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SCORPION-xisa [38]

3 years ago

15

Un paracaidista desciende desde 6 000 m de altura. Si la masa, con su equipo, es de 65 kg, ¿cuánto valdrá su energía potencial e

n el momento de abrir el paracaídas si lo abre cuando ha descendido 3500m?

1 answer:

iris [78.8K]3 years ago

3 0

Explanation:

PRIMERO ENCUENTRAS EL PESO DEL PARACAIDISTA

$F_{peso}$ = 65 kg( $9.80 m/s^{2}$ ) = 637 N

CON LA FÓRMULA DE LA ENERGÍA POTENCIAL

U = 637 N(6000 m - 3500 m) = 1592500 J

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Phys-1A Horizontal Practice 2

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3 years ago

Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T

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Answer:

$v_{2}$ will be less than $v_{1}$ and $P_{2}$ will be greater than $P_{1}$ .

Explanation:

As we know from the conservation of mass, the rate at which any amount of fluid mass ( $m_{1}$ ) is entering in a system is equal to the rate at which the same amount of fluid mass ( $m_{2}$ ) is leaving the system.

Rate of mass flow can be written as,

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where $\rho$ is the density of the fluid, $A$ is the area through which the fluid is flowing and $v$ is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

$&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}$

where $A_{1}$ and $A_{2}$ are the cross-sectional areas through which the fluid is passing and $v_{1}$ and $v_{2}$ are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, $A_{2} > A_{1}$ , so from the above formula $v_{2} < v_{1}$ .

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case $v_{2} < v_{1}$ the pressure in the large cross-sectional area $P_{2}$ will be greater than the pressure $P_{1}$ in the small cross sectional area, i.e.,

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4 years ago

A particle with charge 3.01 µC on the negative x axis and a second particle with charge 6.02 µC on the positive x axis are each

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Answer:

The third particle should be at 0.0743 m from the origin on the negative x-axis.

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Let's assume that the third charge is on the negative x-axis. So we have:

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$E=k\frac{q}{r^{2}}$

Where:

k is the Coulomb constant
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So, we have:

$k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{3}}{r_{3}^{2}}-k\frac{q_{2}}{r_{2}^{2}}=0$

Let's solve it for r(3).

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I hope it helps you!

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