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3 years ago

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According to Newtons second law of motion, which is equal to the acceleration of an object?A: net force + massB: net force x mas

sC: net force - massD: net force divides mass

2 answers:

Assoli18 [71]3 years ago

5 0

D: net force divides mass

Application of F = ma.

mart [117]3 years ago

4 0

Answer:

The acceleration of an object is equal to the net force divided by the mass of the object.

(D) is correct option.

Explanation:

According to Newton's second law of motion

The net force is equal to the product of the mass of the object and the acceleration of the object.

In mathematically terms,

$F = ma$

The acceleration is

$a = \dfrac{F}{m}$

Where, a = acceleration

F = net force

m = mass

Hence, The acceleration of an object is equal to the net force divided by the mass of the object.

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ZanzabumX [31]

Answer:

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3 years ago

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Bogdan [553]

Answer:

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2 years ago

How long was a 60 W light bulb turned on if it used a total of 580 J of energy?

icang [17]

Here's the tool you need. You can't answer the question without this:

   "1 watt"
means
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So 60 watts = 60 joules per second

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3 years ago

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2 years ago

Read 2 more answers

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)

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

$A=\pi DL=5.144$

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

$L=109.16 m$

6 0

3 years ago