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3 years ago

14

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]3 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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Here, mA = 3.65 kg and mB = 7.05 kg. The string connecting the two objects is of negligible mass and the pulley is frictionless.

mr_godi [17]

Answer:

The magintude of the acceleration for both objects is $3.11m/s^2$

Explanation:

Drawing a free body diagram on the two boxes we can analyze the system more easily.

we can take the acceleration going up as positive for reference purposes.

for mA let's suppose that is ascending so:

$T_A-m_A*g=m_A*a$

and for mB (descending):

$T_B-m_B*g=-m_B*a$

$T_A=T_B$

because the two boxes has the same acceleration because they are attached together:

$m_B*g-m_B*a-m_a*g=m_a*a\\(m_B-m_A)*g=(m_a+m_B)*a\\a=\frac{(7.05kg-3.65kg)*9.8m/s^2}{3.65kg+7.05kg}\\\\a=3.11m/s^2$

So the magintude of the acceleration for both objects is $3.11m/s^2$

5 0

3 years ago

An engine absorbs 1749 J from a hot reservoir and expels 539 J to a cold reservoir in each cycle. a. What is the engine’s effici

Masja [62]

Answer:

a)η = 69.18 %

b)W= 1210 J

c)P=3967.21 W

Explanation:

Given that

Q₁ = 1749 J

Q₂ = 539 J

From first law of thermodynamics

Q₁ = Q₂ +W

W=Work out put

Q₂=Heat rejected to the cold reservoir

Q₁ =heat absorb by hot reservoir

W= Q₁- Q₂

W= 1210 J

The efficiency given as

$\eta=\dfrac{W}{Q_1}$

$\eta=\dfrac{1210}{1749}$

$\eta=0.6918$

η = 69.18 %

We know that rate of work done is known as power

$P=\dfrac{W}{t}$

$P=\dfrac{1210}{0.305}\ W$

P=3967.21 W

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3 years ago

The barometer of a mountain hiker reads 930 mbars at the beginning of a hiking trip and 780 mbars at the end. Neglecting the eff

lara31 [8.8K]

We could use the change of pressure to calculate for the height climbed by the mountain hiker. The change of pressure is given by

p = rho * g * h, where p is the change of pressure, rho is the air density, g is the acceleration due to gravity, and h is the height.

Using the conversion 1 mbar = 100 Pa,

(930 - 780)(100) = (1.20)(9.80)h
15000 = 1.20*9.80*h

h = 1.28 km

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3 years ago

What is the scientific revolution??

Basile [38]

The scientific revolution is a concept which explains how the developments of science (biology, chemistry, physics and etc.) changed the way we (society) think about nature. I hope this helps! :)

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4 years ago

You wiggle a string,that is fixed to a wall at the other end, creating a sinusoidalwave with a frequency of 2.00 Hz and an ampli

FinnZ [79.3K]

Answer:

Explanation:

A general wave function is given by:

$f(x,t)=Acos(kx-\omega t)$

A: amplitude of the wave = 0.075m

k: wave number

w: angular frequency

a) You use the following expressions for the calculation of k, w, T and λ:

$\omega = 2\pi f=2\pi (2.00Hz)=12.56\frac{rad}{s}$

$k=\frac{\omega}{v}=\frac{12.56\frac{rad}{s}}{12.0\frac{m}{s}}=1.047\ m^{-1}$

$T=\frac{1}{f}=\frac{1}{2.00Hz}=0.5s\\\\\lambda=\frac{2\pi}{k}=\frac{2\pi}{1.047m^{-1}}=6m$

b) Hence, the wave function is:

$f(x,t)=0.075m\ cos((1.047m^{-1})x-(12.56\frac{rad}{s})t)$

c) for x=3m you have:

$f(3,t)=0.075cos(1.047*3-12.56t)$

d) the speed of the medium:

$\frac{df}{dt}=\omega Acos(kx-\omega t)\\\\\frac{df}{dt}=(12.56)(1.047)cos(1.047x-12.56t)$

you can see the velocity of the medium for example for x = 0:

$v=\frac{df}{dt}=13.15cos(12.56t)$

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3 years ago