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2 years ago

11

A 0.56 kg model rocket produces 53 newtons of upward thrust as it shoots upward off the launch pad. With what acceleration meter

s per second-squared does the rocket launch?

1 answer:

Yanka [14]2 years ago

6 0

resultant force = thrust – weight

acceleration = resultant force (newtons, N) divided by mass (kilograms, kg).

Acceleration = resultant force divided by mass

53N/0.56

=94.64 approximately 95

= 95m/s^2

This means that, every second, the speed of the rocket increases by 95m/s2

the S.I unit of Acceleration is meter per second square.

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This time particle A starts from rest and accelerates to the right at 65.5 cm/s

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Answer:

t = 4 s

Explanation:

As we know that the particle A starts from Rest with constant acceleration

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Now we know that B moves with constant speed so in the same time B will move to another distance

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now we know that B is already 349 cm down the track

so if A and B will meet after time "t"

then in that case

$d_1 = 349 + d_2$

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on solving above kinematics equation we have

$t = 4 s$

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2 years ago

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Answer:

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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

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

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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

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

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$160-T=36$

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As heat exchanger is counter flow therefore

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

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

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

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$LMTD=91.49^{\circ}C$

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$A=\pi DL=5.144$

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

$L=109.16 m$

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2 years ago