Answer:
No, i disagree.
Explanation:
If the car is moving, it only has a velocity with a component in the horizontal direction. If we use galilean relativity, the velocity of the ball observed by my friend standing in the ground should only be affected in the horizonal direction, while the vertical stays the same for both observers.
Answer:
The speed of the cart after 8 seconds of Low fan speed is 72.0 cm/s
The speed of the cart after 3 seconds of Medium fan speed is 36.0 cm/s
The speed of the cart after 6 seconds of High fan speed is 96.0 cm/s
Explanation:
took the test on edgenuity
Answer:
The fraction of mass that was thrown out is calculated by the following Formula:
M - m = (3a/2)/(g²- (a²/2) - (ag/2))
Explanation:
We know that Force on a moving object is equal to the product of its mass and acceleration given as:
F = ma
And there is gravitational force always acting on an object in the downward direction which is equal to g = 9.8 ms⁻²
Here as a convention we will use positive sign with acceleration to represent downward acceleration and negative sign with acceleration represent upward acceleration.
Case 1:
Hot balloon of mass = M
acceleration = a
Upward force due to hot air = F = constant
Gravitational force downwards = Mg
Net force on balloon is given as:
Ma = Gravitational force - Upward Force
Ma = Mg - F (balloon is moving downwards so Mg > F)
F = Mg - Ma
F = M (g-a)
M = F/(g-a)
Case 2:
After the ballast has thrown out,the new mass is m. The new acceleration is -a/2 in the upward direction:
Net Force is given as:
-m(a/2) = mg - F (Balloon is moving upwards so F > mg)
F = mg + m(a/2)
F = m(g + (a/2))
m = F/(g + (a/2))
Calculating the fraction of the initial mass dropped:
![M-m = \frac{F}{g-a} - \frac{F}{g+\frac{a}{2} }\\M-m = F*[\frac{1}{g-a} - \frac{1}{g+\frac{a}{2} }]\\M-m = F*[\frac{(g+(a/2)) - (g-a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{g+(a/2) - g + a)}{(g-a)(g+(a/2))} ]\\M-m = F*[\frac{(3a/2)}{g^{2}-\frac{a^{2}}{2}-\frac{ag}{2}} ]](https://tex.z-dn.net/?f=M-m%20%3D%20%5Cfrac%7BF%7D%7Bg-a%7D%20-%20%5Cfrac%7BF%7D%7Bg%2B%5Cfrac%7Ba%7D%7B2%7D%20%7D%5C%5CM-m%20%3D%20F%2A%5B%5Cfrac%7B1%7D%7Bg-a%7D%20-%20%5Cfrac%7B1%7D%7Bg%2B%5Cfrac%7Ba%7D%7B2%7D%20%7D%5D%5C%5CM-m%20%3D%20F%2A%5B%5Cfrac%7B%28g%2B%28a%2F2%29%29%20-%20%28g-a%29%7D%7B%28g-a%29%28g%2B%28a%2F2%29%29%7D%20%5D%5C%5CM-m%20%3D%20F%2A%5B%5Cfrac%7Bg%2B%28a%2F2%29%20-%20g%20%2B%20a%29%7D%7B%28g-a%29%28g%2B%28a%2F2%29%29%7D%20%5D%5C%5CM-m%20%3D%20F%2A%5B%5Cfrac%7B%283a%2F2%29%7D%7Bg%5E%7B2%7D-%5Cfrac%7Ba%5E%7B2%7D%7D%7B2%7D-%5Cfrac%7Bag%7D%7B2%7D%7D%20%5D)
Answer:
Fe= 2579.68 P
α= 24.8°
Explanation:
Look at the attached graphic
we take the forces acting on the x-y plane and applied at the origin of coordinates
FX = 1260 P , horizontal (-x)
FY = 1530 P , forming 45° with positive x axis
x-y components FY
FYx= - 1530*cos(45)° = - 1081.87 P
FYy= - 1530*sin(45)° = - 1081.87 P
Calculation of the components of net force (Fn)
Fnx= FX + FYx
Fnx= -1260 P -1081.87 P
Fnx= -2341.87 P
Fny=FYy
Fny= -1081.87 P
Calculation of the components of equilibrant force (Fe)
the x-y components of the equilibrant force are equal in magnitude but in the opposite direction to the net force components:
Fnx= -2341.87 P, then, Fex= +2341.87 P
Fny= -1081.87 P P, then, Fex= +1081.87 P
Magnitude of the equilibrant (Fe)


Fe= 2579.68 P
Calculation of the direction of equilibrant force (α)


α= 24.8°
Look at the attached graphic
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