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3 years ago

7

¿Qué características debe tener el movimiento de una persona para que el valor del desplazamiento sea igual al de la distancia r

ecorrida?

1 answer:

SIZIF [17.4K]3 years ago

3 0

Explanation:

What characteristics must the movement of a person have so that the value of the displacement is equal to the distance traveled?

Displacement is equal to the shortest path covered by an object. It is given by the difference of final position and the initial position.

Distance is equal to the total path covered by an object during the journey.

When an object moves in a straight line path, in this case, the displacement is equal to the distance traveled.

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A roller coaster car rapidly picks up speed as it rolls down a slope as it starts down the slope its speed is 4m/s but 3 seconds

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Answer:

The acceleration is 6 [m/s^2]

Explanation:

We can find the acceleration of the roller coaster using the kinematic equation for uniformly accelerated motion.

$v_{f} =v_{i} + a*t\\where:\\v_{f} = final velocity = 22 [m/s]\\v_{i} = initial velocity = 4 [m/s]\\t = time = 3 [s]\\$

Now replacing the values we have:

$a=\frac{v_{f} - v_{i} }{t} \\a=\frac{22 - 4 }{3}\\a = 6 [m/s^{2} ]$

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2 years ago

a heavy jar sits on top of a 3.4 m shelf with a gravitational potential energy of 180 j. What is the mass of the jar?

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g.p.e= mgh

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2 years ago

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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 $d_i=1.5 cm$

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

balancing energy

Heat lost by hot fluid=heat gained by cold Fluid

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

$160-T=36$

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As heat exchanger is counter flow therefore

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

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

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

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

$LMTD=91.49^{\circ}C$

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

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