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4 years ago

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Un bloque de 2.5kg de masa es empujado 2.2m a lo largo de una mesa horizontal sin fricción por una fuerza constante de 16.0 N di

rigida a 25° debajo de la horizontal. Encuentre el trabajo efectuado por: (a) la fuerza aplicada, (b) la fuerza normal ejercida por la mesa, (c) la fuerza de la gravedad, y (d) la fuerza neta sobre el bloque

1 answer:

Tanzania [10]4 years ago

8 0

Answer:

Explanation:

(a) The applied force has two components Fx and Fy. The Fx component is the only one that does work

$W_{x}=F_{x}x=(16N)cos(25)(2.2m)=31.9J$

(b) There in no net force in the vertical component

$F_{N}-F_{y}-F_{g}=0\\F_{N}=F_{y}+F_{g}=(16N)sin(25)+(2.5kg)(9.8\frac{m}{s^{2}})=31.26N$

(c)

$F_{g}=Mg=(2.5kg)(9.8\frac{m}{s^{2}})=24.5N$

(d)

$F_{T}=F_{x}=(16N)cos(25)=14.5N$

I attached an scheme of the force diagram

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A dinner plate falls vertically to the floor and breaks up into three pieces, which slide horizontally along the floor. Immediat

Olegator [25]

Answer:

Explanation:

There will be conservation of momentum along horizontal plane because no force acts along horizontal plane.

momentum of first piece = .320 kg x 2 m/s

= 0.64 kg m/s along x -axis.

momentum of second piece = .355 kg x 1.5 m/s

= 0.5325 kg m/s along y- axis .

Let the velocity of third piece be v and it is making angle of θ with x -axis .

Horizontal component of its velocity = .100 kg x v cosθ = .1 v cosθ

vertical component of its velocity = .100 kg x v sinθ = .1 v sinθ

For making total momentum in the plane zero

.1 v cosθ = 0.64 kg m/s

.1 v sinθ = 0.5325 kg m/s

Dividing

Tanθ = .5325 / .64 = .83

θ = 40⁰.

The angle will be actually 180 + 40 = 220 ⁰ from positive x -axis.

6 0

3 years ago

Read 2 more answers

an optician uses a plane mirror to help him. suppse a patient sits in a chair 2.5m away from him. He views the image of a chart

viva [34]

Answer:

I think 75 m

Explanation:

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5 0

3 years ago

A recent study found that electrons that have energies between 3.45 eV and 19.9 eV can cause breaks in a DNA molecule even thoug

vlada-n [284]

Answer:

The Minimum wavelength is $\lambda_{min}= 382.2nm$

The Maximum wavelength is $\lambda_{max}= 624.2nm$

Explanation:

From the question we are told that

The energy range is $E_r = 3.25eV \ and \ 19.9eV$

Considering $E = 19.9eV$

When a single photon is transferred to to an electron the energy obtained can be calculated as follows

$E = 19.9eV = 19.9 *1.6 *10^{-19}J$

This energy is mathematically represented as

$E = \frac{hc}{\lambda_{max}}$

Here h is the Planck's constant with value of $h= 6.625*10^{-34}J\cdot s$

c is the speed of light with value of $c = 3*10^8 m/s$

Substituting values and making $\lambda$ the subject of the formula

$\lambda_{max} = \frac{hc}{E}$

$= \frac{6.625*10^{-34} * 3.0*10^{8}}{19.9*1.6*10^{-19}}$

$\lambda_{max}= 624.2nm$

Considering $E = 3.25eV$

When a single photon is transferred to to an electron the energy obtained can be calculated as follows

$E = 19.9eV = 3.25 *1.6 *10^{-19}J$

This energy is mathematically represented as

$E = \frac{hc}{\lambda_{min}}$

Substituting values and making $\lambda$ the subject of the formula

$\lambda_{min} = \frac{hc}{E}$

$= \frac{6.625*10^{-34} * 3.0*10^{8}}{3.25*1.6*10^{-19}}$

$\lambda_{min}= 382.2nm$

3 0

3 years ago

A machine is supplied energy at a rate of 4,000 W and does useful work at a rate of 3,760 W. What is the efficiency of the machi

Dmitry_Shevchenko [17]

= (3,760 joule/sec) / (4,000 joule/sec)

= 3,760 / 4,000 = 0.94 = 94%

8 0

4 years ago

Suppose that the Millennium Falcon orbits the Death Star 20,000m above the Star’s surface. Calculate the circular orbit velocity

Nataly [62]

Answer:

18.4 m/s

Explanation:

The gravitational force between the Death Star and the Millenium Falcon is equal to the centripetal force that keeps the Millenium Falcon in circular orbit:

$G\frac{Mm}{(R+h)^2}=m\frac{v^2}{R+h}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$M=5.1\cdot 10^{17} kg$ is the mass of the planet

$m=1.36\cdot 10^6 kg$ is the mass of the Millennium Falcon

$v$ is the orbital velocity of the Millennium Falcon

$R=\frac{160,000 m}{2}=80,000 m$ is the radius of the Death Star

$h=20,000 m$ is the altitude of the Millennium Falcon above the planet's surface

Solving the equation for v, we find the orbital velocity:

$v=\sqrt{\frac{GM}{R+h}}=\sqrt{\frac{(6.67\cdot 10^{-11})(5.1\cdot 10^{17} kg)}{80,000 m+20,000 m}}=18.4 m/s$

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3 years ago