Answer:
x = 0.0685 m
Explanation:
In this exercise we can use the relationship between work and energy conservation
W = ΔEm
Where the work is
W = F x
The energy can be found in two points
Initial. Just when the block with its spring spring touches the other spring
Em₀ = K = ½ m v²
Final. When the system is at rest
= 
= ½ k₁ x² + ½ k₂ x²
We can find strength with Newton's second law
∑ F = F - fr
Axis y
N- W = 0
N = W
The friction force has the equation
fr = μ N
fr = μ W
The job
W = (F – μ W) x
We substitute in the equation
(F - μ W) x = ½ m v² - (½ k₁ x² + ½ k₂ x²)
½ x² (k₁ + k₂) + (F - μ W) x - ½ m v² = 0
We substitute values and solve
½ x² (20 + 40) + (15 -0.2 2) x - ½ (2/32) 6² = 0
x² 30 + 14.4 x - 1,125 = 0
x² + 0.48 x - 0.0375 = 0
We solve the second degree equation
x = [-0.48 ±√(0.48 2 + 4 0.0375)] / 2
x = [-0.48 ± 0.617] / 2
x₁ = 0.0685 m
x₂ = -0.549 m
The first result results from compression of the spring and the second torque elongation.
The result of the problem is x = 0.0685 m
Answer:
the order of importance must be b e a f c
Explanation:
Modern theories indicate that the moon was formed by the collision of a bad plant with the Earth during its initial cooling period, due to which part of the earth's material was volatilized and as a ring of remains that eventually consolidated in Moon.
Based on the aforementioned, let's analyze the statements in order of importance
b) True. Since the moon is material evaporated from Earth, its compassion is similar
e) True. If the moon is material volatilized from the earth it must train a finite receding speed
a) True. The solar system was full of small bodies in erratic orbits that wander between and with larger bodies
f) False. The moon's rotation and translation are equal has no relation to its formation phase
c) false. The amount of vaporized material on the moon is large
Therefore, the order of importance must be
b e a f c
Answer: Normal force, N = 141.64 Newton
Explanation:
All the forces acting on the system and described in free body diagram are:
1) gravitational pull in downward direction
2) Normal force in upward direction
3) External force of 40 N acting at an angle of 37° with the horizontal can be resolved in two rectangular components:
i) F Cos 37° along the horizontal plane in forward direction and
ii) F Sin 37° along the vertical plane in downward direction
Applying the Newton's second law, net forces in the vertical plane are:
Net force, f = N - (mg + F Sin 37°)
As there is no acceleration in the vertical plane hence, net force f = 0.
So,
N - (mg + F Sin 37°) = 0
Adding (mg + F Sin 37°) both the sides in above equation, we get
N = mg + F Sin 37°
N = 12 
9.8 + 40 0.601 because (Sin 37° = 0.601)
N = 117.6 + 24.04
N = 141.64 Newton
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
Answer:
455165.278 m
Explanation:
P = Power = 3.7 W
v = Velocity = 10.7 m/s
Amount of fat = 4 g
1 gram of fat provides about 9.40 (food) Calories
Energy given by 4 g of fat
Time required to burn the fat
Distance traveled by the bird
The bird will fly 455165.278 m
