PHYSICS HIGH SCHOOL

Which event can be excluded from a list of interactions among Earth's spheres?A solar flares
B volcanic eruption
C fracking for natural gas
D hurricane

Answers

Answer 1
Answer:

Final answer:

Solar flares can be excluded from a list of interactions among Earth's spheres because they are not directly related to interactions among the geosphere, hydrosphere, atmosphere, and biosphere.

Explanation:

The event that can be excluded from a list of interactions among Earth's spheres is A solar flares. Solar flares are not directly related to interactions among Earth's spheres, which include the geosphere, hydrosphere, atmosphere, and biosphere.

On the other hand, volcanic eruptions, fracking for natural gas, and hurricanes are all examples of interactions among Earth's spheres. Volcanic eruptions involve the release of gases and materials from the geosphere into the atmosphere. Fracking for natural gas involves extracting resources from the geosphere and potentially impacting the hydrosphere. Hurricanes involve interactions between the atmosphere and the hydrosphere, resulting in powerful storms.

Therefore, the correct answer is A solar flares.

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Answer 2
Answer: Hurricane is the answer

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An object is released from rest at time t = 0 and falls through the air, which exerts a resistive force such that the acceleration a of the object is given by a = g bv, where v is the object's speed and b is a constant. If limiting cases for large and small values of t are considered, which of the following is a possible expression for the speed of the object as an explicit function of time? A) v = g(1-e^-bt)/b B) v = (ge^bt)/b C) v = (g+a)t/b

Answers

Answer:

A) (g)/(b)(1-e^(-bt))

Explanation:

Since a = g - bv,

We can substitute a = dv/dt into the equation.

Then, the equation will be like dv/dt = g - bv.

So we got first order differential equation.

As known, v = 0 at t = 0, and v = g/b at t = ∞.

Since (dv)/(dt)= g - bv = b( (g)/(b) - v) ⇒ (dv)/( (g)/(b) - v)= bdt

So take the integral of both side.

- ln ((g)/(b) - v) = bt + C

Since for t=0, v = 0 ⇒ C =- ln ((g)/(b))

v = (g)/(b) + e^{-bt-ln((g)/(b))} = (g)/(b)- (g)/(b)e^(-bt) = (g)/(b)(1-e^(-bt))

The correct option for the expression of speed as an explicit function of time is option A

A) v = g·(1 - e^{-b \cdot t)/b

The reason why option A is correct is given as follows;

Known:

The initial velocity of the object at time t = 0 is v = 0 (object at rest)

The function that represents the acceleration is a = g - b·v

Where;

v = The speed of the object at the given instant

b = A constant term

By considering the limiting case for time t, we have;

At very large values of t, the velocity will increase such that we have;

\lim \limits_(t \to \infty) a = 0 therefore,  \lim \limits_(t \to \infty) g - b\cdot v = 0 and \lim \limits_(t \to \infty) \left( v_(max) = (g)/(b) \right)

The given equation can be rewritten as follows, to express the equation in terms the velocity;

a = b \cdot \left((g)/(b) - v \right) = b \cdot \left(v_(max) - v \right)

Acceleration, \ a = (dv)/(dt)

Therefore;

(dv)/(dt) = b \cdot \left((g)/(b) - v \right)

The above differential equation gives;

(dv)/( \left((g)/(b) - v \right)) = b \cdot dt

Which gives;

\displaystyle \int\limits {(dv)/( \left((g)/(b) - v \right)) } = \int\limits {b \cdot dt} = b \cdot t + C

\displaystyle \int\limits {(dv)/( \left((g)/(b) - v \right)) } = -\ln \left((g)/(b) - v \right) and  \displaystyle\int\limits{b \cdot dt} = b \cdot t + C

Therefore

\displaystyle -\ln \left((g)/(b) - v \right) =b \cdot t + C

At t = 0, v = 0, therefore;

\displaystyle -\ln \left((g)/(b) - 0 \right) =b * 0 + C

C = \displaystyle -\ln \left((g)/(b) \right)

Which gives;

\displaystyle -\ln \left((g)/(b) - v \right) =b \cdot t \displaystyle -\ln \left((g)/(b) \right)

\displaystyle \ln \left((g)/(b) - v \right) =-b \cdot t \displaystyle +\ln \left((g)/(b) \right)

\displaystyle (g)/(b) - v = e^{-b \cdot t \displaystyle +\ln \left((g)/(b) \right)} = e^(-b \cdot t) * e^\ln \left((g)/(b) \right)} = e^(-b \cdot t) * (g)/(b)

\displaystyle (g)/(b) - e^(-b \cdot t) * (g)/(b) = v

\displaystyle (g)/(b) \cdot \left(1 - e^(-b \cdot t) \right) = v

∴ v = g·(1 - e^{-b \cdot t)/b

The correct option is option (A)

Learn more about differential equation here;

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which of the following statements about black holes is false. a. Astronomers have never seen a black hole. b. A black hole sucks in matter like a vacuum cleaner. c. No light can escape from a black hole. d. All of the mass in a black hole is concentrated in a single point.

Answers

I love this question, because it gets at one of the greatest misconceptions about black holes.

Choice-B is false. A black hole does not suck matter in like a vacuum cleaner.

Answer:

a. Astronomers have never seen a black hole.

Explanation:

A black hole is 1-Dimensional point in space-time having density is extremely large such that space-time curvature bends enough for even light cannot escape. All the mass is concentrated at this point. A black hole is formed when a very high mass star dies. It goes supernova and collapses to form a black hole. It sucks in matter like a vacuum cleaner.

Astronomers have seen a black hole by actually viewing the event horizon which is a bright disk surrounding a black hole. Recently an image of a black hole was also released.

A basic observation of a star is how bright it appears. This brightness is known as the star'sA. luminosity.
B. absent magnitude.
C. apparent magnitude.
D. absolute magnitude.

Answers

The luminosity is how bright an celestial body is.

75 Centimeters to decameters and Kilometer​

Answers

Answer - 75cm=0.075 decametre (divide the length value by 1000.)

75cm=0.00075 kilometer (divide the length value by 100000).

Electronic signals that are smoothly varying signals produced by continuously changing the voltage or current in a circuit area. digital signals.
b. cathode rays.
c. diode signals.
d. analog signals.

Answers

Answer: The correct answer is "analog signals".

Explanation:

Digital signal: This signal varies as 0 or 1. It is discrete signal.

Cathode rays: They are the streams of the negatively charged particles.

Diode signals: Diode is a semiconductor device. It changes the alternating signal to the direct signal.

Analog signals: In this signal, the signal is varying by continuously changing the direction voltage or current with time. It is a continuously varying signal.

In the given problem, electronic signals that are varying signals produced by continuously changing the voltage or current in a circuit are analog signals.

Therefore, the correct option is (d).
The right answer for the question that is being asked and shown above is that: "c. diode signals." Electronic signals that are smoothly varying signals produced by continuously changing the voltage or current in a circuit are diode signals.

What does the object on the screen below model?A. Molecules
B. The orbit of the moon around the Earth
C. The orbits of the planets around the Sun
D. Electronic circuits

Answers

Answer:

A, molecules

Explanation:

Molecules were viewable around the time period of the computers shown on the screen

Answer:

correct option is (A)........
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