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lorasvet [3.4K]
2 years ago
15

Q15.17 Both wave intensity and gravitation obey inverse-square laws. Do they do so for the same reason?Discuss the reason for ea

ch of these inverse-square laws as well as you can. block is attached to an ideal spring and is mov ing in SHM on a horizontalfrictionless surfaceThe amplitude of the motion is 0.120 m.The maximum speed of the block is 3.90 m/s.What is the maximum magnitude of the acceleration of the block?
Physics
1 answer:
s2008m [1.1K]2 years ago
7 0

(a) Intensity obeys inverse square law from basis of light passing through a given surface.

(b) Gravitation obeys inverse square law from the basis of force between two masses.

(c) The maximum magnitude of the acceleration of the block is 126.75 m/s².

<h3>What is intensity?</h3>

Intensity is the ratio is ratio of power to area of a given surface.

I = P/A (W/m²)

where;

  • P is power
  • A is area
<h3>Universal gravitation law</h3>

F = \frac{Gm_1m_2}{r^2}

Intensity and gravitation do not obey inverse square law for same reason;

  • Intensity obeys inverse square law from basis of light passing through a given surface.
  • Gravitation obeys inverse square law from the basis of force between two masses.
<h3>Acceleration of the block</h3>

a = v²/A

a = (3.9²)/0.12

a = 126.75 m/s²

Learn more about acceleration here: brainly.com/question/605631

#SPJ1

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A wooden block of mass m = 9 kg starts from rest on an inclined plane sloped at an angle θ = 30° from the horizontal. The block
jenyasd209 [6]

Answer:

Fk = 21.645N

Explanation:

Let Fb be Force of block and thus;

Fb= mg where m is mass of block and g is acceleration due to gravity

Thus Fb= 9kg x 9.81N/kg = 88.29 N

Now, the question says this force Fb rests at an inclined plane [email protected] 30° angle

Thus;

Force parallel to inclined plane = 88.29 x sin30° = 44.145 N.

Force perpendicular to the inclined plane = 88.29 x cos30 = 76.46 N

Now, when an object is falling freely, we know that

h = (1/2)at^(2)

From the question, the height is 5m and t= 2 seconds

Thus;

5 = (1/2)a(2)^(2)

2a = 5 and thus,

a = 5/2 = 2.5 m/s^(2)

Now, in inclined planes, perpendicular force - kinetic friction force = Resultant force

Thus let perpendicular be Fp and kinetic friction force be Fk and so;

Fp - Fk = F

F= ma = 9 x 2.5 = 22.5N

Thus, 44.145 - Fk = 22.5

Thus, Fk = 44.145 - 22.5 = 21.645N

5 0
3 years ago
A rocket travels at a speed of 14,000 m/s. What is the distance travelled by the rocket in 150s?
kati45 [8]

1) The distance travelled by the rocket can be found by using the basic relationship between speed (v), time (t) and distance (S):

v=\frac{S}{t}

Rearranging the equation, we can write

S=vt

In this problem, v=14000 m/s and t=150 s, so the distance travelled by the rocket is

S=vt=(14000 m/s)(150 s)=2.1 \cdot 10^6 m


2) We can solve the second part of the problem by using the same formula we used previously. This time, t=300 s, so we have:

S=vt=(14000 m/s)(300 s)=4.2 \cdot 10^6 m/s

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The latent heat of fusion of alcohol is 50 kcal/kg and its melting point is -54oC. It has a specific heat of 0.60 in its liquid
sashaice [31]

Answer:

C

Explanation:

To melt the alcohol

Heat needed = M . L  = 2 . 25  = 50 kcal

To warm up the alcohol

Heat needed = M . sp. ht. . ∆t   = 2 . 0.6 . 100  = 120 kcal

Total heat needed = 170 kcal

Assuming that 0.6 kcal/ kg / ˚C  is the specific heat and that the answer is wanted in kcal ( a rather odd unit to be in use here.)

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3 years ago
NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA
kkurt [141]

Answer:

The answer is "q=0.0945\,C".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}

Potential energy shifts:

= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\   =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\

=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5}  }{ 4,228 \times10^{5}} \right ) \\\\

=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\

\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0
3 years ago
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