Karina strikes a match to light a candle. Explain what type of reaction the burning match represents in terms of energy
2 answers:
You might be interested in
188m - 556m is the wavelength range for AM radio.
2.77 - 3.40m is the wavelength range for FM radio.
1) For AM radio
f(max) = 1600 × 10³ Hz
f(min) = 540 ×10³Hz
c = fλ
λ = c/f where,
c = speed of light
f = frequency
λ= wavelength
So,
λ(min) = 3 × / 1600 × 10³ = 188 m
λ(max) = 3 × 10⁸/540 ×10³ = 556 m
So wavelength range of AM radio is 188 m - 556m
2) For FM radio
f(max) = 108 × 10⁶ Hz
f(min) = 88 × 10⁶ Hz
λ(min) = 3 × 10⁸ / 108 × 10⁶ = 2.77 m
λ(max) = 3 × 10⁸ / 88 × 10⁶ = 3.40 m
So wavelength range of FM radio is 2.77m - 3.40m
Learn more about frequency here brainly.com/question/10728818
#SPJ1
Answer:
Part 1
The angular speed is approximately 1.31947 rad/s
Part 2
The change in kinetic energy due to the movement is approximately 675.65 J
Explanation:
The given parameters are;
The rotation rate of the merry-go-round, n = 0.21 rev/s
The mass of the man on the merry-go-round = 99 kg
The distance of the point the man stands from the axis of rotation = 2.8 m
Part 1
The angular speed, ω = 2·π·n = 2·π × 0.21 rev/s ≈ 1.31947 rad/s
The angular speed is constant through out the axis of rotation
Therefore, when the man walks to a point 0 m from the center, the angular speed ≈ 1.31947 rad/s
Part 2
Given that the kinetic energy of the merry-go-round is constant, the change in kinetic energy, for a change from a radius of of the man from 2.8 m to 0 m, is given as follows;
I = m·r²
Where;
m = The mass of the man alone = 99 kg
r = The distance of the point the man stands from the axis, r = 2.8 m
v = The tangential velocity = ω/r
ω ≈ 1.31947 rad/s
Therefore, we have;
I = 99 × 2.8² = 776.16 kg·m²
= 1/2 × 776.16 kg·m² × (1.31947)² ≈ 675.65 J