Answer:
630cm/s
Explanation:
In simple harmonic motion, the tangential velocity is expressed mathematically as v = ὦr
ὦ is the angular velocity = 2πf
r is the radius of the disk
f is the frequency
Given the radius of disk = 10cm
frequency = 10Hz
v = 2πfr
v = 2π×10×10
v = 200π
v = 628.32 cm/s
The tangential velocity = 630cm/s ( to 2 significant figures)
Answer:

Explanation:
For this problem, we can use Boyle's law, which states that for a gas at constant temperature, the product between pressure and volume remains constant:

which can also be rewritten as

In our case, we have:
is the initial pressure
is the initial volume
is the final pressure
Solving for V2, we find the final volume:

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
<span> Let’s determine the initial momentum of each car.
#1 = 998 * 20 = 19,960
#2 = 1200 * 17 = 20,400
This is this is total momentum in the x direction before the collision. B is the correct answer. Since momentum is conserved in both directions, this will be total momentum is the x direction after the collision. To prove that this is true, let’s determine the magnitude and direction of the total momentum after the collision.
Since the y axis and the x axis are perpendicular to each other, use the following equation to determine the magnitude of their final momentum.
Final = √(x^2 + y^2) = √(20,400^2 + 19,960^2) = √814,561,600
This is approximately 28,541. To determine the x component, we need to determine the angle of the final momentum. Use the following equation.
Tan θ = y/x = 19,960/20,400 = 499/510
θ = tan^-1 (499/510)
The angle is approximately 43.85˚ counter clockwise from the negative x axis. To determine the x component, multiply the final momentum by the cosine of the angle.
x = √814,561,600 * cos (tan^-1 (499/510) = 20,400</span>