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AleksandrR [38]
3 years ago
6

(D) 13. Describe the differences between an impact socket and a conventional socket.

Engineering
1 answer:
docker41 [41]3 years ago
3 0

Answer:

The wall of an impact socket is around 50% thicker than that of a regular socket, making it suitable for use with pneumatic impact tools, whereas regular sockets should only be used on hand tools.

Explanation:

This allows the socket to remain securely attached to the impact wrench anvil, even under high stress situations.

You might be interested in
Tahir travel twice as far as ahmed, but onley one third as fast. Ahmed starts travel on tuesday at noon at point x to point z 30
shepuryov [24]

Answer:

6:00 pm the next day

Explanation:

Given that

Tahir traveled twice as far as Ahmed. We say,

Ahmed traveled a distance, D

Tahir would travel a distan, 2D

Tahir traveled 1/3 as fast as Ahmed, so we say

Ahmed traveled at a speed, S

Tahir would travel at a speed, S/3

If Ahmed starts travel on tuesday at noon at point x to point z 300km, by 9:00pm,

Time taken by Ahmed to travel is

9:00 pm - 12:00 pm = 9 hours

Ahmed, traveled 300 km in 9 hours, meaning he traveled at 33.3 km in an hour.

Speed, S that Ahmed traveled with is 33.3 km/h

Remember, we stated that Tahir travels at a speed of S/3, that is, The speed of Tahir is

33.3/3 = 11.1 km/h.

300 km would then be traveled in 300 km/11.1 km/h = 27 hours.

Tahir started traveling, 3 hours after Ahmed, that is 12:00 pm + 3:00 hrs = 3:00 pm, and if he's to spend 27 hours on the journey he would reach destination z at 6:00 pm the next day

7 0
3 years ago
What are two factors that determine the thermal energy of a substance
Sav [38]
Mass and chemical composition
6 0
3 years ago
A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of Syt = 60 kpsi and Syc = 75 kpsi. Using
kow [346]

Answer:

2.135

Explanation:

Lets make use of these variables

Ox 16.5 kpsi, and Oy --14,5 kpsi

To determine the factor of safety for the states of plane stress. We have to first understand the concept of Coulomb-Mohr theory.

Mohr–Coulomb theory is a mathematical model describing the response of brittle materials such as concrete, or rubble piles, to shear stress as well as normal stress.

Please refer to attachment for the step by step solution.

4 0
3 years ago
What is the De Broglie wavelength of an electron under 150 V acceleration?
yanalaym [24]

Answer:

0.1 nm

Explanation

Potential deference of the electron is given as V =150 V

Mass of electron m=9.1\times 10^{-31}

Let the velocity of electron = v

Charge on the electron =1.6\times 10^{-19}C

plank's constant h =6.67\times 10^{-34}

According to energy conservation eV =\frac{1}{2}mv^2

v=\sqrt{\frac{2eV}{M}}=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 150}{9.1\times 10^{-31}}}=7.2627\times 10^{-6}m/sec

Now we know that De Broglie wavelength \lambda =\frac{h}{mv}=\frac{6.67\times 10^{-34}}{9.1\times 10^{-31}\times 7.2627\times10^6 }=0.100\times 10^{-9}m=0.1nm

4 0
3 years ago
Read 2 more answers
Given: A graphite-moderated nuclear reactor. Heat is generated uniformly in uranium rods of 0.05m diameter at the rate of 7.5 x
sineoko [7]

Answer:

The maximum temperature at the center of the rod is found to be 517.24 °C

Explanation:

Assumptions:

1- Heat transfer is steady.

2- Heat transfer is in one dimension, due to axial symmetry.

3- Heat generation is uniform.

Now, we consider an inner imaginary cylinder of radius R inside the actual uranium rod of radius Ro. So, from steady state conditions, we know that, heat generated within the rod will be equal to the heat conducted at any point of the rod. So, from Fourier's Law, we write:

Heat Conduction Through Rod = Heat Generation

-kAdT/dr = qV

where,

k = thermal conductivity = 29.5 W/m.K

q = heat generation per unit volume = 7.5 x 10^7 W/m³

V = volume of rod = π r² l

A = area of rod = 2π r l

using these values, we get:

dT = - (q/2k)(r dr)

integrating from r = 0, where T(0) = To = Maximum center temperature, to r = Ro, where, T(Ro) = Ts = surface temperature = 120°C.

To -Ts = qr²/4k

To = Ts + qr²/4k

To = 120°C + (7.5 x 10^7 W/m³)(0.025 m)²/(4)(29.5 W/m.°C)

To = 120° C + 397.24° C

<u>To = 517.24° C</u>

5 0
3 years ago
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