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eduard
3 years ago
14

A link in a mechanism is to be subjected to a tensile force that varies from 3500 to 500 N in a cyclical fashion as the mechanis

m runs. It has been decided to use SAE 1040 cold-drawn steel. Complete the design of the link, specifying a suitable cross-sectional shape and dimensions.
Physics
1 answer:
shtirl [24]3 years ago
5 0

Answer:

Design explained below

Explanation:

We are designing a link which is able to sustain 3500 N force because it will automatically sustain any lower force.

 

The tensile strength of AISI 1040 steel is : 585 MPa

Shear Strength = 338.15 MPa from data book

Now, 585 * 10^6 = 3500 / [Area]

(from pressure = force / area)

So, area = 5.98*10^-6= pi / 4 * d^2

So, diameter = 2.76 mm

Also, 338.15 *10^6 = 3500 / (pi * d * L)

So, L = 37.5 mm

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What happens to light waves at the interface between different media?
nevsk [136]

Answer:

Refractive motion is the impact of a light wave that travels from medium to medium in an angle away from normal, where the direction of light varies. Light is refracted when it crosses the air-to-glass interface and moves slower.

Explanation:

Refractive motion is the impact of a light wave that travels from medium to medium in an angle away from normal, where the direction of light varies. Light is refracted when it crosses the air-to-glass interface and moves slower.

Hope this helps.

6 0
3 years ago
Hicham El Guerrouj of Morocco holds the world record in the 1500 m running race. He ran the final 400 m in a time of 51.9 s.
konstantin123 [22]

Answer: His average speed in mph over the last 400 m is 7.7 m/s.

Explanation:

Given: Hicham El Guerrouj of Morocco holds the world record in the 1500 m running race. He ran the final 400 m in a time of 51.9 s.

We know that , speed = \dfrac{distance}{time}

Here , distance = 400m and time = 51.9 s

Then, speed =  \dfrac{400}{51.9}\approx7.7\ m/s

Hence, his average speed in mph over the last 400 m is 7.7 m/s.

5 0
3 years ago
An electric field of 2.09 kV/m and a magnetic field of 0.358 T act on a moving electron to produce no net force. If the fields a
My name is Ann [436]

Answer:

The velocity is  v =  5838 \ m/s

Explanation:

From the question we are told that

   The electric field is E  =  2.09 kV/m =  2.09 *10^{3} \ V/m

    The magnetic field is  B  =  0.358 \ T

     

Generally the force experienced by the electron due to the magnetic field is

         F_m  =  qvB

Generally the force experienced by the electron due to the electric  field is

       F_e =  qE

Since from the question the net force is zero  then

     F_e =  F_m

=>    v =  \frac{E}{B}

Substituting values

      v =  \frac{2.09*10^{3}}{0.358 }

    v =  5838 \ m/s

     

8 0
3 years ago
A tellurium-sapphire laser can produce light at wavelength of 800 nm in ultrashort pulses that last only 4.00x10-15s (4.00 femto
natka813 [3]

The speed of light and the propagation of errors allows to find the results on the questions of the radiation emitted by the laser are:

    a) The frequency is: f = 3.7 10¹⁴ Hz

    b) The energy with its uncertainty is: E = (2.465 ± 0.004) 10⁻¹⁹ J

a) The speed of a wave is related to its wavelength and frequency.

           c = λ f

           f = \frac{c}{\lambda}

Where c is the speed of light, λ the wavelength and f the frequency.

They indicate that the wavelength is λ = 800 nm = 800 10⁻⁹ m, the speed of light is a constant c = 2.99 10⁸ m/s.

         f = \frac{2.99 \ 10^8}{800 \ 10^{-9}}  

         F = 3.7 10¹⁴ Hz

b) Planck's equation states that the energy is proportional to the frequency of the radiation.

         E = h f

Where E is the energy, h the Planck constant and f the frequency.

         E = 6.63 10⁻³⁴  3.7 10¹⁴

         E = 2.46467 10⁻¹⁹ J

The uncertainty or error is the fluctuation that a magnitude may have due to the precision in the measurements, when the magnitude is calculated by some formula, the propagation of these uncertainties must be carried out.

         

        Δm = ∑   \sum \frac{dm}{dx_i}  | \Delta x_I|  

the expression for energy is:

        E = \frac{hc}{\lambda }  

        \Delta E = \frac{dE}{d \lambda} |D\lambda |  

        \Delta E = \frac{h c }{\lambda^2 } |\Delta \lambda |

When the error in the measured magnitude is not explicitly indicated, we assume that the error is in the last digit written, therefore

         Δλ = ± 1 nm = ± 1 10⁻⁹ m

We calculate.

        \Delta E = \frac{6.63 \ 10^{-34} \ 2.99 \ 10^8 }{(800 \ 10^{-9})^2} 1 \ 10^{-9}  

        ΔE = 3.1 10⁻²² J

the error is given with a significant figure.

        ΔE = 3 10⁻²² J = 0.004 10⁻¹⁹ J

The result of the energy is:

        E = (2.465 ± 0.004) 10⁻¹⁹ J

In conclusion, using the speed of light and the propagation of errors, we can find the results on the questions of the radiation emitted by the laser are:

    a) The frequency is; f = 3.7 1014 Hz

    b) The energy with its uncertainty is: E = (2.465 ± 0.004) 10⁻¹⁹ J

Learn more here:  brainly.com/question/15557220

3 0
3 years ago
Please help!!!!!!!!!!
pshichka [43]

Answer:

D

Explanation:

5 0
3 years ago
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