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3 years ago

Who was pedro infante?????

2 answers:

5 0

Answer: Pedro Infante Cruz (Spanish: [ˈpeðɾo iɱˈfante]; 18 November 1917 – 15 April 1957) was a Mexican actor and singer. Hailed as one of the greatest actors of the Golden Age of Mexican cinema, he is considered an idol in Mexico and other Latin American countries such as Venezuela, Guatemala and Peru.

Explanation:

Anettt [7]3 years ago

3 0

Answer:

Pedro Infante Cruz was a Mexican actor and singer. Hailed as one of the greatest actors of the Golden Age of Mexican cinema, he is considered an idol in Mexico and other Latin American countries such as Venezuela, Guatemala and Peru. He along with Jorge Negrete and Javier Solís are considered the best ranchera singers. Infante was born in Mazatlán, Sinaloa, but raised in nearby Guamúchil. He died on 15 April 1957 in Mérida, Yucatán, in a plane crash during a flight en route to Mexico City, after an engine failed on takeoff. His remains were later identified by a bracelet found.

His film career began in 1939 with him appearing in more than 60 films - 30 of them with his brother Ángel Infante, and starting in 1943 he recorded about 350 songs. For his performance in the movie Tizoc, he was awarded the Silver Bear for Best Actor at the 7th Berlin International Film Festival.

Resources: https://en.wikipedia.org › wiki › Pedro_Infante

Hope this helps : D

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g You drop a 3.6-kg ball from a height of 3.5 m above one end of a uniform bar that pivots at its center. The bar has mass 9.9 k

Salsk061 [2.6K]

Answer:

h = 3.5 m

Explanation:

First, we will calculate the final speed of the ball when it collides with a seesaw. Using the third equation of motion:

$2gh = v_f^2 - v_i^2\\$

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 3.5 m

vf = final speed = ?

vi = initial speed = 0 m/s

Therefore,

$(2)(9.81\ m/s^2)(3.5\ m) = v_f^2 - (0\ m/s)^2\\v_f = \sqrt{68.67\ m^2/s^2}\\v_f = 8.3\ m/s$

Now, we will apply the law of conservation of momentum:

$m_1v_1 = m_2v_2$

where,

m₁ = mass of colliding ball = 3.6 kg

m₂ = mass of ball on the other end = 3.6 kg

v₁ = vf = final velocity of ball while collision = 8.3 m/s

v₂ = vi = initial velocity of other end ball = ?

Therefore,

$(3.6\ kg)(8.3\ m/s)=(3.6\ kg)(v_i)\\v_i = 8.3\ m/s$

Now, we again use the third equation of motion for the upward motion of the ball:

$2gh = v_f^2 - v_i^2\\$

where,

g = acceleration due to gravity = -9.81 m/s² (negative for upward motion)

h = height = ?

vf = final speed = 0 m/s

vi = initial speed = 8.3 m/s

Therefore,

$(2)(9.81\ m/s^2)h = (0\ m/s)^2-(8.3\ m/s)^2\\$

h = 3.5 m

6 0

2 years ago

A ship's wheel has a moment of inertia of 0.930 kilogram·meters squared. The inner radius of the ring is 26 centimeters, and the

Vikki [24]

We can use the formula of the moment of inertia given by:

$r\cdot F=I\alpha$

Where:

r = Distance from the point about which the torque is being measured to the point where the force is applied

F = Force

I = Moment of inertia

α = Angular acceleration

So:

$\begin{gathered} r\cdot F=(-0.26\times314+290\times0.32)=92.8-81.64=11.16 \\ I=0.930 \\ so,_{\text{ }}solve_{\text{ }}for_{\text{ }}\alpha: \\ \alpha=\frac{r\cdot F}{I} \\ \alpha=\frac{11.16}{0.930} \\ \alpha=\frac{12rad}{s^2} \end{gathered}$

Answer:

12 rad/s²

8 0

1 year ago

A loop of area 0.250 m^2 is in a uniform 0.020 0-T magnetic field. If the flux through the loop is 3.83 × 10-3T· m2, what angle

dangina [55]

Answer:

40.0⁰

Explanation:

The formula for calculating the magnetic flux is expressed as:

$\phi = BAcos\theta$ where: