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2 years ago

The tropic of cancer is to the tropic of capricorn as the arctic circle is to the

Physics

1 answer:

Y_Kistochka [10]2 years ago

7 0

Answer:

Antarctic Circle

Explanation:

The Tropic of Cancer, which is also referred to as the Northern Tropic, is the most northerly circle of latitude on Earth at which the Sun can be directly overhead. This occurs on the June solstice, when the Northern Hemisphere is tilted toward the Sun to its maximum extent.

Tropic of Capricorn Is it Southern Hemisphere counterpart, marking the most southerly position at which the Sun can be directly overhead.

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Let us remember your previous lesson on Physical Press Components! Directions: Analyze the folawna fitness components. Put a che

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Answer:

5 Muscular and strength

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2 years ago

If the position of a particle on the x-axis at time t is âˆ’5t2, then the average velocity of the particle for 0 â‰¤ t â‰¤ 3 is

Drupady [299]

Answer:

v = 15 m / s

Explanation:

In this exercise we are given the position function

x = 5 t²

and we are asked for the average velocity in an interval between t = 0 and t= 3 s, which is defined by the displacement between the time interval

$v= \frac{v_{f} - v_{o} }{t_{f} - t_{o} }$

let's look for the displacements

t = 0 x₀ = 0 m

t = 3 $x_{f}$ = 5 3 2

x_{f} = 45 m

we substitute

$v = \frac{45 -0}{3 - 0}$

v = 15 m / s

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2 years ago

3. How does the mass of a proton compare to the mass of an electron?

o-na [289]

Answer:

Proton, stable subatomic particle that has a positive charge equal in magnitude to a unit of electron charge and a rest mass of 1.67262 × 10−27 kg, which is 1,836 times the mass of an electron.

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2 years ago

In the first direct detection of gravitational waves by LIGO in 2015, the waves came from A. the collapse of a nearby star into

diamong [38]

In the first direct detection of gravitational waves by LIGO in 2015, the waves came from the merger of two black holes. Option B is correct. This is further explained below.

<h3>What are gravitational waves?</h3>

A gravitational wave is simply defined as a ripple in space that is unseen though extremely rapid. Gravitational waves move at light speed. As they pass past, these waves compress and stretch everything in their path.

In conclusion, the merger of two black holes is the first direct detection of gravitational waves.

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1 year ago

At the moment t = 0, a 20.0 V battery is connected to a 5.00 mH coil and a 6.00 Ω resistor. (a) Immediately thereafter, how does

insens350 [35]

(a) On the coil: 20 V, on the resistor: 0 V

The sum of the potential difference across the coil and the potential difference across the resistor is equal to the voltage provided by the battery, V = 20 V:

$V = V_R + V_L$

The potential difference across the inductance is given by

$V_L(t) = V e^{-\frac{t}{\tau}}$ (1)

where

$\tau = \frac{L}{R}=\frac{0.005 H}{6.00 \Omega}=8.33\cdot 10^{-4} s$ is the time constant of the circuit

At time t=0,

$V_L(0) = V e^0 = V = 20 V$

So, all the potential difference is across the coil, therefore the potential difference across the resistor will be zero:

$V_R = V-V_L = 20 V-20 V=0$

(b) On the coil: 0 V, on the resistor: 20 V

Here we are analyzing the situation several seconds later, which means that we are analyzing the situation for

$t >> \tau$

Since $\tau$ is at the order of less than milliseconds.

Using eq.(1), we see that for $t >> \tau$ , the exponential becomes zero, and therefore the potential difference across the coil is zero:

$V_L = 0$

Therefore, the potential difference across the resistor will be

$V_R = V-V_L = 20 V- 0 = 20 V$

(c) Yes

The two voltages will be equal when:

$V_L = V_R$ (2)

Reminding also that the sum of the two voltages must be equal to the voltage of the battery:

$V=V_L +V_R$

And rewriting this equation,

$V_R = V-V_L$

Substituting into (2) we find

$V_L = V-V_L\\2V_L = V\\V_L=\frac{V}{2}=10 V$

So, the two voltages will be equal when they are both equal to 10 V.

(d) at $t=5.77\cdot 10^{-4}s$

We said that the two voltages will be equal when

$V_L=\frac{V}{2}$

Using eq.(1), and this last equation, this means

$V e^{-\frac{t}{\tau}} = \frac{V}{2}$

And solving the equation for t, we find the time t at which the two voltages are equal:

$e^{-\frac{t}{\tau}}=\frac{1}{2}\\-\frac{t}{\tau}=ln(1/2)\\t=-\tau ln(0.5)=-(8.33\cdot 10^{-4} s)ln(0.5)=5.77\cdot 10^{-4}s$

(e-a) -19.2 V on the coil, 19.2 V on the resistor

Here we have that the current in the circuit is

$I_0 = 3.20 A$

The problem says this current is stable: this means that we are in a situation in which $t>>\tau$ , so the coil has no longer influence on the circuit, which is operating as it is a normal circuit with only one resistor. Therefore, we can find the potential difference across the resistor using Ohm's law

$V=I_0 R = (3.20 A)(6.0 \Omega)=19.2 V$

Then the battery is removed from the circuit: this means that the coil will discharge through the resistor.

The voltage on the coil is given by

$V_L(t) = -V e^{-\frac{t}{\tau}}$ (1)

which means that it is maximum at the moment when the battery is disconnected, when t=0:

$V_L(0)=.V$

And V this time is the voltage across the resistor, 19.2 V (because the coil is now connected to the resistor, not to the battery). So, the voltage across the coil will be -19.2 V, and the voltage across the resistor will be the same in magnitude, 19.2 V (since the coil and the resistor are connected to the same points in the circuit): however, the signs of the potential difference will be opposite.

(e-b) 0 V on both

After several seconds,

$t>>\tau$

If we use this approximation into the formula

$V_L(t) = -V e^{-\frac{t}{\tau}}$ (1)

We find that

$V_L = 0$

And since now the resistor is directly connected to the coil, the voltage in the resistor will be the same as the coil, so 0 V. This means that the coil has completely discharged, and current is no longer flowing through the circuit.

7 0

2 years ago