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pantera1 [17]
3 years ago
10

Una pelota de tenis de 100 g de masa lleva una rapidez de 20 m/s. Al ser golpeada por una raqueta, se mueve en sentido contrario

con una rapidez de 40 m/s.
Calcular el impulso. Si le pelota permanece en contacto con la raqueta 〖10〗^(-2) sg, ¿Cuál es el módulo de la fuerza media del golpe?
Physics
1 answer:
REY [17]3 years ago
7 0

Answer:

Explanation:

El impulso aplicado a la pelota produce una variación en su momento lineal.

J = m (V -Vo)

Conviene elegir positivo el sentido de la velocidad final.

J = 0,100 kg [40 - (- 20)] m/s = 6 kg m/s

Saludos Herminio

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3 years ago
The objective lens and the eyepiece of a telescope are spaced 85 cm apart. If the eyepiece is123 D what is the total magnificati
dedylja [7]

Answer:

The magnification would be "103.55". A further explanation is given below.

Explanation:

The given values are:

Distance between lens and eyepiece,

L = 85 cm

Eyepiece is,

= 123 D

Now,

The refractive power of eye piece will be:

⇒ \frac{1}{f_e}=123D

   f_e=\frac{1}{123D}

   f_e=0.813 \ cm

The length of the telescope will be:

⇒ L=f_0+f_e

⇒ f_0=L-f_e

On substituting the values, we get

⇒     =85-0.813

⇒     =84.187 \ cm

Now,

The magnification of the telescope will be:

⇒ M=\frac{f_0}{f_e}

⇒      =\frac{84.187}{0.813}

⇒      =103.55

5 0
3 years ago
. A bead of mass m under the influence of gravity slides along a frictionless and massless wire whose height is given by a funct
eduard

Answer:

y = constant

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Since potential energy depends on height, changes in altitude affect potential energy. Going higher increases this energy, this is accompanied by a reduction of kinetic energy and speed (since kinetic energy is related to speed). If the body goes down potential energy is reduced, but kinetic energy and speed increase.

For speed to remain constant the kinetic energy must remain constant. For the kinetic energy to remanin constant, the potential energy must remain constant, and for the potential energy to remain constant the height must remain constant.

7 0
3 years ago
Help with the one above and the three below. I WILL GIVE BRAINLIEST!
kodGreya [7K]

1. The rocket would experience a downwards force due to gravity from the moon.  Also, the rocket would experience an upwards force from the acceleration of the rocket thrusters.  Assuming you are neglecting the force of air friction, there would be 2 forces.  If you are considering air friction as well, there would be also a downwards force of kinetic air friction acting down on the rocket as it flies, making it 3 forces.  If the rocket has already taken off, then there will be no normal force.  However, if the rocket is still on the surface of the moon, the rocket would experience a normal force from the moon.  Since the question isn't specific I can't provide a concrete answer, but it would range from 2-4 forces depending on the above factors.


2.  Since we are not sure if the person in the question is actively lifting the crate, we have to determine the downwards force of the crate due to gravity and compare it to the normal force.  

F = ma

F = (15.3)(-9.8)

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Since the downwards force of the crate is equivalent to the normal force, it means the person is applying no force in picking up the object.  So to pick up a 150N object from scratch, you would have to exert more force than the weight of the object, so the answer is 294N.


3.  Same idea as question 2.  First determine the weight of the object:

F = ma

F = (30)(-9.8)

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The crate in question is not moving, so the magnitudes of the forces in the upwards and downwards direction has to equal to 0.

-294 + 150N + x = 0

x = 144N

So the person is exerting 144 N.  (Which I hope is a choice that you forgot to type out on the question)


5. Image 3 has the least amount of tension since both tension forces exerted on the cables are shared amongst the two cables and they are parallel to the downwards force of gravity that acts on the object and against them.




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