Answer:
Incomplete question
Check attachment for the given diagram
Explanation:
Given that,
Initial Velocity of drum
u=3m/s
Distance travelled before coming to rest is 6m
Since it comes to rest, then, the final velocity is 0m/s
v=3m/s
Using equation of motion to calculate the linear acceleration or tangential acceleration
v²=u²+2as
0²=3²+2×a×6
0=9+12a
12a=-9
Then, a=-9/12
a=-0.75m/s²
The negative sign shows that the cylinder is decelerating.
Then, a=0.75m/s²
So, using the relationship between linear acceleration and angular acceleration.
a=αr
Where
a is linear acceleration
α is angular acceleration
And r is radius
α=a/r
From the diagram r=250mm=0.25m
Then,
α=0.75/0.25
α =3rad/sec²
The angular acceleration is =3rad/s²
b. Time take to come to rest
Using equation of motion
v=u+at
0=3-0.75t
0.75t=3
Then, t=3/0.75
t=4 secs
The time take to come to rest is 4s
Answer:
Final Velocity = √(eV/m)
Explanation:
The Workdone, W, in accelerating a charge, 2e, through a potential difference, V is given as a product of the charge and the potential difference
W = (2e) × V = 2eV
And this work is equal to change in kinetic energy
W = Δ(kinetic energy) = ΔK.E
But since the charge starts from rest, initial velocity = 0 and initial kinetic energy = 0
ΔK.E = ½ × (mass) × (final velocity)²
(Velocity)² = (2×ΔK.E)/(mass)
Velocity = √[(2×ΔK.E)/(mass)]
ΔK.E = W = 2eV
mass = 4m
Final Velocity = √[(2×W)/(4m)]
Final Velocity = √[(2×2eV)/4m]
Final Velocity = √(4eV/4m)
Final Velocity = √(eV/m)
Hope this Helps!!!
Answer:
<em>C. the blue colour of the Earth's sky</em>
<em></em>
Explanation:
The Pleiades is a cluster of sister stars that are among the closest star cluster to earth.
The reflection nebula of the Pleiades is due to the scattering of the blue light from the hot blue luminous stars that dominate the star cluster. Th blue light is scattered from dust molecules, thought to be predominantly carbon compound like diamond dusts, and other compounds like iron.
The blue colour of the Earth's sky is the closest terrestrial phenomenon to the reflection nebula. On a clear cloudless day, molecules in the air scatter the blue component of light more than the other component colours of white light, giving the sky its characteristic blue coluor.
The common characteristics of the luminous nebula and the Earth's blue sky is that they both have their light scattered by the presence of small particles.
Answer:
a) f = 615.2 Hz b) f = 307.6 Hz
Explanation:
The speed in a wave on a string is
v = √ T / μ
also the speed a wave must meet the relationship
v = λ f
Let's use these expressions in our problem, for the initial conditions
v = √ T₀ /μ
√ (T₀/ μ) = λ₀ f₀
now it indicates that the tension is doubled
T = 2T₀
√ (T /μ) = λ f
√( 2To /μ) = λ f
√2 √ T₀ /μ = λ f
we substitute
√2 (λ₀ f₀) = λ f
if we suppose that in both cases the string is in the same fundamental harmonic, this means that the wavelength only depends on the length of the string, which does not change
λ₀ = λ
f = f₀ √2
f = 435 √ 2
f = 615.2 Hz
b) The tension is cut in half
T = T₀ / 2
√ (T₀ / 2muy) = f = λ f
√ (T₀ / μ) 1 /√2 = λ f
fo / √2 = f
f = 435 / √2
f = 307.6 Hz
Traslate
La velocidad en una onda en una cuerda es
v = √ T/μ
ademas la velocidad una onda debe cumplir la relación
v= λ f
Usemos estas expresión en nuestro problema, para las condiciones iniciales
v= √ To/μ
√ ( T₀/μ) = λ₀ f₀
ahora nos indica que la tensión se duplica
T = 2T₀
√ ( T/μ) = λf
√ ) 2T₀/μ = λ f
√ 2 √ T₀/μ = λ f
substituimos
√2 ( λ₀ f₀) = λ f
si suponemos que en los dos caso la cuerda este en el mismo armónico fundamental, esto es que la longitud de onda unicamente depende de la longitud de la cuerda, la cual no cambia
λ₀ = λ
f = f₀ √2
f = 435 √2
f = 615,2 Hz
b) La tension se reduce a la mitad
T = T₀/2
RA ( T₀/2μ) = λ f
Ra(T₀/μ) 1/ra 2 = λ f
fo /√ 2 = f
f = 435/√2
f = 307,6 Hz
