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3 years ago

the eccentricity of the Moon's orbit is low, medium, or high with respect to most of the planets' orbits around the sun?

Physics

1 answer:

12345 [234]3 years ago

4 0

person sorry answering for points

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For a quantity of gas at a constant temperature, the pressure P is inversely proportional to the volume V. What is the limit of

IceJOKER [234]

Answer:

$lim_{V \to 0^{+}} P = k\lim_{V \to 0^{+}} \frac{1}{V} =\infty$

This limit is not defined.

Explanation:

We need to remember first this law

The Boyle's law states that under a constant temperature when the pressure is inversely proportion to the volume.

So that means: $P \propto \frac{1}{V}$

And we can put an equal if we do this:

$P = \frac{k}{V}$ where k is the proportional constant.

For this case we want to find the following limit:

$lim_{V \to 0^{+}} P = \lim_{V \to 0^{+}} \frac{k}{V}$

And using properties of limits we have:

$lim_{V \to 0^{+}} P = k\lim_{V \to 0^{+}} \frac{1}{V}$

So for this case this limit tend to infinity since we are dividing a constant by a very low number positive near to 0.

So then we can conclude that:

$lim_{V \to 0^{+}} P = k\lim_{V \to 0^{+}} \frac{1}{V} =\infty$

This limit is not defined.

Interpretation: we are seeing that if the volume decrease considerable with the temperature constant by the inverse relation between P and V, the value of P increases to with no limit.

7 0

3 years ago

n Fig. 30.11, R = 15.0 Ω and the battery emf is 6.30 V. With switch S2 open, switch S1 is closed. After several minutes, S1 is o

Alex_Xolod [135]

The inductance of the coil and time is mathematically given as

L=0.1035H
t=12.8 ms

<h3>What is the inductance of the coil and How long after S1 is opened will the current reach 1.00% of its original value?</h3>

Generally, the equation for Current is mathematically given as

$\begin{aligned}I(t) &=I_{0} e^{\frac{-t}{\tau}} \\&I(t) =I_{0} e^{\frac{-R t}{L}}\end{aligned}$

Therefore

$L=\frac{-t R}{\ln \left(\frac{I}{I_{0}}\right)}$

Hence, the Maximum current is,

$\begin{aligned}I_{0} &=\frac{\varepsilon}{R}=\frac{6.30}{15} \\&I_{0} =0.42 \end{aligned}$

Thus,

$L &=\frac{-\left(0.280 * 10^{-3})(13.0)}{\ln \left(0.280}{0.42})} \\$

L=0.1035H

In conclusion, Time taken after S1 is opened is,

$(0.01) I_{0}=I_{0} e^{\frac{-t}{\tau}}$

Time constant is,

$\tau=\frac{L}{R}\\\\\tau=\frac{0.042}{15} \\$

$\tau=0.0028s$

$(0.01) I_{o} &=I_{o} e^{\frac{-t}{(0.00028)}} \\\\t=-\ln (0.01)(0.0028) \\\\t=-(-4.605)(0.0033) \\\\t=0.01289s$

t=12.8 ms

Read more about inductance

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3 0

2 years ago

The car's motion can be divided into three different stages: its motion before the driver realizes he's late, its motion after t

jolli1 [7]

Answer:

B, C, D are valid assumptions

Explanation:

In each stage the acceleration of the car varies because the velocity of car varies in each stage. At first, the acceleration of the car is moving slowly.

Then, driver realizes that he is late so the acceleration of the car increases he hits the gas.

Finally, the acceleration decreases when he sees the police car.

In the first state when driver do not know, that he is late, he will drive with a constant velocity as any one does.

When the driver hits the gas and he does not know about the police vehicle, even then he may drive with constant velocity or he may accelerate the car due to fear of being caught.

In the last part of motion when driver see the police car, even than he may accelerate the car, but acceleration will be constant throughout the motion is not possible, or even than he may continue with constant velocity.

Hence, there are only three assumptions are valid.

B. During each of the three different stages of its motion, the car is moving with constant velocity.

C . The highway is straight.

D. The highway is level.

Note that the rate of change of speed is equal to the acceleration. The acceleration is constant if change in velocity of particle is equal in equal interval of time.

8 0

4 years ago

At time t=0, a particle is located at the point (3,6,9). It travels in a straight line to the point (5,2,7), has speed 8 at (3,6

Elis [28]

The particle has constant acceleration according to

$\vec a(t)=2\,\vec\imath-4\,\vec\jmath-2\,\vec k$

Its velocity at time $t$ is

$\displaystyle\vec v(t)=\vec v(0)+\int_0^t\vec a(u)\,\mathrm du$

$\vec v(t)=\vec v(0)+(2\,\vec\imath-4\,\vec\jmath-2\,\vec k)t$

$\vec v(t)=(v_{0x}+2t)\,\vec\imath+(v_{0y}-4t)\,\vec\jmath+(v_{0z}-2t)\,\vec k$

Then the particle has position at time $t$ according to

$\displaystyle\vec r(t)=\vec r(0)+\int_0^t\vec v(u)\,\mathrm du$

$\vec r(t)=(3+v_{0x}t+t^2)\,\vec\imath+(6+v_{0y}t-2t^2)\,\vec\jmath+(9+v_{0z}t-t^2)\,\vec k$

At at the point (3, 6, 9), i.e. when $t=0$ , it has speed 8, so that

$\|\vec v(0)\|=8\iff{v_{0x}}^2+{v_{0y}}^2+{v_{0z}}^2=64$

We know that at some time $t=T$ , the particle is at the point (5, 2, 7), which tells us

$\begin{cases}3+v_{0x}T+T^2=5\\6+v_{0y}T-2T^2=2\\9+v_{0z}T-T^2=7\end{cases}\implies\begin{cases}v_{0x}=\dfrac{2-T^2}T\\\\v_{0y}=\dfrac{2T^2-4}T\\\\v_{0z}=\dfrac{T^2-2}T\end{cases}$

and in particular we see that

$v_{0y}=-2v_{0x}$

and

$v_{0z}=-v_{0x}$

Then

${v_{0x}}^2+(-2v_{0x})^2+(-v_{0x})^2=6{v_{0x}}^2=64\implies v_{0x}=\pm\dfrac{4\sqrt6}3$

$\implies v_{0y}=\mp\dfrac{8\sqrt6}3$

$\implies v_{0z}=\mp\dfrac{4\sqrt6}3$

That is, there are two possible initial velocities for which the particle can travel between (3, 6, 9) and (5, 2, 7) with the given acceleration vector and given that it starts with a speed of 8. Then there are two possible solutions for its position vector; one of them is

$\vec r(t)=\left(3+\dfrac{4\sqrt6}3t+t^2\right)\,\vec\imath+\left(6-\dfrac{8\sqrt6}3t-2t^2\right)\,\vec\jmath+\left(9-\dfrac{4\sqrt6}3t-t^2\right)\,\vec k$

4 0

3 years ago

Gala2k [10]

Acceleration increases.

4 0

3 years ago

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