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iogann1982 [59]

4 years ago

14

In the physics of motion a force acting over a distance is?

2 answers:

lutik1710 [3]4 years ago

8 0

A force over distance is work the unite is joules

Elan Coil [88]4 years ago

5 0

Simple a force acting over distance is the unite of joules

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Assume that a cloud consists of tiny water droplets suspended (uniformly distributed,

aev [14]

9.8 ms^-2 is acceleration

4 0

4 years ago

1. A heat engine operates between two reservoirs at T2 = 600 K and T1 = 350 K. It takes in 1.00 x 103 J of energy from the highe

Alex73 [517]

Answer:

$\Delta S_u=2.1429\ J.K^{-1}$

$W_c=416.67\ J$

Explanation:

Given:

temperature of source reservoir, $T_H=600\ K$

temperature of sink reservoir, $T_L=350\ K$

energy absorbed from the source, $Q_{in}=1000\ J$

work done, $W=250\ J$

a.

<u>Now change in entropy of the surrounding:</u>

$\Delta S_u=\frac{dQ_L}{T_L}$

<em>Since heat engine is a device that absorbs heat from a high temperature reservoir and does some work giving out heat in the universe as the byproduct.</em>

$\Delta S_u=\frac{Q_H-W}{T_L}$

$\Delta S_u=\frac{1000-250}{350}$

$\Delta S_u=2.1429\ J.K^{-1}$

b.

<u>We know Carnot efficiency is given as:</u>

$\eta_c=1-\frac{T_L}{T_H}$

$\eta_c=1-\frac{350}{600}$

$\eta_c=0.4167$

<u>Now the Carnot work done:</u>

$W_c=Q_H\times \eta_c$

$W_c=1000\times 0.4167$

$W_c=416.67\ J$ .......................(1)

c.

From eq. (1) we have the Carnot work, so the difference:

$\Delta W=W_c-W$

$\Delta W=416.67-250$

$\Delta W=166.67\ J$

Now, we find:

$T_L.\Delta S_u=350\times 2.1429$

5 0

3 years ago

PLEASE HELP! Please get these right

umka2103 [35]

Green. This is because chlorophyll reflects the color green, which is why we see it as this color.

7 0

3 years ago

Given the initial wavefunction Ψ (x, 0) = Axexp (-k x) withx> 0 andk> 0, and Ψ (x, 0) = 0 forx <0, what value must A ta

madam [21]

Answer with explanation:

The Normalization Principle states that

$\int_{-\infty }^{+\infty }f(x)dx=1$

Given

$f(x)=xe^{-kx}(x>0\\\\0(x$

Thus solving the integral we get

$\int_{0 }^{+\infty }A\cdot xe^{-kx}dx=1\\\\A\int_{0 }^{+\infty }\cdot xe^{-kx}dx=1$

The integral shall be solved using chain rule initially and finally we shall apply the limits as shown below

$I=\int xe^{-kx}dx\\\\x\int e^{-kx}dx-\int \frac{d(x)}{dx}\int e^{-kx}dx\\\\-\frac{xe^{-kx}}{k}-\int 1\cdot \frac{-e^{-kx}}{k}\\\\\therefore I=\frac{e^{-kx}}{k}-\frac{xe^{-kx}}{k}$

Applying the limits and solving for A we get

$I=\frac{1}{k}[\frac{1}{e^{kx}}-\frac{x}{e^{kx}}]_{0}^{+\infty }\\\\I=-\frac{1}{k}\\\\\therefore A=-k$

3 0

3 years ago

A closed box is filled with dry ice at a temperature of -94.7°C, while the outside temperature is 26.8°C. The box is cubical, me

Nikitich [7]

Answer:

Explanation:

3.64 x 10⁶ J passes through 6 walls

heat energy passing through 1 wall = 0.606 x 10⁶ J

Surface Area of 1 wall A = .285² = 0.081225 m²

Temperature Difference = T₁ - T₂ = 26.8 + 94.7 = 121.5

Thickness of wall d = 3.75 x 10⁻² m

Rate of heat flow per second R = $\frac{0.606 \times10^6}{24\times60\times60}$

=7.01 J per s.

Formula for rate of heat flow

R = $\frac{KA(T_1-T_2)}{d}$

Where K is thermal conductivity.

7.01 = $\frac{K\times121.5\times.081225}{3.75\times10^{-2}}$

K = 2.66 X 10⁻² W m⁻¹s⁻¹

8 0

3 years ago