Ask question

Ask question

All categories

3 years ago

6

A set of steel reinforcing rods is to be stretched axially in tension to create a tensile stress of 30 ksi prior to being cast i

n concrete to form a beam. Determine how much force will be required to stretch them

1 answer:

yanalaym [24]3 years ago

4 0

Answer:

13.26 kip

Explanation:

Taking diameter of the rods as 0.75 in then considering that stress is Force per unit area

Force is a product of force and area

$F=A\sigma$ where $\sigma$ is stress

$A=\pi \frac {d^{2}}{4}$ and substituting this into the equation of force we get

$F=\pi \frac {d^{2}}{4}$

Force, $F=30*pi*\frac {0.75^{2}}{4}=13.26 kip$

You might be interested in

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

b

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$

$\frac{m(a+b)^2}{3} \omega_3^2 + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0$

$\frac{1.53(0.6+0.6)^2}{3} \omega_3^2 + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0$

$0.7344 \omega_3^2 = 2.128$

$\omega _3 = \sqrt{\frac{2.128}{0.7344} }$

$\omega _3 =1.70 \ rad/s$

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

7 0

3 years ago

You have a set of calipers that can measure thicknesses of a few inches with an uncertainty of ± 0.005 inches. You measure the t

Bond [772]

Answer:

a) x = (0.0114 ± 0.0001) in , b) the number of decks is 5

Explanation:

a) The thickness of the deck of cards (d) is measured and the thickness of a card (x) is calculated

x = d / 52

x = 0.590 / 52

x = 0.011346 in

Let's look for uncertainty

Δx = dx /dd Δd

Δx = 1/52 Δd

Δx = 1/52 0.005

Δx = 0.0001 in

The result of the calculation is

x = (0.0114 ± 0.0001) in

b) You want to reduce the error to Δx = 0.00002, the number of cards to be measured is

#_cards = n 52

The formula for thickness is

x = d / n 52

Uncertainty

Δx = 1 / n 52 Δd

n = 1/52 Δd / Δx

n = 1/52 0.005 / 0.00002

n = 4.8

Since the number of decks must be an integer the number of decks is 5

3 0

4 years ago

A train moves from rest to a speed of 25 m/s in 30.0 seconds. What is the acceleration?

fredd [130]

Answer:

$a = 0.83\ m/s^2$

Explanation:

<u>Uniform Acceleration </u>

When an object changes its velocity at the same rate, the acceleration is constant.

The relation between the initial and final speeds is:

$v_f=v_o+a.t$

Where:

vf = Final speed

vo = Initial speed

a = Constant acceleration

t = Elapsed time

It's known a train moves from rest (vo=0) to a speed of vf=25 m/s in t=30 seconds. It's required to calculate the acceleration.

Solving for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

Substituting:

$\displaystyle a=\frac{25-0}{30}$

$\boxed{a = 0.83\ m/s^2}$

4 0

3 years ago

True.or false A railroad track runs southwest to northeast.

telo118 [61]

Answer:

ns for high-speed rail in the United States date back to the High Speed Ground Transportation Act of 1965. Various state and federal proposals have followed. Despite being one of the world's first countries to get high-speed trains (the Metroliner service in 1969), it failed to spread. Definitions of what constitutes high-speed rail vary, including a range of speeds over 110 mph (180 km/h) and dedicated rail lines. Inter-city railwith top speeds between 90 and 125 mph (140 and 200 km/h) is sometimes referred to in the United States as higher-speed rail.[1]

Amtrak's Acela Express (reaching 150 mph, 240 km/h), Silver Star, Northeast Regional, Keystone Service, Vermonter and certain MARC Penn Line express trains (all five reaching 125 mph, 201 km/h) are the only high-speed services in the country.

As of 2020, the California High-Speed Rail Authority is working on the California High-Speed Rail project and construction is under way on sections traversing the Central Valley. The Central Valley section is planned to open in 2029 and Phase I is planned for completion in 2031.[2]

Contents

1 Definitions in American context

2 History

2.1 Faster inter-city trains: 1920–1941

2.2 Post-war period: 1945–1960

2.3 First attempts: 1960–1992

2.4 Renewed interest: 1993–2008

2.5 Plans for 2008–2013

3 Current state and regional efforts

3.1 The Northeast

3.1.1 Northeast Corridor: Next Generation High-Speed Rail

3.1.1.1 Proposed routes

3.1.2 Northeast Maglev proposal

3.1.3 New Jersey–New York City upgrades

3.1.4 New York

3.1.5 Pennsylvania

3.2 Western States

3.2.1 California

3.2.2 Pacific Northwest

3.2.3 Arizona

3.3 Mid-Atlantic and the South

3.3.1 Florida

3.3.2 Southeast

3.3.3 Texas

3.4 Midwest

3.4.1 Illinois and the Midwest

3.5 The Southwest

4 Federal high-speed rail initiatives

4.1 American Recovery and Reinvestment Act of 2009

4.1.1 Strategic plan

4.2 2009 federal grant funding

4.3 2010 allocation

4.3.1 Cancellation of funds for Wisconsin, Ohio, and Florida

4.4 2011 and 2012 proposals and rejections of funding

5 See also

6 Notes

7 Further reading

8 External links

Explanation:

3 0

3 years ago

if 6000j of energy is supplid to a machine to lift a load of 300N through a vvertical height of 1M calculatework out put

kari74 [83]

Answer:

300J

Explanation:

Work done = Force x the distance travelled in the direction of the force

=300 x 1

=300J

8 0

3 years ago