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4 years ago

8

A 1000kg automobile enters a freeway on-ramp at 20 m/s accelerates uniformly up to 40 m/s in a time of 10 seconds. How far does

the automobile travel during the time it is accelerating?

2 answers:

Tomtit [17]4 years ago

7 0

Answer:

The car went 80 m

Explanation:

iris [78.8K]4 years ago

5 0

Answer:

Hence the distance covered by the object is 80m

Explanation:

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The generation of a magnetic field by an electric current is___ .

Alex787 [66]

Answer:

Electromagnetic induction

Explanation:

The process of generating electric current with a magnetic field. It occurs whenever a magnetic field and an electric conductor move relative to one another so the conductor crosses lines of force in the magnetic field.

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The gold foil experiment led to the conclusion that each atom in the foil was composed mostly of empty space because most alpha

katovenus [111]

Answer:

(1) passed through the foil

Explanation:

Ernest Rutherford conducted an experiment using an alpha particle emitter projected towards a gold foil and the gold foil was surrounded by a fluorescent screen which glows upon being struck by an alpha particle.

When the experiment was conducted he found that most of the alpha particles went away without any deflection (due to the empty space) glowing the fluorescent screen right at the point of from where they were emitted.

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4 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)

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

$A=\pi DL=5.144$

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

$L=109.16 m$

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3 years ago

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zhannawk [14.2K]

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the object is decelerating

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