8240 J
, b) W₁ / W₂ = 1.1
Explanation:
In this exercise you are asked to calculate the work that is defined by
W = F. dy
As the container is rising and the force is vertical the scalar product is reduced to the algebraic product.
W = F dy = F Δy
let's apply this formula to our case
a) Let's use Newton's second law to calculate the force in the first y = 5 m
F - W = m a
W = mg
F = m (a + g)
F = 80 (1 + 9.8)
F = 864 N
The work of this force we will call it W1
We look for the force for the final 5 m, since the speed is constant the force must be equal to the weight (a = 0)
F₂ - W = 0
F₂ = W
F₂ = 80 9.8
F₂ = 784 N
The work of this fura we will call them W2
The total work is
W_total = W₁ + W₂
W_total = (F + F₂) y
W_total = (864 + 784) 5
W_total = 8240 J
b) To find the relationship between work with relate (W1) and work with constant speed (W2), let's use
W₁ / W₂ = F y / F₂ y
W₁ / W₂ = 864/784
W₁ / W₂ = 1.1
Explanation:it is beause they are sharper and also have less surface area and therefore more pressure
