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3 years ago

8

un futbolista patea una pelota que se encuentra en el pasto con un angulo de 30° (medido desde la horizontal) con la intención d

e hacer un gol en un arco que se encuentra a 30m desde su posición. si la altura del arco es de 2m (medido desde el pasto al travesaño) y el jugador patea directamente en dirección del arco, ¿a que velocidad debe patear la pelota para hacer el gol? ¿cuanto tiempo se demora la pelota en llegar al arco?

1 answer:

Rus_ich [418]3 years ago

8 0

Answer:

i dont really know what it is

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When a potassium atom forms an ion, it loses one electron. What is the electrical charge of the potassium ion? *

Ganezh [65]

+1 An electron has a negative charge so losing a charge of -1 from an uncharged, or neutral, atom will leave an ion with a positive charge.

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3 years ago

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A solid sphere of radius R is placed at a height of 30 cm on a15 degree slope. It is released and rolls, without slipping, to th

photoshop1234 [79]

Answer:

The height is $h_c = 42.857$

A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )

Explanation:

From the question we are told that

The height is $h_s = 30 \ cm$

The angle of the slope is $\theta = 15^o$

According to the law of conservation of energy

The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

$mgh = \frac{1}{2} I w^2 + \frac{1}{2}mv^2$

Where I is the moment of inertia which is mathematically represented as this for a sphere

$I = \frac{2}{5} mr^2$

The angular velocity $w$ is mathematically represented as

$w = \frac{v}{r}$

So the equation for conservation of energy becomes

$mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2$

$mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]$

$mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]$

$gh_s =[\frac{7}{10} ] v^2$

$v^2 = \frac{10gh_s}{7}$

Considering a circular hoop

The moment of inertial is different for circle and it is mathematically represented as

$I = mr^2$

Substituting this into the conservation equation above

$mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2$

Where $h_c$ is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

$mgh_c = mv^2$

$gh_c = v^2$

$h_c = \frac{v^2}{g}$

Recall that $v^2 = \frac{10gh_s}{7}$

$h_c= \frac{\frac{10gh_s}{7} }{g}$

$= \frac{10h_s}{7}$

Substituting values

$h_c = \frac{10(30)}{7}$

$h_c = 42.86 \ cm$

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2 years ago

If a body travels half its total path in the last 1.50 s of its fall from rest, find the total time of its fall (in seconds).

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Answer:

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given data

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find the total time of its fall

solution

we consider here s is total distance

so equation of motion for distance

s = ut + 0.5 × at² .........1

here s is distance and u is initial speed that is 0 and a is acceleration due to gravity = 9.8 and t is time

so for last 1.5 sec distance is 0.5 of its distance so equation will be

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and

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so velocity v = 0+ 9.8(t-1.5) ..................2

so first we find time

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solve and we get t

t = 3.37 s

so time to fall is 3.914 seconds

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inna [77]

Answer:

I have fond the answer

Explanation:

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You have a 3.00-liter container filled with N₂ at 25°C and 4.45 atm pressure connected to a 2.00-liter container filled with Ar

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$V_2$ = volume of Ar gas = 2.00 L

P = final pressure of gas = ?

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3 years ago