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4 years ago

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A 1 in diameter solid round bar has a groove 0.1 in deep with a 0.1 in radius machined into it. The bar is made of AISI 1040 CD

steel and is subjected to purely reversed torque of 1800 lbf∙in. Determine the maximum shear stress taking the effect of the groove into account.

1 answer:

Maru [420]4 years ago

3 0

The answer & explanation for this question is given in the attachment below.

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How much energy is required to raise the temperature of 5g of air by 10°C?

Alex777 [14]

You need to know the specific heat capacity of air.
Then energy needed = 0.005 x sp.heat.cap x 10

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3 years ago

Which feature of the sun appears in cycles of about 11 years?

goldfiish [28.3K]

The Sun's magnetic field goes through a cycle, called the solar cycle. Every 11 years or so, the Sun's magnetic field completely flips. This means that the Sun's north and south poles switch places. Then it takes about another 11 years for the Sun's north and south poles to flip back again.

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4 years ago

Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150

Phoenix [80]

Answer:

The temperature of the core raises by $2.8^{o}C$ every second.

Explanation:

Since the average specific heat of the reactor core is 0.3349 kJ/kgC

It means that we require 0.3349 kJ of heat to raise the temperature of 1 kg of core material by 1 degree Celsius

Thus reactor core whose mass is $1.60\times 10^{5}kg$ will require

$0.3349\times 1.60\times 10^{5}kJ\\\\=0.53584\times 10^{5}kJ$

energy to raise it's temperature by 1 degree Celsius in 1 second

Hence by the concept of proportionately we can infer 150 MW of power will increase the temperature by

$\frac{150\times 10^{6}}{0.53584\times 10^{8}}=2.8^{o}C/s$

5 0

3 years ago

A spring with a spring constant value of 125 N/m is compressed 12.2 cm by pushing on it with a 215 g block. When the block is re

allsm [11]

Answer:

v = 2.94 m/s

Explanation:

When the spring is compressed, its potential energy is equal to (1/2)kx^2, where k is the spring constant and x is the distance compressed. At this point there is no kinetic energy due to there being no movement, meaning the net energy in the system is (1/2)kx^2.

Once the spring leaves the system, it will be moving at a constant velocity v, if friction is ignored. At this time, its kinetic energy will be (1/2)mv^2. It won't have any spring potential energy, making the net energy (1/2)mv^2.

Because of the conservation of energy, these two values can be set equal to each other, since energy will not be gained or lost while the spring is decompressing. That means

(1/2)kx^2 = (1/2)mv^2

kx^2 = mv^2

v^2 = (kx^2)/m

v = sqrt((kx^2)/m)

v = x * sqrt(k/m)

v = 0.122 * sqrt(125/0.215) <--- units converted to m and kg

v = 2.94 m/s

3 0

3 years ago

Ann (mass 50 kg) is standing at the left end of a 15-m-long, 500 kg cart that has frictionless wheels and rolls on a frictionles

ivanzaharov [21]

Answer:

13.5 m

Explanation:

M = Mass of cart = 500 kg

m = Ann's mass = 50 kg

$v_m$ = Velocity of Ann relative to cart = 5 m/s

$v_M$ = Velocity of Cart relative to Ann

As the linear momentum of the system is conserved

$Mv_M+mv_m=0\\\Rightarrow v_M=-\frac{mv_m}{M}\\\Rightarrow v_M=-\frac{50\times 5}{500}\\\Rightarrow v_M=-0.5\ m/s$

Time taken to reach the right end by Ann

$Time=\frac{Distance}{Speed}\\\Rightarrow Time=\frac{15}{5}=3\ s$

Distance the cart will move in the 3 seconds

$Distance=Speed\times Time\\\Rightarrow Distance=-0.5\times 3=-1.5\ m$

The negative sign indicates opposite direction

Movement of Ann will be the sum of the distances

$15+(-1.5)=13.5\ m$

The net movement of Ann is 13.5 m

5 0

3 years ago