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3 years ago

A system is initially at conditions of

Physics

1 answer:

adelina 88 [10]3 years ago

5 0

Answer:

a)W=20 KJ

b) ΔQ= 220 KJ

Explanation:

Given:

V₁=0.1 m^3, P₁=200 kPa and heat is added to the system such that system expands under constant pressure.

Therefore V₂= 2V₁= 0.2 m^3

a) Work transfer W= P(V₂-V₁)= $200\times(0.2-0.1)\times10^{5} = 2\times10^4 joules$

W=20 KJ

b) internal energy change ΔU= 200 KJ

from first law we know that ΔQ(net heat transfer)= ΔU + W

ΔQ= $200\times10^3 +2\times10^4$

ΔQ= $22\times10^4 J$

ΔQ= 220 KJ

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the half-life of carbon-14 is 5,730 years. After 11,460 year, how much of original carbon-14 remains?

Inessa [10]

Via the half-life equation:

$A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h}}$

Where the time elapse is 11,460 year and the half-life is 5,730 years.

$A_{final}=A_{initial}(\frac{1}{2})^\frac{11460}{5730} \\\\A_{final}=A_{initial}\frac{1}{4} \\\\A_{final}=\frac{1}{4}A_{initial}$

Therefore after 11,460 years the amount of carbon-14 is one fourth (1/4) of the original amount.

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3 years ago

You push a 1.30 kg physics book 2.80 m along a horizontal tabletop with a horizontal push of 1.55 N while the opposing force of

Rzqust [24]

Answer:

Explanation:

Step one:

given data

mass m= 1.3kg

distance moved s= 2.8m

opposing frictional force= 0.34N

assume g= 9.81m/s^2

we know that work done= force *distance moved

1. work done to push the book= 1.55*2.8=4.34J

2. Work against friction = force of friction x distance

= 0.34*2.8=0.952J

Step two:

the work done on the book is the net work, which is

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Network done= 4.32-0.952=3.36J

<u>Therefore the work of the 1.55N 3.36J</u>

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Seja uma carga Q=1.2×10(-8)C no vácuo, distando 40 cm de um ponto M e 50 cm de um ponto N . Determine os pontencias no ponto M e

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<span>Americano, por favor?</span>

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The graph in the accompanying figure (Figure 1) shows the magnitude of the force exerted by a given spring as a function of the

serious [3.7K]

The elastic potential energy stored in the stretched spring is 1 J.

<h3>What is Hooke's law?</h3>

Hooke's law states that; provided the elastic limit is not exceeded, the extension of the spring is directly proportional to the force on the spring.

Given that;

Force on the spring = 350 Newton

Distance stretched = 7 centimeters or 0.07 m

Hence;

F = ke

k = F/e = 350 Newton/0.07 m = 5000 N/m

Work done in stretching a spring = 1/2ke^2

= 0.5 × 5000 × (2 × 10^-2)^2 =1 J

Learn more about elastic potential energy: brainly.com/question/156316

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2 years ago