A stone is dropped from a bridge and hits the pavement below in two seconds. What is the velocity of the stone when it hits the pavement?A. 0.20 m/s
B. 19.6 m/s
C. 4.9 m/s
D. 9.8 m/s

Answers

Answer 1

Answer: We have: a = v/t
Here, t = 2 s [ Given ]
a = 9.8 m/s² [constant value for earth system ]

Substitute their values into the expression:
9.8 = v/2
v = 9.8 × 2
v = 19.6 m/s

In short, Your Answer would be Option B

Hope this helps!

Answer 2

Answer:

i just did the quiz and so the only 2 options that made logical sense would be B (19.6 m/s) or D (9.8 m/s) the reason someone could get confused between these two is because 19.6 divided by 2 is 9.8 m/s so for each second it will go down 9.8 m/s but since it wants to know the velocity it would be 19.6 m/s

hope this helps

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An observer on the earth sees a spaceship approaching at 0.54c. The ship then launches an exploration vehicle that, according to the earth observer, approaches at 0.82c. What is the velocity of the exploration vehicle relative to the spaceship? Give your answer as a decimal fraction of c, i.e. a β factor.

Answers

Answer:

Explanation:

Expression for relative velocity

= $(v_1+v_2)/(1+(v_1v_2)/(c^2) )$

= (.54 + .82 )c/ $1+ (.54 *.82)/(1)$

= 1.36 c / 1.4428

= .94 c

β = .94

Newton's Law of Cooling says that the rate of cooling of an object is proportional to the difference between its own temperature and the temperature of its surrounding. Write a differential equation that expresses Newton's Law of Cooling for this particular situation.

Answers

Answer:

$(dQ)/(dt) =-hA\Delta T(t)$

Explanation:Newton.s law of cooling states that the rate of cooling of an object is proportional to the difference between its own temperatures and temperature of its surroundings. Mathematically,

$(dQ)/(dt) =-hA [T(t)-T(s)]\n$

$(dQ)/(dt) =-hA\Delta T(t)$

where $Q$ is the heat transfer

$h$ is heat transfer coefficient

$A$ is the heat transfer surface area

$T$ is the temperature of the object's surface

$T(s)$ is the temperature of the surroundings

A long coaxial cable consists of an inner cylindrical conductor with radius a and an outer coaxial cylinder with inner radius b and outer radius c. The outer cylinder is mounted on insulating supports and has no net charge. The inner cylinder has a uniform positive charge per unit length λCalculate the electric field
(a) At any point between the cylinders a distance r from the axis and
(b) At any point outside the outer cylinder.
(c) Graph the magnitude of the electric field as a function of the distance r from the axis of the cable, from r = 0 to r = 2c.
(d) Find the charge per unit length on the inner surface and on the outer surface of the outer cylinder.

Answers

Answer:

Part a)

$E = (\lambda)/(2\pi \epsilon_0 r)$

Part b)

$E = (\lambda)/(2\pi \epsilon_0 r)$

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

Explanation:

Part a)

By Guass law we know that

$\int E. dA = (q)/(\epsilon_0)$

$E. 2\pi rL = (\lambda L)/(\epsilon_0)$

$E = (\lambda)/(2\pi \epsilon_0 r)$

Part b)

Outside the outer cylinder we will again use Guass law

$\int E. dA = (q)/(\epsilon_0)$

$E. 2\pi rL = (\lambda L)/(\epsilon_0)$

$E = (\lambda)/(2\pi \epsilon_0 r)$

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

Final answer:

The electric field between the cylinders is given by E = λ / (2πε₀r). The electric field outside the outer cylinder is zero due to the absence of net charge. Graph the electric field magnitude using the equation E = λ / (2πε₀r). The inner surface charge of the outer cylinder is -λ and the outer surface charge is 0.

Explanation:

To calculate the electric field between the cylinders at a distance r from the axis, you can use Gauss's Law. Since the charging is uniform, the electric field will also be uniform. Therefore, the electric field at any point between the cylinders is given by E = λ / (2πε₀r), where ε₀ is the permittivity of free space.

To calculate the electric field at any point outside the outer cylinder, you can use the principle of superposition. The electric field due to the outer cylinder is zero because it has no net charge. The electric field due to the inner cylinder can be calculated using the same formula as before.

To graph the magnitude of the electric field as a function of the distance r from the axis, you can plot the equation E = λ / (2πε₀r) for values of r ranging from 0 to 2c.

The charge per unit length on the inner surface of the outer cylinder is -λ, while the charge per unit length on the outer surface of the outer cylinder is 0. This is because the outer cylinder has no net charge and the inner cylinder has a uniform positive charge per unit length λ.

Learn more about Electric Field here:

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What is the metric unit of force? kg kg/cm^3 kg/cm^2 N

Answers

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Solar energy is: expensive renewable inefficient during winter all of the above

Answers

All of the above is the right answer for Odesseyware.

Solar energy is free and renewable.

Its efficiency is determined by the photovoltaic or heat-exchanging equipment used, and the efficiency of the infrastructure built to interconnect them. None of that depends on the time of the year. For a given amount of solar energy absorbed by the collector, the amount of useful output energy doesn't change with the season.

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Answers

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Oh, wait . . .

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