Answer:
1) 50 seconds 2) 100°C
Explanation:
(Follows formula of Power=Energy/Time)
1) 500W x X = 2000J/kg°C x .25kg x 50°C
X = 50 seconds.
2) 2000W x 300s = 1000J/kg°C x 2kg x X
X = 300
Initial temperature => 400°C-300°C = 100°C
Answer:
d=0.137 m ⇒13.7 cm
Explanation:
Given data
m (Mass)=3.0 kg
α(incline) =34°
Spring Constant (force constant)=120 N/m
d (distance)=?
Solution
F=mg
F=(3.0)(9.8)
F=29.4 N
As we also know that
Force parallel to the incline=FSinα
F=29.4×Sin(34)
F=16.44 N
d(distance)=F/Spring Constant
d(distance)=16.44/120
d(distance)=0.137 m ⇒13.7 cm
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
D. Teach the public energy conservation
Answer:
Explanation:
According to the exercise we know the angle which the bait was released and its maximum height
To find the initial y-component of velocity we need to do the following steps:
At maximum height the y-component of velocity is 0 and from the exercise we know that the initial y position is 0
Since the bait is released at 25º
