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elena-14-01-66 [18.8K]
2 years ago
9

happy valentines everyone <3 I dont want a partner i just want a crumb of future PLEASE "A steel bridge is 1000 m long at -20

°C in winter. What is the change in length when the temperature rises to 40°C in summer? The average coefficient of linear expansion of this steel is 11 x 10-6 K-1. *
Physics
2 answers:
pickupchik [31]2 years ago
8 0

Answer:

1000.66m

Explanation:

l 2 = l1(1 +  \alpha \:   \times change \:in \: temperature)

L1=1000m

Temperature 1=-20

L2=?

Temperature 2=40

Temperature difference=40-(-20)

40+20=60

inserting into the formula

l2=l1(1+α×changeintemperature)

L2=1000(1+11×10^-6 ×60)

L2=1000(1+6.6×10^-4)

L2=1000(1.000.66)

L2=1000.66m

notsponge [240]2 years ago
3 0
<h2>Answer :-</h2>

l2 = l1 (1 + α × change in temperature)

L1 = 1000m

Temperature 1 = -20

L2 = ?

Temperature 2 = 40

Temperature difference = 40 - (- 20)

40+20=60

<h3>Inserting into the formula</h3>

  • l2 = l1 (1 + α × change in temperature)

  • L² = 1000(1+11×10^-6 ×60)

  • L² = 1000(1+6.6×10^-4)

  • L² = 1000(1.000.66)

  • L² = 1000.66m Ans.
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A 1650 kg car accelerates at a rate of 4.0 m/s^2. How much force is the car's engine producing?
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F = M A

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If The density of this stainless steel is7.85 g/cm3,specific heatis 0.5 J/g.K, melting pointis 1673K, heat of fusion s0.260J/kg.
kodGreya [7K]

Answer:

\Delta H=687.4 J

Explanation:

Hello!

In this case, for this melting process, we can identify two sub-processes in order to take the stainless steel from solid to liquid:

1. Heat up from 298.15 K to 1673 K.

2. Undergo the phase transition.

Both process have an associated enthalpy as shown below:

\Delta H_1=1g*0.5\frac{J}{g*K} (1673K-298.15K)=687.4J

\Delta H_2=0.001kg*\frac{0.260J}{kg} =0.00026J

Therefore, the required heat is:

\Delta H=\Delta H_1+\Delta H_2\\\\\Delta H=687.4J+0.00026J\\\\\Delta H=687.4J

Notice the problem is not providing neither the mass or volume, that is why we assumed the mass is 1 g; however, it can be changed to the mass you are given.

Best regards!

4 0
2 years ago
a trampoline launches a 50kg person 2m into the air. if the springs push with 1960N of force, how much displacement was there in
Lina20 [59]

Answer: 0.5 m

Explanation:

Given

Mass of the person is m=50\ kg

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Force experience by springs is F=1960\ N

Here, the work done on displacing the springs is equivalent to the Potential energy acquired by the person i.e.

\Rightarrow F\cdot x=mgh\quad [\text{x=displacement of the trampoline}]\\\\\text{Insert the values}\\\\\Rightarrow x=\dfrac{50\times 9.8\times 2}{1960}\\\\\Rightarrow x=\dfrac{980}{1960}\\\\\Rightarrow x=0.5\ m

6 0
2 years ago
Read 2 more answers
2×3.14√(1.0m/(9.8〖ms〗^(-1) )=)
il63 [147K]

This is the period in a simple harmonic motion which is 2 seconds in this question.

<h3>What is Period ?</h3>

The period of an oscillatory object can be defined as the total time taken  by a vibrating body to make one complete revolution about a reference point.

We are given the below question

2×3.14√(1.0m/(9.8〖ms〗^(2) )= T

This question can as well be expressed as

2π√(L/g) which is equal to period T.

In a nut shell, Period T = 2×3.14√(1.0m/9.8)

T = 6.28√0.102

T = 6.28 × 0.32

T = 2.006 s

Therefore, the period T of the oscillation is 2 seconds approximately.

Learn more about Period here: brainly.com/question/12588483

#SPJ1

8 0
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