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Alik [6]
3 years ago
7

Workers load a 500kg safe onto an elevator that lifts it 90m to the twentieth floor of a building. How much work did the elevato

r do on the safe?
Physics
1 answer:
Fantom [35]3 years ago
7 0
Work is Force X Distance. The force is equal to mass X acceleration (F=MA). In this case the acceleration is gravity, which is 9.81 m/s^2. So you multiply 500 by 9.81 to get 4905 Newtons (N). This is the amount of force required to lift the safe. Multiply this force by the distance, which in this case is 90 meters. This gives you 441450 Newton Meters (N*m) of work. 1 N*m is equivalent to 1 Joule (J), so the answer can be written as 441450 Joules (J).
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Parallel rays of monochromatic light with wavelength 571nm illuminate two identical slits and produce an interference pattern on
Natasha2012 [34]

Answer:

8.8\times 10^{-6} W/m^2

Explanation:

We are given that

Wavelength,\lambda=571 nm=571\times 10^{-9} m

1 nm=10^{-9} m

R=75 cm=\frac{75}{100}=0.75 m

1 m=100 cm

d=0.640 mm=0.64\times 10^{-3} m

1 mm=10^{-3} m

a=0.434 mm=0.434\times 10^{-3} m

y=0.830 mm=0.830\times 10^{-3} m

I_0=5.00\times 10^{-4}W/m^2

tan\theta=\frac{y}{R}

\theta=\frac{0.830\times 10^{-3}}{0.75\times 10^{-2}}

\theta=1.1\times 10^{-3}rad

tan\theta\approx \theta

Because \theta is small.

\phi=\frac{2\pi dsin\theta}{\lambda}

\sin\theta\approx \theta,

Therefore

\phi=\frac{2\times\pi\times 0.64\times 10^{-3}\times 1.1\times 10^{-3}}{571\times 10^{-9}}

\phi=7.74 rad

\beta=\frac{2\pi a\theta}{\lambda}

\beta=5.3 rad

I=I_0cos^2(\frac{\phi}{2})(\frac{sin\frac{\beta}{2}}{\frac{\beta}{2}})^2

I=5\times 10^{-4}cos^2(\frac{7.74}{2})(\frac{sin\frac{5.3}{2}}{\frac{5.3}{2}})^2

I=8.8\times 10^{-6} W/m^2

6 0
3 years ago
Exposure to the Sun's harmful infrared radiation should be kept to a minimum.
zysi [14]

Of course! If it's harmful, then your exposure to it should be kept
to a minimum.  That's a no-brainer.  But the sun's infrared radiation
is generally less harmful than its ultraviolet radiation is.

7 0
3 years ago
Ultraviolet light of wavelength 270 nm strikes a metal whose work function is 2.3 eV.What is the shortest de Broglie wavelength
soldier1979 [14.2K]

Answer:

The shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm

Explanation:

Given;

wavelength of ultraviolet light, λ = 270 nm

work function of the metal, φ = 2.3 eV = 2.3 x 1.602 x 10⁻¹⁹ J = 3.685 x 10⁻¹⁹ J

The energy of the ultraviolet light is given by;

E = \frac{hc}{\lambda}\\\\E = \frac{(6.626*10^{-34} )(3*10^{8}) }{270*10^{-9} }\\\\E = 7.362 * 10^{-19} \ J

The energy of the incident light is related to kinetic energy of the electron and work function of the metal by the following equation;

E = φ  + K.E

K.E = E - φ

K.E = (7.362 x 10⁻¹⁹ J) - (3.685 x 10⁻¹⁹ J )

K.E = 3.677 x 10⁻¹⁹ J

K.E = ¹/₂mv²

mv² = 2K.E

velocity of the electron is given by;

V = \sqrt{\frac{2K.E}{m} }\\\\V =  \sqrt{\frac{2(3.677*10^{-19}) }{9.1*10^{-31} } }\\\\V = 8.99*10^{5}  \ m/s

the shortest de Broglie wavelength for the electrons is given by;

\lambda = \frac{h}{mv}\\ \\\lambda = \frac{6.626*10^{-34} }{(9.1*10^{-31})( 8.99*10^{5} )}\\\\\lambda = 8.10*10^{-10} \ m\\\\\lambda = 0.81 \ nm

Therefore, the shortest de Broglie wavelength for the electrons that are produced as photoelectrons is 0.81 nm

7 0
3 years ago
During their physics field trip to the amusement park, Leslie and Maria took a ride on the Whirligig. The Whirligig ride consist
grandymaker [24]

Answer:

a) frequency = 0.1724 Hz

b) Period = 5.8 sec

c) speed = 7.04 m/s

d) acceleration = 7.62 m/s²

Explanation:

Given that;

radius = 6.5m

time period = 5.8 sec every circle

a)  the frequency

frequency is the number of rotation in unit time

frequency = 1 / time period = 1/5.8

frequency = 0.1724 Hz

b)  the period

period is time taken in one rotation

period = total time / rotation = 5.8 / 1

Period = 5.8 sec

c)  the speed

speed = distance/time = circumference/time period = 2πr / t = (2π×6.5) / 5.8

speed = 7.04 m/s

d) acceleration

To find the acceleration we take the linear velocity squared divided by the radius of the circle.

so

acceleration = v² / r = (7.04)² / 6.5 = 49.5616 / 6.5

acceleration = 7.62 m/s²

6 0
3 years ago
An object is dropped from rest from a height of 4.1 107 m above the surface of the Earth. If there is no air resistance, what is
Valentin [98]

Answer:

2.83 x 10^4m/s

Explanation:

First, let us calculate the time taken by the object to hit the surface of the earth.

H = 4.1 x 10^7m

g = 9.8m/s2

t = √(2H/g)

t = √((2x 4.1 x 10^7) /9.8)

t = 2892.64secs

Now, we can find the velocity with which the object strikes the earth as follows:

V = gt

V = 9.8 x 2892.64

V = 2.83 x 10^4m/s

3 0
3 years ago
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