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3 years ago

15

A force of 2N will stretch a rubber band 0.02m. Assuming that Hooke's Law applies, answer the following: How far will a 1600N fo

rce stretch the rubber band? How much work does it take to stretch the rubber band this far?

1 answer:

denis-greek [22]3 years ago

7 0

Hooke's Law states that the extension is directly proportional to the force applied so:
F/x = constant

F₁/x₁ = F₂/x₂
2 / 0.02 = 1600 / x₂
x₂ = 16 m

Elastic work = 1/2 Fx
= 1/2 * 1600 * 16
= 12.8 kJ

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The maximum distance from the Earth to the Sun (at aphelion) is 1.521 1011 m, and the distance of closest approach (at perihelio

LUCKY_DIMON [66]

Answer:

29274.93096 m/s

$2.73966\times 10^{33}\ J$

$-5.39323\times 10^{33}\ J$

$2.56249\times 10^{33}\ J$

$-5.21594\times 10^{33}$

Explanation:

$r_p$ = Distance at perihelion = $1.471\times 10^{11}\ m$

$r_a$ = Distance at aphelion = $1.521\times 10^{11}\ m$

$v_p$ = Velocity at perihelion = $3.027\times 10^{4}\ m/s$

$v_a$ = Velocity at aphelion

m = Mass of the Earth = 5.98 × 10²⁴ kg

M = Mass of Sun = $1.9889\times 10^{30}\ kg$

Here, the angular momentum is conserved

$L_p=L_a\\\Rightarrow r_pv_p=r_av_a\\\Rightarrow v_a=\frac{r_pv_p}{r_a}\\\Rightarrow v_a=\frac{1.471\times 10^{11}\times 3.027\times 10^{4}}{1.521\times 10^{11}}\\\Rightarrow v_a=29274.93096\ m/s$

Earth's orbital speed at aphelion is 29274.93096 m/s

Kinetic energy is given by

$K=\frac{1}{2}mv_p^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(3.027\times 10^{4})^2\\\Rightarrow K=2.73966\times 10^{33}\ J$

Kinetic energy at perihelion is $2.73966\times 10^{33}\ J$

Potential energy is given by

$P=-\frac{GMm}{r_p}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.471\times 10^{11}}\\\Rightarrow P=-5.39323\times 10^{33}$

Potential energy at perihelion is $-5.39323\times 10^{33}\ J$

$K=\frac{1}{2}mv_a^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(29274.93096)^2\\\Rightarrow K=2.56249\times 10^{33}\ J$

Kinetic energy at aphelion is $2.56249\times 10^{33}\ J$

Potential energy is given by

$P=-\frac{GMm}{r_a}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.521\times 10^{11}}\\\Rightarrow P=-5.21594\times 10^{33}$

Potential energy at aphelion is $-5.21594\times 10^{33}\ J$

6 0

3 years ago

Which of the following is not possible? A. Gas flow equals pressure gradient over resistance. B. Resistance equals pressure grad

kaheart [24]

Answer:

C. Pressure gradient equals gas flow over resistance.

Explanation:

As we know that pressure gradient is the driving force for the gas to flow from one point to other point

And we know that the flow rate is directly proportional to the driving force and it inversely depends on the resistance to flow

so we can say

Flow Rate = $\frac{Driving \: force}{Resistance}$

Flow Rate = $\frac{Pressure \: Gradient}{Resistance}$

so we can say that correct statements are as below

A. Gas flow equals pressure gradient over resistance.

B. Resistance equals pressure gradient over gas flow.

D. The amount of gas flowing in and out of the alveoli is directly proportional to the difference in pressure or pressure gradient between the external atmosphere and the alveoli.

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3 years ago

Glycerin at a temperature of 30 degrees celcius flows at a rate of 8×10−6m3/sthrougha horizontal tube with a 30mmdiameter. what

marusya05 [52]

The pressure drop in pascal is 3.824*10^4 Pascals.

To find the answer, we need to know about the Poiseuille's formula.

<h3>How to find the pressure drop in pascal?</h3>

We have the Poiseuille's formula,

$Q=\frac{\pi r^4P}{8\beta l}$

where, Q is the rate of flow, P is the pressure drop, r is the radius of the pipe, is the coefficient of viscosity (0.95Pas-s for Glycerin) and l being the length of the tube.

By substituting values and rearranging we will get the pressure drop as,

$P=3.284Pascals$

Thus, we can conclude that, the pressure drop in pascal is 3.824*10^4.

Learn more about the Poiseuille's formula here:

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the pressure difference between an oil and water pipe is measured by double fluid manometer as shown in figure below for the giv

Ahat [919]

The answer above me is correct

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jeka94

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