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 1

Answer:

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

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a painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How far from the wall should the observer stand to get the best view?

Answers

Solution:

With reference to Fig. 1

Let 'x' be the distance from the wall

Then for $\Delta$ DAC:

$tan\theta = (d)/(x)$

⇒ $\theta = tan^(-1) (d)/(x)$

Now for the $\Delta$ BAC:

$tan\theta = (d + h)/(x)$

⇒ $\theta = tan^(-1) (d + h)/(x)$

Now, differentiating w.r.t x:

$(d\theta )/(dx) = (d)/(dx)[tan^(-1) (d + h)/(x) - tan^(-1) (d)/(x)]$

For maximum angle, $(d\theta )/(dx)$ = 0

Now,

0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}][/tex]

0 = $(-(d + h))/((d + h)^(2) + x^(2)) -(-d)/(x^(2) + d^(2))$

$(-(d + h))/((d + h)^(2) + x^(2)) = \frac{{d}{x^(2) + d^(2)}$

After solving the above eqn, we get

x = $\sqrt{(d)/(d + h)}$

The observer should stand at a distance equal to x = $\sqrt{(d)/(d + h)}$

Final answer:

For optimum viewing of a painting in a gallery, an observer should position themselves a distance away from the painting calculated using Pythagoras theorem, forming a right-angled triangle with the painting and the floor. This distance can be expressed as c = √[(h/2 + d)² + (h/2)²], where h is the height of the painting and d is the height from the observer's eye to the bottom of the painting.

Explanation:

In the physics of optics, the viewer should position themselves to where they form a right-angled triangle with the ceiling and the painting leading to the best viewing experience. This is widely known as the 'normal viewing distance'.

Given that the painting has a height h and its lower edge is at a distance d above the observer's eye, the observer should stand a distance away from the wall, which can be calculated using Pythagoras' theorem in right triangles, which states that the square of the hypotenuse (c) is equal to the sum of the square of the other two sides (a and b), i.e., c² = a² + b²

Since the painting height and viewer height forms the right-angle in this case, we have: a = (h/2 + d), and b = h/2. Substituting a and b in Pythagoras equation, we can solve for c which is the required distance: c = √[(h/2 + d)² + (h/2)²]

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g Adjacent rows in the first part of the experiment are found to have potentials of 3.66 V and 4.22 V. If the distance between rows is found to be 0.4 cm, what is the magnitude of the electric field at the location between the rows

Answers

Answer:

$E=140V/m$

Explanation:

If the electric field is uniform, the electric field between two points at potentials $V_1$ and $V_2$ which are separated by a distance d will be given by the formula:

$E=(\Delta V)/(d)$

So in our case, we have $E=(4.22V-3.66V)/(0.004m)=140V/m$

A 4,667 kHz AM radio station broadcasts with a power of 84 kW. How many photons does the transmitting antenna emit each second.

Answers

Answer:

Explanation:

Frequency( ν ) 4667 x 10³ Hz = 4.667 x 10⁶ Hz

Energy of one photon = hν [ h is plank's constant ]

= 6.6 x 10⁻³⁴ x 4.667 x 10⁶ = 30.8 x 10⁻²⁸ J

Power = 84 x 10³ J/s

No of photons emitted = Power / energy of one photon

= 84 x 10³ / 30.8 x 10⁻²⁸ =2.727 x 10³¹ per second .

1. Compare and contrast the SI and the English systems of measurement.

Answers

Answer:The SI system is based on the number 10 as well as multiples and products of 10. This makes it much easier to use, and so it has been the accepted system in scientific and technical applications. The English system is more complicated as relationships between units of the same quantity aren't uniform.

Explanation:

Answer:

The metric system is an internationally agreed decimal system of measurement while The International System of Units (SI) is the official system of measurement in almost every country in the world

A piston-cylinder device initially contains 1.4 kg saturated liquid water at 200oC. Now heat is transferred to the water until the volume quadruples and the cylinder contains saturated vapor only. Determine (a) the volume of the cylinder, (b) the final temperature.

Answers

Answer:

Explanation:

Given

mass of saturated liquid water $m=1.4\ kg$

at $200^(\circ)$ specific volume is $\nu =0.001157\ m^3\kg$ (From Table A-4,Saturated water Temperature table)

$V_1=m\nu _1$

$V_1=1.4* 0.001157$

$V_1=1.6198* 10^(-3)\ m^3$

Final Volume $V_2=4V_1$

$V_2=4* (1.6198* 10^(-3))$

$V_2=6.4792* 10^(-3)\ m^3$

Specific volume at this stage

$\nu _2=(V_2)/(m)$

$\nu _2=(6.4792* 10^(-3))/(1.4)$

$\nu _2=0.004628\ m^3/kg$

Now we see the value and find the temperature it corresponds to specific volume at vapor stage in the table.

$T_2=T_1^(*)+(T_2^(*)-T_1^(*))/(\alpha _2^(*)-\alpha _1^(*))* (\alpha _2-\alpha _1^(*))$

$T_2=370^(\circ)+(373.95-370)/(0.003106-0.004953)* (0.004628-0.004953)$

$T_2=370.7^(\circ) C$

Final answer:

The problem in the question is solved using the principles of thermodynamics. The volume of the device after the heat transfer is 6311.2 cm³. The final temperature inside the cylinder, when the water reached the state of saturated vapor, is approximately 240°C.

Explanation:

The subject question is a thermodynamics problem; more specifically dealing with changes of state, volume, and temperature in a system under certain conditions.

For solving part (a), one would first need to find the specific volume (v) at the initial state, which is saturated liquid at 200°C. Looking up in the property tables, we see that v = 1.127 cm³/g for saturated water at 200°C. Then, the initial volume (V) is mass times specific volume, so V = 1.4 kg x 1.127cm³/g x 1000g/kg = 1577.8 cm³. Because volume quadrupled, the final volume is 4 x 1577.8 cm³ = 6311.2 cm³.

For part (b), at the final state, the water is a saturated vapor. The specific volume at the final state is the final volume divided by the mass, which equals to 6311.2 cm³ / 1.4kg / 1000g/kg = 4.507 cm³/g. Look this value up in the property table to find the corresponding temperature. We get a final temperature of about 240°C.

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wo charged spheres are 1.5 m apart and are exerting an electrostatic force (Fo) on each other. If the charge on each sphere decreases by a factor of 9, determine (in terms of Fo) how much electrostatic force each sphere will exert on the other.