PHYSICS COLLEGE

A 3.00N rock is thrown vertically into the air from the ground. At h=15.0m, v=25m/s upward. Use the work-energy theorem to find the initial speed of the rock.a. 3m/s
b. 30.3 m/s
c. None of the above

Answers

Answer 1
Answer:

Answer:

so initial speed of the rock is 30.32 m/s

correct answer is b. 30.3 m/s

Explanation:

given data

h = 15.0m

v = 25m/s

weight of the rock m = 3.00N  

solution

we use here work-energy theorem that is express as here

work = change in the kinetic energy    ..............................1

so it can be written as

work = force × distance     ...................2

and

KE is express as

K.E = 0.5 × m × v²  

and it can be written as

F × d = 0.5 × m × (vf)² - (vi)²      ......................3

here

m is mass and vi and vf is initial and final velocity

F = mg = m  (-9.8)  , d = 15 m and v{f} = 25 m/s

so put value in equation 3 we get

m  (-9.8) × 15 = 0.5 × m × (25)² - (vi)²

solve it we get

(vi)² =  919

vi = 30.32 m/s

so initial speed of the rock is 30.32 m/s

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What is the wavelength of the photons emitted by hydrogen atoms when they undergo n =5 to n =3 transitions?

Answers

Answer:

\lambda=1282nm

Explanation:

The wavelength of the photons emitted due to an  atomic electron transition in a hydrogen atom, is given by the Rydberg formula:

(1)/(\lambda)=R_H((1)/(n_1^2)-(1)/(n_2^2)})

Here R_H is the Rydberg constant for hydrogen and n_1,n_2 are the lower and higher quantum number for the energy levels of the  atomic electron transition, respectively. Replacing the given values and solving for \lambda

(1)/(\lambda)=1.097*10^7m^(-1)((1)/(3^2)-(1)/(5^2)})\n(1)/(\lambda)=7.81*10^5m^(-1)\n\lambda=(1)/(7.81*10^5m^(-1))\n\lambda=1.282*10^(-6)m\n\lambda=1.282*10^(-6)m*(1nm)/(10^(-9)m)\n\lambda=1282nm

A friend tells you that a lunar eclipse will take place the following week, and invites you to join him to observe the eclipse through a high-powered telescope he owns. You are curious what the eclipse might look like from different perspectives in space. If the moon has a diameter of 2,159.14 miles, what is the maximum distance that it could be observed by the naked eye with enough detail that you could distinguish it from other celestial bodies (assuming that you have 20/20 vision)

Answers

Answer:

y = 80.2 mille

Explanation:

The minimum size of an object that can be seen is determined by the diffraction phenomenon, if we use the Rayleigh criterion that establishes that two objects can be distinguished without the maximum diffraction of a body coincides with the minimum of the other body, therefore so much for the pupil of the eye that it is a circular opening

          θ = 1.22 λ/ d

in a normal eye the diameter of the pupils of d = 2 mm = 0.002 m, suppose the wavelength of maximum sensitivity of the eye λ = 550 nm = 550 10⁻⁹ m

         θ = 1.22 550 10⁻⁹ / 0.002

         θ = 3.355 10⁻⁴ rad

Let's use trigonometry to find the distance supported by this angle, the distance from the moon to the Earth is L = 238900 mille = 2.38900 10⁵ mi

       tan θ = y / L

       y = L tan θ

       y = 2,389 10⁵ tan 3,355 10⁻⁴

       y = 8.02 10¹ mi

       y = 80.2 mille

This is the smallest size of an object seen directly by the eye

Final answer:

An individual with 20/20 vision can observe the moon from a maximum distance of around 6200 km or 3850 miles. Beyond this distance, it might be difficult to distinguish the moon from other celestial objects without using a telescope. The use of a telescope can expand this range significantly.

Explanation:

The detailed observation of a lunar eclipsed, when viewed without any form of optical aid like a telescope, is contingent on many factors, one of which is the human eye's angular resolution—the eye's ability to differentiate between two separate points of light. For an average human eye with 20/20 vision, the angular resolution is approximately 0.02 degrees.

To calculate the maximum distance at which the moon could be observed clearly with the eye, the formula for small angle approximation can be used, which in this context is: Distance = Size / Angle = (2159.14 miles) / (0.02 degrees in radians). This calculates to a distance of approximately 6200 km or 3850 miles.

Beyond this distance, distinguishing the moon from other celestial bodies might be challenging using just the eye. Utilizing a high-powered telescope would significantly extend this range by magnifying the image, allowing clearer detail over much greater distances.

Learn more about Observing Moon here:

brainly.com/question/34031036

#SPJ3

Callisto, one of Jupiter's moons, has an orbital period of 16.69 days and its distance from Jupiter is 1.88*10^6 km. What is Jupiter's mass?

Answers

Answer:

The Jupiter´s mass is approximately 1.89*10²⁷ kg.

