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irga5000 [103]
3 years ago
12

What is the significance of the fact that the gravitational constant, G, is a very small number and that Coulomb’s constant, k,

is a very large number?
The comparison indicates that the electromagnetic force is a much weaker force than the gravitational force.

>>>>>>>>>>>>>>>>The comparison indicates that the gravitational force is a much weaker force than the electromagnetic force.

The comparison does not give enough information to determine which force is weaker because the forces are in different units.
Physics
2 answers:
WITCHER [35]3 years ago
7 0

The comparison indicates that the gravitational force is a much weaker force than the electromagnetic force. This is the significance of the fact that the gravitational constant, G, is a very small number and that Coulomb’s constant, k, is a very large number

Answer: The comparison indicates that the gravitational force is a much weaker force than the electromagnetic force.

<u>Explanation: </u>

We know that the universal formula for the estimation of gravitational force is,

                       F=G \times\left(\frac{m_{1} \times m_{2}}{r^{2}}\right)

Where,

F, Gravitational force, m_{1} \text { and } m_{2}- objects masses r- The relative distance between the two objects  

G, Gravitational Constant 6.67 \times 10^{-11} \frac{N m^{2}}{k g^{2}}

And, the electrostatic force between two scalar charges according to the Coulomb’s law is,

                  F=k \times\left(\frac{q_{1} \times q_{2}}{r^{2}}\right)

F, Electrostatic force between two charges,q_{1} \text { and } q_{2}- Two scalar charges, r - The relative distance between two scalar charges, k- Coulomb’s constant 8.987 \times 10^{9} \frac{N m^{2}}{c^{2}}

Now, on comparing the values of G and k, we can easily evaluate that eventually, the gravitational force will be lesser than the coulomb’s force.  

Besides this, we can also judge this fact through various examples such as, a balloon rubbed with a cloth, easily sticks to the wall for some time. Opposing the gravitational force of the Earth which is not the case with the normal balloon.

It drops without having the electrostatic force between the wall and the balloon.  This shows that the gravitational force draws lesser impact on objects as compared to the electrostatic force.

ankoles [38]3 years ago
6 0

Answer: The comparison indicates that the gravitational force is a much weaker force than the electromagnetic force.

Explanation: Edge

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Answer:

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You are traveling in a car toward a hill at a speed of 36.4 mph. The car's horn emits sound waves of frequency 231 Hz, which mov
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Answer:

<em>a. The frequency with which the waves strike the hill is 242.61 Hz</em>

<em>b. The frequency of the reflected sound wave is 254.23 Hz</em>

<em>c. The beat frequency produced by the direct and reflected sound is  </em>

<em>    11.62 Hz</em>

Explanation:

Part A

The car is the source of our sound, and the frequency of the sound wave it emits is given as 231 Hz. The speed of sound given can be used to determine the other frequencies, as expressed below;

f_{1} = f[\frac{v_{s} }{v_{s} -v} ] ..............................1

where f_{1} is the frequency of the wave as it strikes the hill;

f is the frequency of the produced by the horn of the car = 231 Hz;

v_{s} is the speed of sound = 340 m/s;

v is the speed of the car = 36.4 mph

Converting the speed of the car from mph to m/s we have ;

hint (1 mile = 1609 m, 1 hr = 3600 secs)

v = 36.4 mph *\frac{1609 m}{1 mile} *\frac{1 hr}{3600 secs}

v = 16.27 m/s

Substituting into equation 1 we have

f_{1} =  231 Hz (\frac{340 m/s}{340 m/s - 16.27 m/s})

f_{1}  = 242.61 Hz.

Therefore, the frequency which the wave strikes the hill is 242.61 Hz.

Part B

At this point, the hill is the stationary point while the driver is the observer moving towards the hill that is stationary. The frequency of the sound waves reflecting the driver can be obtained using equation 2;

f_{2} = f_{1} [\frac{v_{s}+v }{v_{s} } ]

where f_{2} is the frequency of the reflected sound;

f_{1}  is the frequency which the wave strikes the hill = 242.61 Hz;

v_{s} is the speed of sound = 340 m/s;

v is the speed of the car = 16.27 m/s.

Substituting our values into equation 1 we have;

f_{2} = 242.61 Hz [\frac{340 m/s+16.27 m/s }{340 m/s } ]

f_{2}  = 254.23 Hz.

Therefore, the frequency of the reflected sound is 254.23 Hz.

Part C

The beat frequency is the change in frequency between the frequency of the direct sound  and the reflected sound. This can be obtained as follows;

Δf = f_{2} -  f_{1}  

The parameters as specified in Part A and B;

Δf = 254.23 Hz - 242.61 Hz

Δf  = 11.62 Hz

Therefore the beat frequency produced by the direct and reflected sound is 11.62 Hz

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According to Newton's Universal Law of Gravitation, when the distance between two interacting objects doubles, the gravitational
maksim [4K]

Answer:

<em>If the distance doubles, the gravitational force is divided by 4</em>

Explanation:

<u>Newton’s Universal Law of Gravitation </u>

Objects attract each other with a force that is proportional to their masses and inversely proportional to the square of the distance.

\displaystyle F=G{\frac {m_{1}m_{2}}{r^{2}}}

Where:

m1 = mass of object 1

m2 = mass of object 2

r     = distance between the objects' center of masses

G   = gravitational constant: 6.67\cdot 10^{-11}~Nw*m^2/Kg^2

If the distance between the interacting objects doubles to 2r, the new force F' is:

\displaystyle F'=G{\frac {m_{1}m_{2}}{(2r)^{2}}}

Operating:

\displaystyle F'=\frac{1}{4}G{\frac {m_{1}m_{2}}{r^{2}}}

Substituting the original value of F:

\displaystyle F'=\frac{1}{4}F

If the distance doubles, the gravitational force is divided by 4

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3 years ago
A bucket of water with total mass 23 kg is attached to a rope, which in turn is wound around a 0.050-m radius pulley at the top
adoni [48]

Answer:

I = 0.083 kg m^2

Explanation:

Mass of the bucket, m = 23 kg

Radius of the pulley, r = 0.050 m

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The time taken to fall, t = 2 s

Speed, v = 8.0 m/s

Moment of Inertia of the pulley, I = ?

Using the equation of motion:

v = u + at

8 = 0 + 2a

a = 8/2

a = 4 m/s²

The relationship between the linear and angular accelerations is given by the equation:

a = \alpha r

Angular acceleration, \alpha = a/r

\alpha = 4/0.050\\\alpha = 80 rad/s^2

Since the bucket is falling, it can be modeled by the equation:

mg - T = ma

T = mg - ma = m(g-a)

T = 23(9.8 - 4)

The tension, T = 133.4 N

The equation for the pulley can be modeled by:

T* r = I * \alpha\\133.4 * 0.050 = I * 80\\6.67 = 80 I\\I = 6.67/80\\I = 0.083 kg m^2

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3 years ago
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