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2 years ago

12

Un pintor de 75.0 kg sube por una escalera de 2.75 m que está inclinada contra una pared vertical. La escalera forma un ángulo d

e 30.0° con la pared. ¿Cuánto trabajo (en Julios) realiza la gravedad sobre el pintor? *

1 answer:

dezoksy [38]2 years ago

6 0

Answer:

Work done, W = 1786.17J

Explanation:

The question says "A 75.0-kg painter climbs a 2.75-m ladder that is leaning against a vertical wall. The ladder makes an angle of 30.0 ° with the wall. How much work (in Joules) does gravity do on the painter? "

Mass of a painter, m = 75 kg

He climbs 2.75-m ladder that is leaning against a vertical wall.

The ladder makes an angle of 30 degrees with the wall.

We need to find the work done by the gravity on the painter.

The angle between the weight of the painter and the displacement is :

θ = 180 - 30

= 150°

The work done by the gravity is given by :

$W=Fd\cos\theta\\\\=75\times 10\times 2.75\times \cos30\\\\=1786.17\ J$

Hence, the required work done is 1786.17 J.

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A visible light has a wavelength of 727.3 nm. Determine its frequency, energy per photon, and color.

Snezhnost [94]

Answer:

$f=4.12\times 10^{14}\ Hz$ and $E=2.73\times 10^{-19}\ J$

Explanation:

The wavelength of a visible light is 727.3 nm.

$727.3\ nm=727.3 \times 10^{-9}\ m$

The formula is as follows :

$c=f\lambda$

f is the frequency of the visible light

$f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{727.3 \times 10^{-9}}\\\\f=4.12\times 10^{14}\ Hz$

Energy of a photon is given by :

E = hf, h is Planck's constant

$E=6.63\times 10^{-34}\times 4.12\times 10^{14}\\\\E=2.73\times 10^{-19}\ J$

Red color has a frequency of $4.12\times 10^{14}\ Hz$ and energy per photon is $2.73\times 10^{-19}\ J$ .

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3 years ago

Consider the cross section to the right with an overall height h 4.2 inch, bottom flange plate width b1-2.5 inch, top flange pla

ICE Princess25 [194]

Answer: a) 1.98inch

b) 2.737 2.44

c) 0.497inch

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Explanation:

Solution is attached in the pictures below

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3 years ago

Choose the correct statement regarding a compound microscope being used to look at a mite. The eyepiece lens forms a virtual ima

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Answer:

a) True. The image of the mite is virtual

e) True. The image must be within the focal length of the eyepiece len

Explanation:

Let's review the general characteristics of compound microscopes

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Magnification is

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Where fo is the focal length of the objective lens and fe is the focal length of the ocular lens, L is the tube length

Let's review the claims

a) True. The image of the mite is virtual

b) False. The effect is the opposite of the magnification equation

c) False. The objective lens forms a real image

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3 years ago

An electron in a cathode-ray beam passes between 2.5cm long parallel-plate electrodes that are 6.0mm apart. A 2.1mT, 2.5-cm-wide

Dmitry_Shevchenko [17]

Answer:

(a). The speed of electron is $1.56\times10^{7}\ m/s$ .

(b). The radius of electron is $4.2\ cm$

Explanation:

Given that,

Length = 2.5 cm

Distance = 6.0 mm

Magnetic field = 2.1 T

Potential difference = 700 V

(a). We need to calculate the electron's speed

Using formula of speed

$v=\sqrt{\dfrac{2eV}{m}}$

Put the value into the formula

$v=\sqrt{\dfrac{2\times1.6\times10^{-19}\times700}{9.1\times10^{-31}}}$

$v=15689290.81\ m/s$

$v=1.56\times10^{7}\ m/s$

(b). We need to calculate the radius of electron

Using formula of centripetal force

$\dfrac{mv^2}{r}=qvB$

$r=\dfrac{mv}{qB}$

Where,

m = mass of electron

v = speed of electron

r = radius

q = charge of electron

B = magnetic field

Put the value into the formula

$r=\dfrac{9.1\times10^{-31}\times1.56\times10^{7}}{1.6\times10^{-19}\times2.1\times10^{-3}}$

$r=0.042\ m$

$r=4.2\ cm$

Hence, (a). The speed of electron is $1.56\times10^{7}\ m/s$ .

(b). The radius of electron is 4.2 cm

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3 years ago

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tresset_1 [31]

Given:

Height of tank = 8 ft

and we need to pump fuel weighing 52 lb/ $ft^{3}$ to a height of 13 ft above the tank top

Solution:

Total height = 8+13 =21 ft

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Area of cross-section = $\pi r^{2}$ = $\pi 4^{2}$ =16 $\pi$ $ft^{2}$

Now,

Work done required = $\int_{0}^{8} 52\times 16\pi (21 - y)dy$

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= 832 $\pi($ $[ 21y ]_{0}^{8} - [\frac{y^{2}}{2}]_{0}^{8}\\$ )

= 113152 $\pi$ = 355477 ft-lb

Therefore work required to pump the fuel is 355477 ft-lb

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3 years ago