Explanation:

The only force acting on Calisto while is rotating around Jupiter, is the gravitational force, as defined by the Newton´s Universal Law of Gravitation:

Fg = G*mc*mj / rcj²

where G = 6.67*10⁻¹¹ N*m²/kg², mc= Callisto´s mass, mj= Jupiter´s mass, and rcj = distance from Jupiter for Callisto= 1.88*10⁹ m.

At the same time, there exists a force that keeps Callisto in orbit, which is the centripetal force, that actually is the same gravitational force we have already mentioned.

This centripetal force is related with the period of the orbit, as follows:

Fc = mc*(2*π/T)²*rcj.

In order to be consistent in terms of units, we need to convert the orbital period, from days to seconds, as follows:

T = 16.69 days* 86,400 (sec/day) = 1.44*10⁶ sec.

We have already said that Fg= Fc, so we can write the following equality:

G*mc*mj / rcj² = mc*(2*π/T)²*rcj

Simplifying common terms, and solving for mj, we get:

mj = 4*π²*(1.88*10⁹)³m³ / ((1.44*10⁶)² m²*6.67*10⁻11 N*m²/kg²)

mj = 1.89*10²⁷ kg.

Answer: Mass of Jupiter ~= 1.89 × 10^23 kg

Explanation:

Given:

Period P= 16.69days × 86400s/day= 1442016s

Radius of orbit a = 1.88×10^6km × 1000m/km

r = 1.88 × 10^9 m

Gravitational constant G= 6.67×10^-11 m^3 kg^-1 s^-2

Applying Kepler's third law, which is stated mathematically as;

P^2 = (4π^2a^3)/G(M1+M2) .....1

Where M1 and M2 are the radius of Jupiter and callisto respectively.

Since M1 >> M1

M1+M2 ~= M1

Equation 1 becomes;

P^2 = (4π^2a^3)/G(M1)

M1 = (4π^2a^3)/GP^2 .....3

Substituting the values into equation 3 above

M1 = (4 × π^2 × (1.88 × 10^9)^3)/(6.67×10^-11 × 1442016^2)

M1 = 1.89 × 10^27 kg

What minimum distance would you have to hit a baseball from the center of the earth so that it would eventually reach the moon? Assume you can hit the ball directly along the line that connects the centers of the earth and moon. The distance between the centers of the earth and moon is ???? = 3.82 × 108 m.

Answers

Answer:

d = 3.44 x 10⁸ m

Explanation:

The minimum distance required will be the distance from the centre of the earth to a point where gravitational intensity due to both earth and moon becomes equal . Once this point is reached , moon will attract the baseball on its own .

Let this distance be d from the centre of the earth

So GM / d² = G m / ( 3.82 x 10⁸ - d )²

M is mass of the earth , m is mass of the moon

M / m =  ( d / 3.82 x 10⁸ - d )²

5.972 x 10²⁴ / 7.34 x 10²² = ( d / 3.82 x 10⁸ - d )²

81.36 = ( d / 3.82 x 10⁸ - d )²

9.02 = d / 3.82 x 10⁸ - d

34.45 x 10⁸ - 9.02 d = d

34.45 x 10⁸ = 10.02 d

d = 3.44 x 10⁸ m

A 140 W lightbulb emits 4% of its energy as electromagnetic radiation. What is the radiation pressure (in N/m2) on a perfectly absorbing sphere of radius 14 m that surrounds the bulb

Answers

Answer: 7.578x10^-12

Explanation:

First, we find the power:

Power P = 140x4/100 =5,6W

Distance r = 14m

Then,

Intensity I = P/4πr2

= 5.6/(4π x 14 x 14)

=. 2.27 x10^3 W/m2

Radiation pressure:

P(rad) = I/c =0.00227÷{3 x 10^8)

=7.578x10^-12 N/m2

Answer:

Pr=7.57*10^{11}Pa

Explanation:

We can solve this problem by taking into account the expression

P_r=(IA)/(c)

where I is the irradiance, c is the speed of light and A is the area.

We have that the power is 140W, but only 4% is electromagnetic energy, that is

P=140W=140(J)/(s)\n0.4*140J=56J

56J is the electromagnetic energy.

The area of the bulb is

A_b=4\pi r^2=4\pi (14m)^2=2463m^2

The radiation pressure is

P_r=(56)/(2463m^2*3*10^8m/s^2)=7.57*10^(-11)Pa

hope this helps!!

When water freezes, its volume increases by 9.05% (that is, ΔV / V0 = 9.05 × 10-2). What force per unit area is water capable of exerting on a container when it freezes? (It is acceptable to use the bulk modulus of water, B = 2.2 × 109 N/m2, for this problem.) Give your answer in N/cm2.

Answers

Answer:

P = 1.99 10⁸ Pa

Explanation:

The definition of the bulk module is

      B = - P / (ΔV / V)

The negative sign is included for which balk module is positive, P is the pressure and V that volume

They tell us that the variation in volume is 9.05%, that is

    ΔV / V = ​​9.0Δ5 / 100 = 0.0905

    P = - B DV / V

    P = 2.2 10⁹ (0.0905)

    P = 1.99 10⁸ Pa
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