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3 years ago

10

On a caterpillars map all distances are marked in kilometers . The caterpillars map shows the distance between two milkweed plan

ts is 4,012 milimeters what is the distance in kilometers

1 answer:

frutty [35]3 years ago

5 0

Answer:

The distance in kilometers is 4012 × $10^{-6}$ km.

Explanation:

We know that the conversion of 1 millimeters is equal to $10^{-3}$ meter. And then the conversion of 1 meter is equal to $10^{-3}$ km. Then the conversion of 1 millimeter to km will be

1 mm = $10^{-3}$ m

1 m = $10^{-3}$ km

So, 1 mm = $10^{-3}$ × $10^{-3}$ km = $10^{-6}$ km.

As here the the distance is 4012 mm, then the distance in km will be

4012 mm = 4012 × $10^{-6}$ km.

So the distance is 4012 × $10^{-6}$ km.

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In which medium would sound travel the fastest? a)across a room b)in a swimming pool c)through outer space d)through a railroad

lora16 [44]

d)through a railroad track

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3 years ago

A counterflow double-pipe heat exchanger is used to heat water from 20°C to 80°C at a rate of 1.2 kg/s. The heating is to be com

bixtya [17]

Answer:L=109.16 m

Explanation:

Given

initial temperature $=20^{\circ}C$

Final Temperature $=80^{\circ}C$

mass flow rate of cold fluid $\dot{m_c}=1.2 kg/s$

Initial Geothermal water temperature $T_h_i=160^{\circ}C$

Let final Temperature be T

mass flow rate of geothermal water $\dot{m_h}=2 kg/s$

diameter of inner wall $d_i=1.5 cm$

$U_{overall}=640 W/m^2K$

specific heat of water $c=4.18 kJ/kg-K$

balancing energy

Heat lost by hot fluid=heat gained by cold Fluid

$\dot{m_c}c(T_h_i-T_h_e)= \dot{m_h}c(80-20)$

$2\times (160-T)=1.2\times (80-20)$

$160-T=36$

$T=124^{\circ}C$

As heat exchanger is counter flow therefore

$\Delta T_1=160-80=80^{\circ}C$

$\Delta T_2=124-20=104^{\circ}C$

$LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}$

$LMTD=\frac{80-104}{\ln \frac{80}{104}}$

$LMTD=91.49^{\circ}C$

heat lost or gain by Fluid is equal to heat transfer in the heat exchanger

$\dot{m_c}c(80-20)=U\cdot A\cdot$ (LMTD)

$A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2$

$A=\pi DL=5.144$

$L=\frac{5.144}{\pi \times 0.015}$

$L=109.16 m$

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3 years ago

Simple Pendulum: A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light

SIZIF [17.4K]

Answer:

The correct answer is "4.443 sec".

Explanation:

Given:

Mass of child,

= 34 kg

Mass of swing,

= 18 kg

Length,

= 4.9 m

The time period of pendulum will be:

T = $2 \pi \sqrt{4g}$

= $2 \pi \sqrt{\frac{4.9}{9.8} }$

= $4.443 \ sec$

5 0

3 years ago

Read 2 more answers

How long does it take for a train to increase its velocity from 10m/s to 40m/s if it accelerates at 3 m/s

DIA [1.3K]

Answer:

Explanation:

Givens

Vi = 10 m/s

Vf = 40 m/s

a = 3 m/s^2

Formula

a = (vf - vi) /t Substitute the givens into this formuls

Solution

3 = (40 - 10) / t Multiply both sides by t

3*t = t(40 - 10)/t Combine. Cancel t's on the right

3*t = 30 Divide by 3

3t/3 = 30 / 3

Answer: t = 10 seconds.

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2 years ago

a child pulls on a string that is attached to a car. if the child does 80.2 J of work while pulling the car 25.0 m, with what fo

12345 [234]

Answer:

F = 3.20 N

Explanation:

Given:

Work done by child = 80.2 j

Distance that the car moves = 25.0 m

We need to find the force acting on the car.

Solution:

Using work done formula as.

$W = F\times d$

Where:

W = Work done by any object.

F = Force (push or pull)

d = distance that the object moves.

Substitute $W = 80.2\ J\ and\ d =25.0\ m$ in work done formula.

$80.2 = F\times 25$

$F=\frac{80.2}{25}$

F = 3.20 N

Therefore, force acting on the car F = 3.20 N

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3 years ago