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3 years ago

5

The best way to become a better reader is to study longer and harder in every subject. engage in a variety of extracurricular ac

tivities. enrich background knowledge and vocabulary. limit reading to books that are interesting.

2 answers:

harina [27]3 years ago

7 0

Answer:

C

Explanation:

I just did the test on enginuity and it also is the only one that makes sence

MatroZZZ [7]3 years ago

6 0

Answer:

c

Explanation:

got it right on edge

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Torque is a twisting force. If the required torque applied on a 3 ft wrench is 45 ft·lb, what is the force that must be applied?

natka813 [3]

Answer:

15 lbs

Explanation:

assuming you push from the end of the wrench (3ft)

torque = force(distance)

force = torque/distance

(45 ft·lb)/(3 ft)= 15 lbs

8 0

4 years ago

For the same cross-sectional area, which column provides the higher buckling load: a circular bar or a circular tube?

juin [17]

Answer:

Circular tube

Explanation:

Now for better understanding lets take an example

Lets take

Diameter of solid bar= $4\sqrt{2}$ cm

Outer diameter of tube =6 cm

Inner diameter of tube=2 cm

So from we can say that both tubes have equal cross sectional area.

We know that buckling load is given as $P = \dfrac{\pi ^2EI}{L_e^2}$

If area moment of inertia(I) is high then buckling load will be high.

We know that area moment of inertia(I)

For circular tube $I = \dfrac{\pi }{64}(D_o^4-D_i^4)$

For circular bar $I = \dfrac{\pi }{64}D^4$

Now by putting the values

For circular tube $I=62.83 cm^4$

For circular bar $I=50.26 cm^4$

So we can say that for same cross sectional area the area moment of inertia(I) is high for tube as compare to bar.So buckling load will be higher in tube as compare to bar.

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3 years ago

Good night. I need to go to bed. Byeeeeeeeeeeeee.

dimaraw [331]

Answer:

BYEEEEEEEEEEEE3EEEEEEEEEE

Explanation:

dawg

8 0

3 years ago

A commuter train traveling at 50 mi/h is 3 mi from a station. The train then decelerates so that its speed is 15 mi/h when it is

jonny [76]

Answer:

a) $t = 277.477\,s\,(4.625\min)$ , b) $v_{f} = 0\,\frac{mi}{h}$ , c) $a = -0.128\,\frac{ft}{s^{2}}$

Explanation:

a) The deceleration experimented by the commuter train in the first 2.5 miles is:

$a=\frac{[(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}-[(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot (\frac{1\,h}{3600\,s} )]^{2}}{2\cdot (2.5\,mi)\cdot (\frac{5280\,ft}{1\,mi} )}$

$a = -0.185\,\frac{ft}{s^{2}}$

The time required to travel is:

$t = \frac{(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )-(50\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,fi} )\cdot(\frac{1\,h}{3600\,s} )}{-0.185\,\frac{ft}{s^{2}} }$

$t = 277.477\,s\,(4.625\min)$

b) The commuter train must stop when it reaches the station to receive passengers. Hence, speed of train must be $v_{f} = 0\,\frac{mi}{h}$ .

c) The final constant deceleration is:

$a = \frac{(0\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )-(15\,\frac{mi}{h} )\cdot (\frac{5280\,ft}{1\,mi} )\cdot(\frac{1\,h}{3600\,s} )}{(2.875\,min)\cdot (\frac{60\,s}{1\,min} )}$

$a = -0.128\,\frac{ft}{s^{2}}$

7 0

3 years ago

What's the monomer? Show the structure.

ivolga24 [154]

In order to understand a monomer let´s first see the structure of a polymer. As an example, in the first figure polyethylene (or polyethene) is shown. This polymer, like every other one, is composed of many repeated subunits, these subunits are called monomer. In the second figure, polyethylene's monomer is shown.

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4 years ago