Answer:
α = 5 rad / s²
Explanation:
This is a rotational kinematics exercise.
They indicate the initial velocity wo = 10 rad / s
w = w₀ + α t
α =
let's calculate
α =
α = 5 rad / s²
The velocity, the angular relation are the same in all the points of the wheel, the velocities and linear accelerations are the ones that change
a = α r
v = w r
Answer: radio waves, microwaves, infrared waves.
Explanation:
Electromagnetic spectrum increases in wavelength and decreases with frequency in the following order
Gamma ray, X ray, Ultraviolet rays, Visible light, Infrared rays, Radio waves.
Since 397nm is greater than the value of visible spectrum, this shows that radio waves, microwaves and infrared wave falls in the spectrums of the wavelength value.
Answer:3.92475J
Explanation:
ideal gas equation
PV=nrT
P=pressure
V=volume 9*5*14=630m^3
n=mole
Mole=mass/molecular mass
molecular mass of air=28.1g/mol
R=gas constant
T=temperature in kelvin
1.01*10^5*630=(m/28)*8.314j/molK *273K
784959.5677356082g
784.95kg
Kinetic energy=0.5*m*v^2
Velocity of air
0.5*784.95kg*0.1^2
K.E=3.92475J
6
Magnetic field B is produced when a current I Amphere passes through a solenoid. B is parallel to its axis.
B=U N/I I. N is number of turns in the solenoid of lm length
N=200, l= 20cm= 0.2m, I = 1₀sin (2πft) where f is equal to 60Hz
B= 4π × 10⁻⁷(200/0.2) l₀ sin (2πft) T
=1.256 × 10⁻³ l₀ sin(2πft) Tesla
Area of the coil is πr² = π (1.5cm)² = 2.25π ×10⁴m²
magnetic flux which is through the coil is given by
Ф = B.A = BA cosФ
ФФ = O since B is in direction of A
A= 40 ×π×2.25 ×⁻⁴m² which is the number of turns being 40.
Flux Ф through the coil is,
1.256 ×10⁻³ l₀ sin (2πft) ×9π ××10⁻³m²
=35.5 ×10⁻⁶ l₀ sin ₀(2πft)ab
Ф is time-varying emf will be generated in the coil
∈= dФ/dt
∈ = d/dt [35.5 × 10⁻d l₀ sin (2πft) ab]
∈ = 35.5 ×10⁻⁶ l₀ 2πf cos 2πftV
f = 60Hz
∈∈ 13376.4 ×10⁻⁶ l₀ cos 2πftV
Current I amp shall be induced in the cell of resistance Rohm so
I= E/R
I = 13376.4 ×10⁻⁶ l₀ cos 2πft)V/0.4∩
=33441 ×10⁻⁶ I₀ cos 2πft A
I = 3344q × 10 ⁻⁶ l₀
But I = 0.2A
l₀ = (0.2)(10⁶)/33441 = 6.0A