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andrey2020 [161]

2 years ago

12

If the 20-mm-diameter rod is made of A-36 steel and the stiffness of the spring is k = 61 MN/m , determine the displacement of e

nd A when the 60-kN force is applied. Take E = 200 GPa.

1 answer:

ryzh [129]2 years ago

5 0

Answer:

σ = 1.09 mm

Explanation:

Step 1: Identify the given parameters

rod diameter = 20 mm

stiffness constant (k) = 55 MN/m = 55X10⁶N/m

applied force (f) = 60 KN = 60 X 10³N

young modulus (E) = 200 Gpa = 200 X 10⁹pa

Step 2: calculate length of the rod, L

K = \frac{A*E}{L}K=

L

A∗E

L = \frac{A*E}{K}L=

K

A∗E

A=\frac{\pi d^{2}}{4}A=

4

πd

2

d = 20-mm = 0.02 m

A=\frac{\pi (0.02)^{2}}{4}A=

4

π(0.02)

2

A = 0.0003 m²

L = \frac{A*E}{K}L=

K

A∗E

L = \frac{(0.0003142)*(200X10^9)}{55X10^6}L=

55X10

6

(0.0003142)∗(200X10

9

)

L = 1.14 m

Step 3: calculate the displacement of the rod, σ

\sigma = \frac{F*L}{A*E}σ=

A∗E

F∗L

\sigma = \frac{(60X10^3)*(1.14)}{(0.0003142)*(200X10^9)}σ=

(0.0003142)∗(200X10

9

)

(60X10

3

)∗(1.14)

σ = 0.00109 m

σ = 1.09 mm

Therefore, the displacement at the end of A is 1.09 mm

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2 years ago

Match each label to the boundary it describes. convergent boundary new crust forms transform boundary crust submerges into the m

Paraphin [41]

The answers would be:

CONVERGENT boundary - Crust submerges into the mantle

TRANSFORM boundary - neither forms nor submerges

DIVERGENT boundary - new crust forms

If you'd like to know more about the different boundaries, read on:

Convergent boundaries occur when two plates move TOWARDS each other. The event where crust submerges into the mantle is called <em><u>subduction</u></em> and this occurs when an oceanic plate and a continental plate collide. The oceanic plate is more dense and thinner than the continental plate, so it slides under it.

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7 0

2 years ago

Read 2 more answers

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

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1 year ago

The price of coffee fell sharply last month, while the quantity sold remained the same. Five people suggest various explanations

miss Akunina [59]

Answer:

France. True Supply increased and demand did not adapt

Explanation:

If the price of coffee decreases it may be due to two causes

- Increase of the offer

- decrease in demand

- both simultaneously.

Let's examine the statement. They tell us that the amount sold remained constant, so there is no decrease in demand, therefore the only cause may be the increase in supply.

Let's check the answers

Antonio False. The demand is constant.

Carolina. False. The demand is constant.

Dmitri False. The demand is constant.

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1 year ago

What is the change in internal energy (in J) of a system that does 4.50 ✕ 105 J of work while 3.20 ✕ 106 J of heat transfer occu

Dmitrij [34]

Answer:

$-3.25\times 10^6 J$

Explanation:

We are given that

Work done by the system= $4.5\times 10^5$ J

Heat transfer into the system= $U_1=3.2\times 10^6$ J

Heat transfer to the environment= $U_2=6\times 10^6$ J

We have to find the change in internal energy

By first law of thermodynamics

$\Delta Q=\Delta U+w$

$\Delta Q=U_1-U_2=3.2\times 10^6-6\times 10^6=-2.8\times 10^6J$

Substitute the values then we get

$-2.8\times 10^6=\Delta U+4.5\times 10^5$

$\Delta U=-2.8\times 10^6-4.5\times 10^5=-28\times 10^5-4.5\times 10^5=-32.5\times 10^5=-3.25\times 10^6 J$

Hence, the change in internal energy = $-3.25\times 10^6 J$

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2 years ago