How much time would it take for an airplane to reach its destination. If it traveled at an average velocity of 790km/hr east for a distance of 4,700 kilometres

Answers

Answer 1

Answer: It would be 5.94 hours but you can round it to just 6 hours.

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Which of the following best defines the science of separating color wavelengths?a. spectroscopy b. color-detection c. photo-chemistry d. spectrophotometry
A change that occurs when a solid melts or a liquid freezes is an example of a phase change.a. Trueb. False

A charge of 0.2 C experiences an electric force of 5 N. What is the strength of the electric field in N/C?

Answers

YOUR ANSWER SHOULD BE AS FOLLOWS

25N/C

$\vec{E}=\frac{\vec{F}}{q}\n \n \vec{E}=(5 \ N)/(0.2 \ C)=25 \ N/C$

Which model of the atom has electrons traveling in specific paths around the nucleus? Bohr's model Rutherford's model Thomson's model Dalton's model

Answers

The model of the atom that has electrons traveling in specific paths around the nucleus is Rutherford's model.

Therefore, the correct answer is option B

J. J Thomson proposed that the atom is a sphere of positively charged matter in which negatively charged electron are embedded.

In Rutherford's model of atomic structure, the electrons moves in orbits by electrostatic attraction to the positively charged nucleus.

According to Dalton's atomic theory, atom were indestructible and indivisible solid particle.

Bohr's model made assumption that electron can only exist in circular orbits of definite quantum energy.

According to the question, the model of the atom that has electrons traveling in specific paths around the nucleus is Rutherford's model.

Therefore, the correct answer is option B

Learn more here : brainly.com/question/16776207

Answer:

A, Bohr's model

Explanation:

Took the quiz, hope it's right for you.

(The other three did have something to do with it they just didn't create that specific model)

The force between two point charges is described by _____ law.Newton's
Ohm's
Coulomb's
Gilbert's

Answers

Coulombs is the correct answer.

One student did an experiment on the rock cycle. The steps of the experiment are shown below.

Collect five pieces of wax of different colors.
Use a plastic knife to scrape off thin shavings of the pieces of wax.
Pile the wax shavings of different colors one over the other.
Hold the pile of wax shavings in the palm and squeeze it for two to three minutes. Fold the pile and squeeze it in the palm again for a minute.
What does the student's experiment most likely demonstrate about the rock cycle?

a. Weathering can change igneous rocks into sediments.
b. Weathering can change metamorphic rocks into sediments.
c. Heat and pressure can change igneous rocks into metamorphic rocks.
d. Heat and pressure can change sedimentary rocks into metamorphic rocks.

Answers

First, when the student added the layers of wax over each other, this became a representation of sedimentary rocks.

Then the student folded his/her palm and squeezed the layers of wax. This means that the student applied heat and pressure on the wax (sedimentary rocks)

Referring to the diagram below which represents the rock cycle, we will find that applying heat and pressure on sedimentary rocks would convert these rocks into metamorphic rocks.

Based on the above, the best choice would be:
d. Heat and pressure can change sedimentary rocks into metamorphic rocks.

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;

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

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

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